{"id":2121,"date":"2016-04-13T03:18:14","date_gmt":"2016-04-13T03:18:14","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=2121"},"modified":"2024-08-21T14:58:01","modified_gmt":"2024-08-21T14:58:01","slug":"resolucao-comentada-dos-exercicios-de-vestibulares-sobre-primeiro-principio-da-termodinamica-ou-principio-da-conservacao-da-energia","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/fisica-termica\/termodinamica\/primeiro-principio-da-termodinamica-ou-principio-da-conservacao-da-energia\/resolucao-comentada-dos-exercicios-de-vestibulares-sobre-primeiro-principio-da-termodinamica-ou-principio-da-conservacao-da-energia\/","title":{"rendered":"Primeiro Princ\u00edpio da Termodin\u00e2mica ou Princ\u00edpio da Conserva\u00e7\u00e3o da Energia"},"content":{"rendered":"

Resolu\u00e7\u00e3o comentada dos exerc\u00edcios de vestibulares sobre<\/b><\/span><\/span><\/span><\/p>\n

Primeiro Princ\u00edpio da Termodin\u00e2mica ou Princ\u00edpio da Conserva\u00e7\u00e3o da Energia<\/b><\/span><\/span><\/span><\/p>\n

01-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Se a temperatura \u00e9 a mesma para gases diferentes, a energia cin\u00e9tica das mol\u00e9culas deve ser a mesma, independente da massa de cada g\u00e1s. Como E<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=mV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 e a energia cin\u00e9tica \u00e9 a mesma o g\u00e1s de maior massa deve ter menor velocidade para que essas energias se igualem\u00a0 —\u00a0\u00a0R- C<\/b><\/span><\/span><\/span><\/p>\n

02-<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

A press\u00e3o exercida por um g\u00e1s \u00e9 o resultado das colis\u00f5es das mol\u00e9culas do g\u00e1s entre si e contra as paredes do recipiente\u00a0 —\u00a0\u00a0R- C<\/b><\/span><\/span><\/span><\/p>\n

03-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>01. Falsa\u00a0 —\u00a0 O postulado b\u00e1sico da teoria cin\u00e9tica dos gases \u00e9 que as dire\u00e7\u00f5es e as intensidades das velocidades das mol\u00e9culas est\u00e3o distribu\u00eddas ao acaso, ou seja, s\u00e3o diferentes para cada mol\u00e9cula do g\u00e1s.<\/b><\/span><\/span><\/span><\/p>\n

02. Verdadeira\u00a0 —\u00a0 veja teoria.<\/b><\/span><\/span><\/span><\/p>\n

03- Verdadeira\u00a0 — veja teoria<\/b><\/span><\/span><\/span><\/p>\n

04. Verdadeira\u00a0 —\u00a0 veja teoria.<\/b><\/span><\/span><\/span><\/p>\n

R- 01-F,=; 02-V; 3-V; 4-V\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n

04-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>(01) Verdadeira\u00a0 —\u00a0 como est\u00e3o em equil\u00edbrio t\u00e9rmico com o ambiente, possuem a mesma temperatura, consequentemente suas mol\u00e9culas ter\u00e3o a mesma energia cin\u00e9tica m\u00e9dia\u00a0 —\u00a0 m<\/b><\/span><\/span><\/span>am<\/b><\/span><\/span><\/sub><\/span>.(V<\/b><\/span><\/span><\/span>am<\/b><\/span><\/span><\/sub><\/span>)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2= m<\/b><\/span><\/span><\/span>ar<\/b><\/span><\/span><\/sub><\/span>.(V<\/b><\/span><\/span><\/span>ar<\/b><\/span><\/span><\/sub><\/span>)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 observe na express\u00e3o anterior que, para que a igualdade se mantenha, quanto maior a massa das mol\u00e9culas, menor ser\u00e1 sua velocidade.<\/b><\/span><\/span><\/span><\/p>\n

(02) Falsa\u00a0 —\u00a0 o som transporta energia, n\u00e3o transporta mat\u00e9ria.<\/b><\/span><\/span><\/span><\/p>\n

(04) Verdadeira\u00a0 —\u00a0 quanto maior a temperatura, maior a velocidade de agita\u00e7\u00e3o das mol\u00e9culas de am\u00f4nia.<\/b><\/span><\/span><\/span><\/p>\n

(08) Falsa\u00a0 —\u00a0 a perturba\u00e7\u00e3o de press\u00e3o corresponde a uma onda, cuja velocidade de propaga\u00e7\u00e3o depende do meio.<\/b><\/span><\/span><\/span><\/p>\n

(16) Verdadeira\u00a0 —\u00a0 veja teoria.<\/b><\/span><\/span><\/span><\/p>\n

R- (01 + 04 + 16) = 21<\/b><\/span><\/span><\/span><\/p>\n

05-<\/b><\/span><\/span><\/span> <\/b>R- B\u00a0\u00a0—\u00a0 veja teoria<\/b><\/span><\/span><\/span><\/p>\n

06-<\/b><\/span><\/span><\/span> <\/b>R- E\u00a0 —\u00a0 veja teoria\u00a0<\/b><\/span><\/span><\/span><\/p>\n

07-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Quanto maior a velocidade de agita\u00e7\u00e3o das mol\u00e9culas, elas escapar\u00e3o com maior facilidade\u00a0 —\u00a0\u00a0R- C<\/b><\/span><\/span><\/span><\/p>\n

08-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>01. Falsa\u00a0 —\u00a0 calor \u00e9 energia que \u00e9 transferida de um corpo para outro devido \u00e0 diferen\u00e7a de temperatura entre eles e \u00e9 transferida sempre do corpo de maior temperatura para o de menor temperatura. Portanto n\u00e3o \u00e9 energia armazenada.<\/b><\/span><\/span><\/span><\/p>\n

02. Falsa\u00a0 —\u00a0 veja 01.<\/b><\/span><\/span><\/span><\/p>\n

04. Falsa\u00a0 —\u00a0 veja 01.<\/b><\/span><\/span><\/span><\/p>\n

08. Correta\u00a0 —\u00a0 veja 01.<\/b><\/span><\/span><\/span><\/p>\n

16. Falsa\u00a0 —\u00a0 veja 01<\/b><\/span><\/span><\/span><\/p>\n

32. Verdadeira\u00a0 —\u00a0 como a temperatura do term\u00f4metro aumenta, sua energia interna tamb\u00e9m aumentar\u00e1.<\/b><\/span><\/span><\/span><\/p>\n

R- (08 + 32) = 40<\/b><\/span><\/span><\/span><\/p>\n

09-<\/b><\/span><\/span><\/span> <\/b>R- C e E\u00a0\u00a0—\u00a0\u00a0veja teoria.<\/b><\/span><\/span><\/span><\/p>\n

