{"id":2099,"date":"2016-03-28T23:16:44","date_gmt":"2016-03-28T23:16:44","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=2099"},"modified":"2016-03-28T23:17:49","modified_gmt":"2016-03-28T23:17:49","slug":"resolucao-comentada-dos-exercicios-de-vestibulares-sobre-tipos-de-transformacoes-gasosas","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/fisica-termica\/termodinamica\/tipos-de-transformacoes-gasosas\/resolucao-comentada-dos-exercicios-de-vestibulares-sobre-tipos-de-transformacoes-gasosas\/","title":{"rendered":"Resolu\u00e7\u00e3o comentada dos exerc\u00edcios de vestibulares sobre Tipos de transforma\u00e7\u00f5es gasosas"},"content":{"rendered":"

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

Tipos de transforma\u00e7\u00f5es gasosas<\/b><\/span><\/span><\/span><\/p>\n

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

02-<\/b><\/span><\/span><\/span>\u00a0Isot\u00e9rmica\u00a0 —\u00a0 P.V=constante\u00a0 —\u00a0 P \u00e9 inversa a V\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- D<\/b><\/span><\/span><\/p>\n

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

04-<\/b><\/span><\/span><\/span> R- A\u00a0<\/b><\/span><\/span>\u00a0—\u00a0 veja teoria<\/b><\/span><\/span><\/p>\n

05-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Observe que a press\u00e3o sobre o \u00eambolo \u00e9 constante, press\u00e3o atmosf\u00e9rica que \u00e9 constante\u00a0 mais a press\u00e3o devido ao peso, que tamb\u00e9m \u00e9 constante. Assim trata-se de uma transforma\u00e7\u00e3o isob\u00e1rica onde o g\u00e1s se expande com aumento de temperatura\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- B<\/b><\/span><\/span><\/p>\n

06-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Trata-se de uma transforma\u00e7\u00e3o isot\u00e9rmica\u00a0 —\u00a0 P.V=constante\u00a0 — inversamente proporcionais\u00a0 —\u00a0 se voc\u00ea dobra a press\u00e3o, o volume cai pela metade\u00a0 —\u00a0 massa espec\u00edfica\u00a0 —\u00a0 \u03bc=m\/V\u00a0 —\u00a0 \u03bc e V s\u00e3o inversamente proporcionais\u00a0 —\u00a0 se V cai pela metade, a massa espec\u00edfica \u03bc duplicar\u00e1\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- B<\/b><\/span><\/span><\/p>\n

07-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Trata-se de uma transforma\u00e7\u00e3o isot\u00e9rmica\u00a0 —\u00a0 P.V=K\u00a0 — \u00a0\u00a0(V=1, P=2)\u00a0 —\u00a0 2.1=K\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- D<\/b><\/span><\/span>\u00a0\u00a0<\/b><\/span><\/span><\/p>\n

08-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Isot\u00e9rmica\u00a0 —\u00a0 P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>.V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=P.V\u00a0 —\u00a0 P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>.1=0,9P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>.V\u00a0 —\u00a0 V=1\/0,9\u00a0 —\u00a0 V=1,111\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- D<\/b><\/span><\/span><\/p>\n

09-<\/b><\/span><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/span><\/sub>.V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/span><\/sub>\/T<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/span><\/sub>=P.V\/T\u00a0 —\u00a0 (3P\/2).V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/span><\/sub>\/0,93.T=P.V\/T\u00a0 —\u00a0 3V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/span><\/sub>\/1,86=V\/1\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/span><\/sub>\/V=1,86\/3\u00a0 —\u00a0<\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/span><\/sub>\/V=0,62<\/b><\/span><\/span><\/span><\/p>\n

10-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Observe que a primeira transforma\u00e7\u00e3o o g\u00e1s recebe calor e sua temperatura e volume aumentam sob press\u00e3o constante (press\u00e3o atmosf\u00e9rica + press\u00e3o devido ao peso do \u00eambolo), portanto trata-se de uma transforma\u00e7\u00e3o \u00e0 press\u00e3o constante (isob\u00e1rica)\u00a0 —\u00a0 na segunda transforma\u00e7\u00e3o o g\u00e1s est\u00e1 em equil\u00edbrio t\u00e9rmico com o ambiente tendo a mesma temperatura que ele (isot\u00e9rmica)\u00a0 —<\/b><\/span><\/span>R- A<\/b><\/span><\/span><\/p>\n

11-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Devido \u00e0 redu\u00e7\u00e3o da press\u00e3o externa o volume do bal\u00e3o aumenta para que ocorra redu\u00e7\u00e3o da press\u00e3o interna\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- C<\/b><\/span><\/span><\/p>\n

12-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>R- (1 + 2 + 8) = 11\u00a0<\/b><\/span><\/span><\/p>\n

13-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Como a temperatura de ebuli\u00e7\u00e3o \u00e9 crescente, com o aumento de press\u00e3o m\u00e1xima o \u00fanico diagrama que representa isto \u00e9 o II.\u00a0<\/b><\/span><\/span><\/p>\n

14-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>.V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\/T<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=P.V\/T\u00a0 —\u00a0 P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\/T<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=P.(6\/5)V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\/(9\/10)T<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>.V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\/T<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=(P.6V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>)\/5 x 10\/(9T<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>)\u00a0 —\u00a0 P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=60P\/45\u00a0 —\u00a0 P=3\/4P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- B<\/b><\/span><\/span>
\n<\/b><\/span><\/span><\/p>\n

15-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>A transforma\u00e7\u00e3o \u00e9 isot\u00e9rmica com diminui\u00e7\u00e3o da press\u00e3o e aumento de volume\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- D<\/b><\/span><\/span><\/p>\n

16-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>\u00a0a) Observe na \u00faltima coluna que o produto\u00a0<\/b><\/span><\/span>\u00a0PxV \u00e9 praticamente constante\u00a0<\/b><\/span><\/span>(Lei de Boyle para transforma\u00e7\u00f5es isot\u00e9rmicas)<\/b><\/span><\/span><\/p>\n

b) Supondo que cada 8 unidades tenha 0,5cm\u00a0 —\u00a0 regra de tr\u00eas\u00a0 —\u00a0 8u – 0,5m\u00a0 —\u00a0 24u \u2013 h\u00a0 —\u00a0 h=1,5m\u00a0 —\u00a0 P=P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0+\u00a0<\/b><\/span><\/span>dgh=1,0.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>\u00a0+ 14.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>.10.1,5\u00a0 —\u00a0 P=3,1.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>N\/m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>(Pa)<\/b><\/span><\/span><\/p>\n

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

18-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Como o volume do cilindro que cont\u00e9m o g\u00e1s se dilata muito pouco com a temperatura, pode-se dizer que se trata de uma transforma\u00e7\u00e3o gasosa isom\u00e9trica ou isovolum\u00e9trica (volume constante)\u00a0 —\u00a0 como o carro fica exposto ao sol, a temperatura do g\u00e1s aumenta e consequentemente a press\u00e3o exercida pelo g\u00e1s tamb\u00e9m aumenta, pois P\/T=constante e P \u00e9 diretamente proporcional a T\u00a0 —\u00a0 R- D<\/b><\/span><\/span><\/p>\n

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

20-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>\u00c0 temperatura constante (isot\u00e9rmica), se a press\u00e3o diminui, no caso a externa, o volume aumenta, pois PV=constante\u00a0 —\u00a0 se a bola aumenta o volume, a press\u00e3o interna da bola diminui para igualar a press\u00e3o externa que tamb\u00e9m diminuiu\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- A<\/b><\/span><\/span><