10-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>W=P.(V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>)\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 W=0\u00a0 —\u00a0\u00a0R- C<\/b><\/span><\/span><\/span><\/p>\n

11-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>W=P.(V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>)\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span><V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0(compress\u00e3o)\u00a0 —\u00a0 W<0\u00a0 —\u00a0\u00a0R- C<\/b><\/span><\/span><\/span><\/p>\n

12-<\/b><\/span><\/span><\/span> <\/b>R- C\u00a0 —\u00a0 veja teoria<\/b><\/span><\/span><\/span><\/p>\n

13-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Isot\u00e9rmica\u00a0 —\u00a0 Q=W\u00a0 —\u00a0 Q=W=150J\u00a0 —\u00a0 adiab\u00e1tica\u00a0 —\u00a0 mesmo trabalho\u00a0 —\u00a0 W=150J\u00a0 —\u00a0 \u0394U=-W\u00a0 —\u00a0\u00a0\u0394U=-150J<\/b><\/span><\/span><\/span><\/p>\n

14-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a) W=P.\u0394V=50.8\u00a0 —\u00a0\u00a0W=400J<\/b><\/span><\/span><\/span><\/p>\n

b) \u0394U=Q<\/b><\/span><\/span><\/span>\u00b4g\u00e1s<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 W\u00a0 —\u00a0 \u0394U=(2.000 \u2013 1.500) \u2013 400\u00a0 —\u00a0 \u0394U=500 \u2013 400\u00a0 —\u00a0\u00a0\u0394U=100J<\/b><\/span><\/span><\/span><\/p>\n

c)\u00a0aumenta, pois o g\u00e1s teve aumento de temperatura, devido ao aumento de energia interna.<\/b><\/span><\/span><\/span><\/p>\n

15-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>W=P.(V \u2013 V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>)=20.(10 \u2013 5)\u00a0 —\u00a0 W=100J\u00a0 —\u00a0 Q=250J\u00a0 —\u00a0 \u0394U=Q \u2013 W=250 \u2013 100=150J\u00a0 —\u00a0\u00a0R- B<\/b><\/span><\/span><\/span><\/p>\n

16-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>\u00c8 poss\u00edvel um g\u00e1s receber calor e sua temperatura n\u00e3o sofrer varia\u00e7\u00e3o desde que seja uma transforma\u00e7\u00e3o isot\u00e9rmica onde todo calor recebido \u00e9 transformado em trabalho, pois, sendo \u0394U=0\u00a0 —\u00a0 Q=W\u00a0 —\u00a0\u00a0R- E<\/b><\/span><\/span><\/span><\/p>\n

17-<\/b><\/span><\/span><\/span> <\/b>Mesma varia\u00e7\u00e3o de temperatura (\u0394T)\u00a0 —\u00a0 mesma varia\u00e7\u00e3o de energia interna (\u0394U)\u00a0 —\u00a0 \u0394U<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>= \u0394U<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>= \u0394U\u00a0 —\u00a0 isovolum\u00e9trica\u00a0 —\u00a0 \u0394U<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=Q<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>= \u0394U\u00a0 —\u00a0 isob\u00e1rica\u00a0 —\u00a0 \u0394U<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=Q<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 W\u00a0 —\u00a0 \u0394U=Q<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 W\u00a0 —\u00a0 Q<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=Q<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 W\u00a0 —\u00a0 Q<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 Q<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=W\u00a0 —\u00a0 Q<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>>Q<\/b><\/span><\/span><\/span>1\u00a0\u00a0<\/b><\/span><\/span><\/sub><\/span>—\u00a0\u00a0R- A<\/b><\/span><\/span><\/span><\/p>\n

18-<\/b><\/span><\/span><\/span> <\/b>R- C\u00a0\u00a0—\u00a0 veja teoria<\/b><\/span><\/span><\/span><\/p>\n

19-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Varia\u00e7\u00e3o de energia interna\u00a0 — \u0394U= U<\/b><\/span><\/span><\/span>final<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 U<\/b><\/span><\/span><\/span>inicial<\/b><\/span><\/span><\/sub><\/span>=20,8T<\/b><\/span><\/span><\/span>final<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 20,8T<\/b><\/span><\/span><\/span>inicial<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 \u0394U=20,8 \u0394T\u00a0 —\u00a0 \u0394U=Q \u2013 W\u00a0 —\u00a0 adiab\u00e1tica Q=0 (n\u00e3o houve trocas de calor)\u00a0 —\u00a0 \u0394U=0 \u2013 W\u00a0 —\u00a0 \u0394U=-W\u00a0 —\u00a0 20,8 \u0394T= -1.664\u00a0 —\u00a0 \u0394T= – 80K\u00a0 —\u00a0\u00a0R- C<\/b><\/span><\/span><\/span><\/p>\n

20-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a) W<\/b><\/span><\/span><\/span>AB<\/b><\/span><\/span><\/sub><\/span>=\u00e1rea=b.h=2.4.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 W<\/b><\/span><\/span><\/span>AB<\/b><\/span><\/span><\/sub><\/span>=8.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span>J\u00a0 —\u00a0 W<\/b><\/span><\/span><\/span>BC<\/b><\/span><\/span><\/sub><\/span>=0 (volume n\u00e3o varia)\u00a0 —\u00a0 W<\/b><\/span><\/span><\/span>total<\/b><\/span><\/span><\/sub><\/span>=W<\/b><\/span><\/span><\/span>AB<\/b><\/span><\/span><\/sub><\/span>\u00a0+ W<\/b><\/span><\/span><\/span>BC<\/b><\/span><\/span><\/sub><\/span>=8.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span>\u00a0+ 0\u00a0 —\u00a0\u00a0W<\/b><\/span><\/span><\/span>total<\/b><\/span><\/span><\/sub><\/span>=8.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span>J<\/b><\/span><\/span><\/span><\/p>\n

b) observe que os pontos A e C est\u00e3o sobre a mesma isoterma e, portanto, possuem a mesma temperatura e consequentemente a mesma energia interna\u00a0 —\u00a0 \u0394U=0\u00a0 —\u00a0 \u0394U=Q \u2013 W\u00a0 —\u00a0 0=Q \u2013 W\u00a0 —\u00a0\u00a0Q=W=8.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span>J<\/b><\/span><\/span><\/span><\/p>\n

21-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Est\u00e1 transformando calor em trabalho\u00a0 —\u00a0\u00a0R- A\u00a0<\/b><\/span><\/span><\/span><\/p>\n

22-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Observe que se trata de uma transforma\u00e7\u00e3o isob\u00e1rica\u00a0 —\u00a0 ocorre \u00e0 press\u00e3o constante\u00a0 —\u00a0 a varia\u00e7\u00e3o de temperatura (\u0394T) provoca uma varia\u00e7\u00e3o de energia interna (\u0394U) do sistema e a varia\u00e7\u00e3o de volume (\u0394V) produz trabalho. Parte do calor (Q) recebido pelo sistema \u00e9 armazenada sob forma de energia interna e parte \u00e9 transformada em trabalho, de modo que \u0394U=Q \u2013 W\u00a0 —\u00a0 observe que W>0 (expans\u00e3o isob\u00e1rica)\u00a0 —\u00a0\u00a0R- D<\/b><\/span><\/span><\/span><\/p>\n

23-<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

24-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Trata-se de uma expans\u00e3o adiab\u00e1tica (muito r\u00e1pida e sem troca de calor com o meio ambiente), onde Q=0 e \u0394U= Q \u2013 W\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

\u0394U= 0 \u2013 W\u00a0 —\u00a0\u00a0\u0394U= \u2013 W\u00a0 —\u00a0 o volume do g\u00e1s aumenta (trabalho positivo) fazendo com que a energia interna (\u0394U) fique negativa e diminua, diminuindo assim, a press\u00e3o e a temperatura e a press\u00e3o —\u00a0 o g\u00e1s resfria\u00a0 —\u00a0\u00a0R- A\u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n

25-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>No trecho AB, como se trata de uma isob\u00e1rica V\/T=K\u00a0 —\u00a0 como o volume aumenta a temperatura tamb\u00e9m deve aumentar provocando um aumento da energia interna\u00a0 —\u00a0 Falsa<\/b><\/span><\/span><\/span><\/p>\n

– O trecho BC corresponde a uma isot\u00e9rmica (mesma temperatura)\u00a0 —\u00a0 Falsa<\/b><\/span><\/span><\/span><\/p>\n

– O trecho CA corresponde a outra isot\u00e9rmica (T constante)\u00a0 —\u00a0 Q=0\u00a0 —\u00a0 \u0394U=Q \u2013 W\u00a0 —\u00a0 \u0394U=0 \u2013 W\u00a0 —\u00a0 \u0394U= \u2013 W\u00a0 — como o volume varia, a existe trabalho e, portanto \u0394U tamb\u00e9m varia\u00a0 —\u00a0 Falsa\u00a0<\/b><\/span><\/span><\/span><\/p>\n

R- E<\/b><\/span><\/span><\/span><\/p>\n

26-<\/b><\/span><\/span><\/span>Trecho AB\u00a0 —\u00a0 isot\u00e9rmica\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=P<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>.0,1=0,5.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span>.0,5\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=2,5.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span>N\/m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0—\u00a0 Verdadeira<\/b><\/span><\/span><\/span><\/p>\n

Trecho AB\u00a0 —\u00a0 isot\u00e9rmica\u00a0 —\u00a0 \u0394U=0\u00a0 —\u00a0\u00a0 Q=W\u00a0 —\u00a0 todo calor Q recebido pelo sistema \u00e9 totalmente transformado em trabalho\u00a0 —\u00a0 Falsa<\/b><\/span><\/span><\/span><\/p>\n

Trecho BC\u00a0 —\u00a0 isob\u00e1rica\u00a0 —\u00a0 W=b.h=(0,1 \u2013 0,5).0,5.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 W= – 2,0.10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span>J\u00a0 —\u00a0 Falsa<\/b><\/span><\/span><\/span><\/p>\n

Trecho CA\u00a0 —\u00a0 isovolum\u00e9trica\u00a0 —\u00a0 W=0\u00a0 —\u00a0 Verdadeira<\/b><\/span><\/span><\/span><\/p>\n

Falsa\u00a0 —\u00a0 observe que o W no trecho CD \u00e9 negativo e que o trabalho no ciclo \u00e9 a soma das \u00e1reas do trabalho de cada trecho.<\/b><\/span><\/span><\/span><\/p>\n

R- V F F V F<\/b><\/span><\/span><\/span><\/p>\n

27-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Nos dois processos a temperatura interna do ar dentro da bola permanece constante\u00a0 —\u00a0 trata-se de uma isoterma, que \u00e9 curva\u00a0 —\u00a0\u00a0R- B<\/b><\/span><\/span><\/span><\/p>\n

28-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Como o volume \u00e9 constante, trata-se de uma isovolum\u00e9trica o —\u00a0 P\/T=constante\u00a0 —\u00a0 mas V tamb\u00e9m \u00e9 constante\u00a0 —\u00a0\u00a0R- D<\/b><\/span><\/span><\/span><\/p>\n

29-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Adiab\u00e1tica\u00a0 —\u00a0 Q=0\u00a0 —\u00a0 \u0394U=Q \u2013 W\u00a0 —\u00a0 \u0394U=-W\u00a0 —\u00a0 W + \u0394U=0\u00a0 —\u00a0 249 + (3\/2)R\u0394T=0\u00a0 —\u00a0 249 + (3\/2).8,3.(T \u2013 T<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>)=0\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

249 + (3\/2).8,3.(T \u2013 333)=0\u00a0 —\u00a0 249 + 12,45.(T \u2013 333)=0\u00a0 —\u00a0 (T -333)=-249\/12,45\u00a0 —\u00a0 T=333 \u2013 20\u00a0 —\u00a0\u00a0T=313K=40<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>C<\/b><\/span><\/span><\/span><\/p>\n

30-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>I. Falsa\u00a0 —\u00a0 a parede n\u00e3o permite trocas de calor (\u00e9 adiab\u00e1tica)<\/b><\/span><\/span><\/span><\/p>\n

II. Verdadeira\u00a0 —\u00a0 o recipiente com maior temperatura faz maior press\u00e3o na parede.<\/b><\/span><\/span><\/span><\/p>\n

III. Falsa<\/b><\/span><\/span><\/span><\/p>\n

R- B<\/b><\/span><\/span><\/span><\/p>\n

31-<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

32-<\/b><\/span><\/span><\/span> <\/b>R- A<\/b><\/span><\/span><\/span><\/p>\n

33-<\/b><\/span><\/span><\/span> <\/b>R- D\u00a0\u00a0—\u00a0 veja teoria<\/b><\/span><\/span><\/span><\/p>\n

34-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Sendo o trabalho realizado numa transforma\u00e7\u00e3o c\u00edclica \u00e9 fornecido pelo c\u00e1lculo da \u00e1rea do ciclo\u00a0 —\u00a0\u00a0R- E<\/b><\/span><\/span><\/span><\/p>\n

35-<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

36-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Trata-se de uma expans\u00e3o adiab\u00e1tica (muito r\u00e1pida e sem troca de calor com o meio ambiente), onde Q=0 e \u0394U= Q \u2013 W\u00a0 —\u00a0 \u0394U= 0 \u2013 W\u00a0 —\u00a0\u00a0\u0394U= \u2013 W\u00a0 —\u00a0 o volume do g\u00e1s aumenta (trabalho positivo) fazendo com que a energia interna (\u0394U) fique negativa e diminua, diminuindo assim, a press\u00e3o e a temperatura e a press\u00e3o, e o g\u00e1s resfria\u00a0 —\u00a0\u00a0R- D\u00a0<\/b><\/span><\/span><\/span><\/p>\n

37-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>01. Falsa\u00a0 —\u00a0 o trabalho total \u00e9 fornecido pela soma das \u00e1reas das duas transforma\u00e7\u00f5es.<\/b><\/span><\/span><\/span><\/p>\n

02. Correta\u00a0 —\u00a0 \u0394U=100J\u00a0 —\u00a0 W=\u00e1rea do trap\u00e9zio + \u00e1rea do ret\u00e2ngulo=(B + b).h\/2 + b.h =(120 + 40).4\/2 + 8.120\u00a0 — W= 320 + 960=1.280J\u00a0 —\u00a0\u00a0 \u0394U=Q \u2013 W\u00a0 —\u00a0 100=Q \u2013 1.280\u00a0 —\u00a0 Q=1.380J.<\/b><\/span><\/span><\/span><\/p>\n

04. Falsa\u00a0 —\u00a0 s\u00e3o duas transforma\u00e7\u00f5es e nenhuma delas isot\u00e9rmica (gr\u00e1fico \u00e9 uma curva).<\/b><\/span><\/span><\/span><\/p>\n

08. Correta\u00a0 —\u00a0 veja 02.<\/b><\/span><\/span><\/span><\/p>\n

16- Falsa\u00a0 —\u00a0 Falsa \u00a0—\u00a0 houve trocas de calor\u00a0 —\u00a0 veja 02.<\/b><\/span><\/span><\/span><\/p>\n

R- (02 + 08) = 10\u00a0<\/b><\/span><\/span><\/span><\/p>\n

38-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>W=P.\u0394V=1,1.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span>.(4.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>\u00a0\u2013 2.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>)\u00a0 —\u00a0 W=2,2.10<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>J\u00a0 —\u00a0 \u0394U=Q \u2013 W=5,5.10<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0\u2013 2,2.10<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 \u0394U=3,3.10<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>J\u00a0 —\u00a0\u00a0R- D<\/b><\/span><\/span><\/span><\/p>\n

39-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>A energia interna do sistema aumentar\u00e1, fazendo com que o g\u00e1s realize trabalho se expandindo e deslocando o \u00eambolo da seringa\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n

R- B
\n<\/b><\/span><\/span><\/span><\/p>\n

40-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Pela 1\u00aa lei da Termodin\u00e2mica\u00a0 — Q = W + DU\u00a0 —\u00a0 como A e B est\u00e3o na mesma isoterma tem a mesma temperatura e ent\u00e3o DU = 0\u00a0 —\u00a0 logo Q = W, onde W \u00e9 o trabalho realizado entre A e B. Este trabalho \u00e9 igual a \u00e1rea do diagrama PV neste intervalo\u00a0 —\u00a0 para o c\u00e1lculo da \u00e1rea \u00e9 necess\u00e1rio determinar a press\u00e3o do ponto B\u00a0 —\u00a0 como A e B est\u00e3o na mesma isoterma\u00a0 —\u00a0\u00a0 PV = constante\u00a0 —\u00a0\u00a0\u00a0 p0.V0 = p.V0\/3\u00a0\"\"\u00a0p = 3.p0\u00a0 —\u00a0 QAB = (p0 + 3p0).(V0 \u2013 V0\/3)\/2 = 4p0.(2V0\/3)\/2\u00a0 —\u00a0 QAB = 4p0V\u00ad0\/3<\/b><\/span><\/span><\/span><\/p>\n

41-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>I. Correta\u00a0 —\u00a0 as temperaturas finais e iniciais de ambos s\u00e3o as mesmas\u00a0 —\u00a0 mesma \u0394T\u00a0 —\u00a0 mesma \u0394U.<\/b><\/span><\/span><\/span><\/p>\n

II. Falsa\u00a0 —\u00a0 isoc\u00f3rica\u00a0 —\u00a0 \u0394U= Q \u2013 W\u00a0 —\u00a0 Q= \u0394U + W\u00a0 —\u00a0 isoc\u00f3rico\u00a0 —\u00a0 W=0\u00a0 — \u0394U= Q \u2013 W\u00a0 —\u00a0 Q= \u0394U.<\/b><\/span><\/span><\/span><\/p>\n

III. Correta\u00a0 —\u00a0 a isoterma que cont\u00e9m T<\/b><\/span><\/span><\/span>x<\/b><\/span><\/span><\/sub><\/span>\u00a0est\u00e1 mais afastada dos eixos que a isoterma que cont\u00e9m T<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

R- C<\/b><\/span><\/span><\/span><\/p>\n

42-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a) Os processos isot\u00e9rmicos (curvas) s\u00e3o os trechos AB e CD\u00a0 —\u00a0 os isoc\u00f3ricos correspondem \u00e0s retas verticais.<\/b><\/span><\/span><\/span><\/p>\n

b) AB\u00a0 —\u00a0 isot\u00e9rmica\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=P<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 3P<\/b><\/span><\/span><\/span>atm<\/b><\/span><\/span><\/sub><\/span>.V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=P<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>.2V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0\u00a0P<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>=3P<\/b><\/span><\/span><\/span>atm<\/b><\/span><\/span><\/sub><\/span>\/2\u00a0 —\u00a0 DC\u00a0 —\u00a0 isot\u00e9rmica\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>D<\/b><\/span><\/span><\/sub><\/span>V<\/b><\/span><\/span><\/span>D<\/b><\/span><\/span><\/sub><\/span>=P<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>V<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>atm<\/b><\/span><\/span><\/sub><\/span>.V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=P<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>.2V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0\u00a0P<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>=P<\/b><\/span><\/span><\/span>atm<\/b><\/span><\/span><\/sub><\/span>\/2<\/b><\/span><\/span><\/span><\/p>\n

c)\u00a0AD\u00a0 —\u00a0 isoc\u00f3rica\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\/T<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=P<\/b><\/span><\/span><\/span>D<\/b><\/span><\/span><\/sub><\/span>\/T<\/b><\/span><\/span><\/span>D<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 3P<\/b><\/span><\/span><\/span>atm<\/b><\/span><\/span><\/sub><\/span>\/T<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=P<\/b><\/span><\/span><\/span>atm<\/b><\/span><\/span><\/sub><\/span>\/T<\/b><\/span><\/span><\/span>D<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0\u00a0T<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\/T<\/b><\/span><\/span><\/span>D<\/b><\/span><\/span><\/sub><\/span>=3<\/b><\/span><\/span><\/span><\/p>\n

43-<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

44-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Nas transforma\u00e7\u00f5es isob\u00e1ricas a press\u00e3o permanece constante, variando assim, o volume e a temperatura. Se a temperatura varia a energia interna varia, o que invalida a op\u00e7\u00e3o A.<\/b><\/span><\/span><\/span><\/p>\n

Nas transforma\u00e7\u00f5es isom\u00e9tricas o volume permanece constante, e desta forma, press\u00e3o e temperatura variar\u00e3o, o que invalida a op\u00e7\u00e3o B.<\/b><\/span><\/span><\/span><\/p>\n

N\u00e3o h\u00e1 trabalho, de fato se o volume n\u00e3o varia, nas transforma\u00e7\u00f5es isom\u00e9tricas ou isovolum\u00e9tricas. O que invalida a op\u00e7\u00e3o D.<\/b><\/span><\/span><\/span><\/p>\n

Nas transforma\u00e7\u00f5es isom\u00e9tricas apenas o volume permanece constante, a press\u00e3o e a temperatura variam, o que invalida a op\u00e7\u00e3o E.<\/b><\/span><\/span><\/span><\/p>\n

Apenas na op\u00e7\u00e3o C temos corre\u00e7\u00e3o, pois de fato nas transforma\u00e7\u00f5es adiab\u00e1ticas n\u00e3o h\u00e1 troca de calor entre o g\u00e1s e o recipiente, al\u00e9m \u00e9 claro do meio externo.<\/b><\/span><\/span><\/span><\/p>\n

\u00a0R- C\u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n

45-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a) trecho AB\u00a0 —\u00a0\u00a0 isob\u00e1rica\u00a0 —\u00a0\u00a0 W=P.(V \u2013 V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>)-10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span>.(70.10<\/b><\/span><\/span><\/span>-4<\/b><\/span><\/span><\/sup><\/span>\u00a0\u2013 20.10<\/b><\/span><\/span><\/span>-4<\/b><\/span><\/span><\/sup><\/span>) —\u00a0 W=500J\u00a0 —\u00a0 \u0394U=Q \u2013 W\u00a0 —\u00a0 1.000=Q \u2013 500\u00a0 —\u00a0\u00a0Q=1.500J<\/b><\/span><\/span><\/span><\/p>\n

b) trecho BC\u00a0 —\u00a0 isovolum\u00e9trica\u00a0 —\u00a0 W=0\u00a0 —\u00a0 o volume n\u00e3o varia<\/b><\/span><\/span><\/span><\/p>\n

c) P<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\/T<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>=P<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>\/T<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span>\/350=P<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>\/700\u00a0 —\u00a0\u00a0P<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>=2.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span>N\/m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0(Pa)<\/b><\/span><\/span><\/span><\/p>\n

46-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Q = U + W<\/b><\/span><\/span><\/span>AB<\/b><\/span><\/span><\/sub><\/span>\u00a0+ W<\/b><\/span><\/span><\/span>BC<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0\u00a0 Q = U + P<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\u00a0(V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>) + W\u00a0\u00a0 —\u00a0\u00a0R- D<\/b><\/span><\/span><\/span><\/p>\n

47-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a) Equa\u00e7\u00e3o geral dos gases perfeitos\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\/T<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=P<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\/T<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 \u00a0isob\u00e1rica\u00a0 —\u00a0\u00a0\u00a0 V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\/T<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\/T<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 2.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>\/300=3,5.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>\/T<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0\u00a0T<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=525K<\/b><\/span><\/span><\/span><\/p>\n

b) Primeira lei da termodin\u00e2mica, a varia\u00e7\u00e3o da energia interna (DU) \u00e9 igual \u00e0 diferen\u00e7a entre o calor recebido (Q) e o trabalho realizado (W)\u00a0 —\u00a0 \u00a0DU = Q \u2013 W\u00a0 —\u00a0 \u00a0isob\u00e1rica, o trabalho realizado \u00e9\u00a0 —\u00a0 W = p DV\u00a0 —\u00a0 DU = Q \u2013 p DV = 375 \u2013 10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span>(3,5 \u00b4 10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>\u00a0\u2013 2 \u00b4 10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>) = 375 \u2013 10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span>(1,5 \u00b4 10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>)\u00a0 —\u00a0 \u00a0DU = 375 \u2013 150\u00a0 —\u00a0\u00a0\u00a0DU = 225 J<\/b><\/span><\/span><\/span><\/p>\n

48-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Observe as figuras abaixo:<\/b><\/span><\/span><\/span><\/p>\n

\u00a0<\/span>Nas figuras ao lado\u00a0 —\u00a0 A: \u00e1rea da sec\u00e7\u00e3o transversal do \u00eambolo\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/sub><\/span>: m\u00f3dulo da for\u00e7a el\u00e1stica\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/sub><\/span>\u00a0= k<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub><\/span>x\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>G<\/b><\/span><\/span><\/sub><\/span>: m\u00f3dulo da for\u00e7a de press\u00e3o exercida pelo g\u00e1s\u00a0 —\u00a0 \u00a0F<\/b><\/span><\/span><\/span>G<\/b><\/span><\/span><\/sub><\/span>\u00a0= P<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub><\/span>A\u00a0 —\u00a0 Dados: P<\/b><\/span><\/span><\/span>0<\/b><\/span><\/span><\/sub><\/span>; V<\/b><\/span><\/span><\/span>0<\/b><\/span><\/span><\/sub><\/span>; V = 2<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub><\/span>V<\/b><\/span><\/span><\/span>0<\/b><\/span><\/span><\/sub><\/span>\u00a0 e\u00a0 n = 1 mol\u00a0 —\u00a0 \u00a0o sistema est\u00e1 termicamente isolado, ou seja, a transforma\u00e7\u00e3o \u00e9 adiab\u00e1tica (Q = 0)\u00a0 —\u00a0 1\u00aa lei da termodin\u00e2mica\u00a0 —\u00a0 DU = Q \u2013 W\u00a0 —\u00a0 \u00a0DU = 0 \u2013<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub><\/span>W \u00a0\u00a0—\u00a0 W = \u2013<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub><\/span>DU \u00a0—\u00a0 W =-(3\/2)nRT = -(3\/2).(1)R.(T \u2013 T<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>) —\u00a0 W=3\/2.R.(T<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 T)\u00a0 —\u00a0 esse trabalho \u00e9 armazenado na mola na forma de energia potencial el\u00e1stica\u00a0 —\u00a0 KX<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=(3\/2).R.(T<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 T)\u00a0 —\u00a0 kx<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=3R.(T<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 T) (I)\u00a0 —\u00a0 volume inicial do \u00eambolo\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=A.x\u00a0 —\u00a0 x=V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\/A (II)\u00a0 —\u00a0 na figura (b), na posi\u00e7\u00e3o de equil\u00edbrio\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/sub><\/span>\u00a0= F<\/b><\/span><\/span><\/span>G<\/b><\/span><\/span><\/sub><\/span>\u00a0\u00de k<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub><\/span>x = P<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub><\/span>A (III)\u00a0 —\u00a0 as equa\u00e7\u00f5es (II) e (III) sugerem que\u00a0 —\u00a0 k<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub><\/span>x<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0= (k<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub><\/span>x)<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub><\/span>(x) = (P<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub><\/span>A).(V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\/A) \u00a0—\u00a0 k<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub><\/span>x<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0= P<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub><\/span>V<\/b><\/span><\/span><\/span>0<\/b><\/span><\/span><\/sub><\/span>(IV)\u00a0 —\u00a0 novamente na figura (b)\u00a0 —\u00a0 PV=nRT\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub><\/span>(2V<\/b><\/span><\/span><\/span>0<\/b><\/span><\/span><\/sub><\/span>) = (1)<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub><\/span>R<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub><\/span>T \u00a0—\u00a0 \u00a0P<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub><\/span>V<\/b><\/span><\/span><\/span>0<\/b><\/span><\/span><\/sub><\/span>\u00a0=RT\/2 (V)\u00a0 —\u00a0\u00a0 (IV) e (V)\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\"\"<\/p>\n

k<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub><\/span>x<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0=RT\/2\u00a0 —\u00a0\u00a0 substituindo essa express\u00e3o na equa\u00e7\u00e3o (I), temos\u00a0 —\u00a0 RT\/2=3R(T<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 T) \u00a0—\u00a0 T = 6(T<\/b><\/span><\/span><\/span>0<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 T) \u00a0—\u00a0 7T = 6<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub><\/span>T<\/b><\/span><\/span><\/span>0<\/b><\/span><\/span><\/sub><\/span>\u00a0\u00a0—\u00a0 T=(6\/7)t<\/b><\/span><\/span><\/span>O<\/b><\/span><\/span><\/sub><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

49-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>(01) Correta\u00a0 —\u00a0\u00a0 da equa\u00e7\u00e3o geral dos gases perfeitos\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\/T<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=P<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\/T<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 como T<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0= T<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0\u00a0—\u00a0 \u00a0p<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0= p<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>.\u00a0<\/b><\/span><\/span><\/span><\/p>\n

(02) Correta\u00a0 —\u00a0\u00a0 ao ser comprimido bruscamente, o ar sofre aquecimento, perdendo calor para o meio externo (\u00e1gua e gelo), provocando fus\u00e3o de certa massa de gelo.<\/b><\/span><\/span><\/span><\/p>\n

(04) Correta\u00a0 —\u00a0\u00a0 como j\u00e1 afirmado na proposi\u00e7\u00e3o anterior, a compress\u00e3o \u00e9 brusca, o g\u00e1s aquece rapidamente, sendo toda a energia transferida na compress\u00e3o transformada em energia interna do ar, pois, nesse intervalo de tempo t\u00e3o pequeno a quantidade de calor que atravessa as paredes do cilindro \u00e9 praticamente nula, o que caracteriza uma transforma\u00e7\u00e3o \u00e9 adiab\u00e1tica\u00a0 —\u00a0 a seguir, o g\u00e1s come\u00e7a a perder calor para o sistema \u00e1gua-gelo, sob press\u00e3o constante, que \u00e9 a press\u00e3o exercida pela for\u00e7a de compress\u00e3o que o tijolo provoca no \u00eambolo, que tem a mesma intensidade de seu peso, constante\u00a0 —\u00a0 ent\u00e3o, a press\u00e3o \u00e9 constante, caracterizando uma transforma\u00e7\u00e3o isob\u00e1rica.<\/b><\/span><\/span><\/span><\/p>\n

(08) Errada\u00a0 —\u00a0 j\u00e1 justificado nas proposi\u00e7\u00f5es anteriores.<\/b><\/span><\/span><\/span><\/p>\n

(16) Errada\u00a0 —\u00a0 \u00a0J\u00e1 justificado nas proposi\u00e7\u00f5es anteriores.\u00a0<\/b><\/span><\/span><\/span><\/p>\n

R- (01+ 02 + 04) = 07<\/b><\/span><\/span><\/span><\/p>\n

50-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>01) Falsa\u00a0 —\u00a0\u00a0 numa transforma\u00e7\u00e3o isot\u00e9rmica, a varia\u00e7\u00e3o da energia interna (DU) \u00e9 nula\u00a0 —\u00a0\u00a0 logo, o calor (Q) recebido \u00e9 transformado integralmente em trabalho (W), como indica a primeira lei da termodin\u00e2mica\u00a0 —\u00a0 Q = DU + W\u00a0 —\u00a0 Q = 0 + W\u00a0 —\u00a0\u00a0Q = W.<\/b><\/span><\/span><\/span><\/p>\n

02) Correta\u00a0 —\u00a0 dados: P<\/b><\/span><\/span><\/span>AB<\/b><\/span><\/span><\/sub><\/span>\u00a0= 8 \u00b4 10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span>\u00a0Pa e DV<\/b><\/span><\/span><\/span>AB<\/b><\/span><\/span><\/sub><\/span>\u00a0= 2 \u00b4 10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>\u00a0m<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 W<\/b><\/span><\/span><\/span>AB<\/b><\/span><\/span><\/sub><\/span>\u00a0= P<\/b><\/span><\/span><\/span>AB<\/b><\/span><\/span><\/sub><\/span>\u00a0DV<\/b><\/span><\/span><\/span>AB<\/b><\/span><\/span><\/sub><\/span>\u00a0=\u00a0 (8 \u00b4 10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span>\u00a0) (2 \u00b4 10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>) = 16 \u00b4 10<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0= 1,6 \u00b4 10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>\u00a0J\u00a0 —\u00a0\u00a0W<\/b><\/span><\/span><\/span>AB<\/b><\/span><\/span><\/sub><\/span>=1,6 kJ.<\/b><\/span><\/span><\/span><\/p>\n

04) Correta\u00a0 —\u00a0\u00a0 aplicando a lei geral dos gases ideais para os estados A e B\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\/T<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=P<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>V<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>\/T<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 5\/900=7\/T<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 T<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=1.260K\u00a0 —\u00a0 \u00a0lei geral dos gases ideais para os estados B e C\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\/T<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=P<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>V<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>\/T<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 \u00a0\u00a08 \u00b4 10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span>\u00a0\u00b4 7 \u00b4 10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>\u00a0= p<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>\u00a0(9 \u00b4 10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>)\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

\u00a0 P<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=56.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span>\/9\u00a0\u00a0 —\u00a0 \u00a0\u00a0p<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>\u00a0@ 6,22.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span>\u00a0Pa.<\/b><\/span><\/span><\/span><\/p>\n

08) Falsa\u00a0 — a temperatura final \u00e9 igual \u00e0 inicial, mas ao longo do ciclo h\u00e1 aquecimentos e resfriamentos.<\/b><\/span><\/span><\/span><\/p>\n

16) Falsa\u00a0 —\u00a0 a transforma\u00e7\u00e3o CD \u00e9 uma expans\u00e3o adiab\u00e1tica, pois V<\/b><\/span><\/span><\/span>D<\/b><\/span><\/span><\/sub><\/span>\u00a0> V<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

32) Correta\u00a0 —\u00a0 a transforma\u00e7\u00e3o EA \u00e9 isoc\u00f3rica (W<\/b><\/span><\/span><\/span>EA<\/b><\/span><\/span><\/sub><\/span>\u00a0= 0)\u00a0 —\u00a0\u00a0 da primeira lei da termodin\u00e2mica\u00a0 —\u00a0 Q = DU + W\u00a0 —\u00a0 \u00a0\u00a0Q = DU + 0\u00a0\u00a0 —\u00a0\u00a0\u00a0Q = DU.\u00a0<\/b><\/span><\/span><\/span><\/p>\n

R- (02 + 04 +32)=38<\/b><\/span><\/span><\/span><\/p>\n

51-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a) Observe os processos indicados no gr\u00e1fico\u00a0 —\u00a0 de para b \u00e9 um processo isoc\u00f3rico\u00a0 —\u00a0 \u2206V = 0\u00a0 —\u00a0 \u2206P= (1,0.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span>\u00a0\u2013 3,0.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span>)=<\/b><\/span><\/span><\/span><\/p>\n

-2.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span>\u00a0Pa \u00a0—\u00a0 de b para c o processo \u00e9 isob\u00e1rico\u00a0 —\u00a0 \u2206V=(6.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>\u00a0\u2013 2.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>)= 4.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>\u00a0m<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 entre a e c\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span>.V<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span>\/T<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span>\u00a0= P<\/b><\/span><\/span><\/span>b<\/b><\/span><\/span><\/sub><\/span>.V<\/b><\/span><\/span><\/span>b<\/b><\/span><\/span><\/sub><\/span>\/T<\/b><\/span><\/span><\/span>b<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

3.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span>.2.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>\/T<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span>\u00a0= 1.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span>.6.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>\/T<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0\u00a0T<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span>\/T<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=1<\/b><\/span><\/span><\/span><\/p>\n

b) Energia interna do g\u00e1s\u00a0 —\u00a0 U=3\/2nRT\u00a0 —\u00a0 T<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span>=T<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>\u00a0(veja \u00edtem anterior)\u00a0 —\u00a0 como n e R s\u00e3o constantes voc\u00ea conclui que\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

U<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span>=U<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>\u00a0e que \u2206U<\/b><\/span><\/span><\/span>ac<\/b><\/span><\/span><\/sub><\/span>=0\u00a0 —\u00a0 \u2206U<\/b><\/span><\/span><\/span>ac<\/b><\/span><\/span><\/sub><\/span>=Q<\/b><\/span><\/span><\/span>ac<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 W<\/b><\/span><\/span><\/span>ac<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 0=Q<\/b><\/span><\/span><\/span>ac<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 W<\/b><\/span><\/span><\/span>ac<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 Q<\/b><\/span><\/span><\/span>ac<\/b><\/span><\/span><\/sub><\/span>=W<\/b><\/span><\/span><\/span>ac<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 W<\/b><\/span><\/span><\/span>ac<\/b><\/span><\/span><\/sub><\/span>=W<\/b><\/span><\/span><\/span>ab<\/b><\/span><\/span><\/sub><\/span>\u00a0+ W<\/b><\/span><\/span><\/span>bc<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 W<\/b><\/span><\/span><\/span>ac<\/b><\/span><\/span><\/sub><\/span>=0 + W<\/b><\/span><\/span><\/span>bc<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 W<\/b><\/span><\/span><\/span>ac<\/b><\/span><\/span><\/sub><\/span>=W<\/b><\/span><\/span><\/span>bc<\/b><\/span><\/span><\/sub><\/span>=<\/b><\/span><\/span><\/span><\/p>\n

1.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span>.(6 \u2013 2).10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>\u00a0—\u00a0\u00a0Q<\/b><\/span><\/span><\/span>ac<\/b><\/span><\/span><\/sub><\/span>=W<\/b><\/span><\/span><\/span>ac<\/b><\/span><\/span><\/sub><\/span>=4,0.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>\u00a0J<\/b><\/span><\/span><\/span><\/p>\n

52-<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

R- C<\/b><\/span><\/span><\/span><\/p>\n

53-<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0<\/span><\/p>\n

54-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Os biocombustiveis de primeira gera\u00e7\u00e3o s\u00e3o produzidos a partir da fermenta\u00e7\u00e3o alco\u00f3lica dos a\u00e7\u00facares vegetais da biomassa que contenha amido ou sacarose, como por exemplo o milho, o trigo, o sorgo, o amendoim, a soja, a beterraba, a batata doce, o girassol e a cana-de-a\u00e7\u00facar, n\u00e3o utilizando a celulose —\u00a0\u00a0 por\u00e9m, para evitar que se atinja o limite da oferta ou venha a ocorrer a competi\u00e7\u00e3o pelo uso da terra para a produ\u00e7\u00e3o de biocombust\u00edveis e de alimentos, \u00e9 necess\u00e1rio investir no desenvolvimento de tecnologias de segunda gera\u00e7\u00e3o para produ\u00e7\u00e3o de etanol com o aproveitamento eficiente da celulose\u00a0 —\u00a0 \u00a0estimativa \u00e9 de que o aproveitamento do baga\u00e7o e parte das palhas e pontas da cana-de-a\u00e7\u00facar eleve a produ\u00e7\u00e3o de \u00e1lcool em 30% a 40%, para uma mesma \u00e1rea plantada\u00a0 — \u00a0demais mat\u00e9rias-primas para as quais se buscam tecnologias de processamento da celulose, tais como capim-elefante, braqui\u00e1rias, panicuns e \u00e1rvores de crescimento r\u00e1pido podem representar alternativas competitivas e eficientes para locais onde n\u00e3o se cultiva ou cultivar\u00e1 cana-de-a\u00e7\u00facar, podendo gerar novos empregos\u00a0 —\u00a0\u00a0R- A.<\/b><\/span><\/span><\/span><\/p>\n

\u00a055- <\/b><\/span><\/span><\/span>Pode-se definir o Segundo Princ\u00edpio da Termodin\u00e2mica da seguinte maneira: \u201c\u00c9 imposs\u00edvel obter uma m\u00e1quina t\u00e9rmica que, operando em ciclos, seja capaz de transformar totalmente o calor por ela recebido em trabalho\u201d\u00a0 — sempre haver\u00e1 energia dissipada pelo motor\u00a0 —\u00a0\u00a0\u00a0R- C.<\/b><\/span><\/span><\/p>\n

\u00a0<\/span><\/p>\n

56- <\/b><\/span><\/span><\/span>Veja a teoria a seguir:<\/b><\/span><\/span><\/span><\/p>\n

\u00c9 um modelo te\u00f3rico. \u00c9 um g\u00e1s que obedece \u00e0s equa\u00e7\u00f5es\u00a0p\u00b7V\/T = k\u00a0e\u00a0p\u00b7V = n\u00b7R\u00b7T\u00a0 —\u00a0 o modelo adotado para um g\u00e1s ideal compreende a\u00a0Teoria Cin\u00e9tica Molecular dos Gases, cujos aspectos fundamentais s\u00e3o:<\/b><\/span><\/span><\/span><\/p>\n

* A press\u00e3o do g\u00e1s \u00e9 exercida igualmente em todos os pontos do recipiente;<\/b><\/span><\/span><\/span><\/p>\n

* Os choques entre as mol\u00e9culas s\u00e3o el\u00e1sticos;<\/b><\/span><\/span><\/span><\/p>\n

* As mol\u00e9culas de um g\u00e1s s\u00e3o pontos materiais, ou seja, possuem massa, mas apresentam um volume praticamente nulo;<\/b><\/span><\/span><\/span><\/p>\n

* Em um g\u00e1s ideal n\u00e3o h\u00e1 atra\u00e7\u00e3o nem repuls\u00e3o entre as mol\u00e9culas.<\/b><\/span><\/span><\/span><\/p>\n

[A] Correta\u00a0 —\u00a0 veja teoria.<\/b><\/span><\/span><\/span><\/p>\n

[B] Falsa\u00a0 —\u00a0 veja teoria.<\/b><\/span><\/span><\/span><\/p>\n

[C] Falsa\u00a0 —\u00a0 se V \u00e9 constante\u00a0 —\u00a0 P.V\/T=k (constante)\u00a0 —\u00a0 observe nessa express\u00e3o que P \u00e9 diretamente proporcional a T.<\/b><\/span><\/span><\/span><\/p>\n

[D] Falsa\u00a0 —\u00a0 P.V=k\u00a0 —\u00a0 P e V s\u00e3o inversamente proporcionais.<\/b><\/span><\/span><\/span><\/p>\n

[E] Falsa\u00a0 —\u00a0 V\/T=k\u00a0 —\u00a0 V e T s\u00e3o diretamente proporcionais.<\/b><\/span><\/span><\/span><\/p>\n

<\/a> R- A.<\/b><\/span><\/span><\/span><\/p>\n

\u00a0<\/span><\/p>\n

57-<\/b><\/span><\/span><\/span> Ocorre \u00e0 press\u00e3o constante. A varia\u00e7\u00e3o de temperatura (\u0394T) provoca uma varia\u00e7\u00e3o de energia interna (\u0394U) do<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

sistema e a varia\u00e7\u00e3o\u00a0de volume (\u0394V) produz trabalho. Parte do calor (Q) recebido pelo sistema \u00e9 armazenada sob forma de energia interna e parte \u00e9 transformada em trabalho, de modo que \u0394U=Q \u2013 W.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

C\u00e1lculo do trabalho realizado pelo ambiente sobre o g\u00e1s\u00a0 —\u00a0 W=P.\u2206V=4.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>.(-2.10<\/b><\/span><\/span><\/span>-1<\/b><\/span><\/span><\/sup><\/span>)\u00a0 —\u00a0 W= – 8.10<\/b><\/span><\/span><\/span>2\u00a0<\/b><\/span><\/span><\/sup><\/span>J (negativo, volume diminui)\u00a0 —\u00a0 Q= – 1,8.10<\/b><\/span><\/span><\/span>3\u00a0<\/b><\/span><\/span><\/sup><\/span>J (negativo, perde calor)\u00a0 —\u00a0 \u00a0\u0394U=Q \u2013 W= – 1,8.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>\u00a0\u2013 ( – 8.10<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>)\u00a0 —\u00a0 \u0394U= – 1,8.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>\u00a0+ 0,8.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 \u0394U= – 1,0.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>\u00a0J negativo, a temperatura diminui)\u00a0 —\u00a0\u00a0R- D.<\/b><\/span><\/span><\/span><\/p>\n

\u00a0<\/span><\/p>\n

Voltar para os exerc\u00edcios<\/span><\/a><\/h3>\n

\u00a0<\/span><\/p>\n","protected":false},"excerpt":{"rendered":"

Resolu\u00e7\u00e3o comentada dos exerc\u00edcios de vestibulares sobre Primeiro Princ\u00edpio da Termodin\u00e2mica ou Princ\u00edpio da Conserva\u00e7\u00e3o da Energia 01-\u00a0Se a temperatura \u00e9 a mesma para gases diferentes, a energia cin\u00e9tica das mol\u00e9culas deve ser a mesma, independente da massa de cada g\u00e1s. Como Ec=mV2\/2 e a energia cin\u00e9tica \u00e9 a mesma o g\u00e1s de maior massa deve ter menor velocidade para<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":2117,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"om_disable_all_campaigns":false,"_monsterinsights_skip_tracking":false,"_monsterinsights_sitenote_active":false,"_monsterinsights_sitenote_note":"","_monsterinsights_sitenote_category":0,"footnotes":""},"class_list":["post-2121","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2121","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/comments?post=2121"}],"version-history":[{"count":3,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2121\/revisions"}],"predecessor-version":[{"id":10742,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2121\/revisions\/10742"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2117"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=2121"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}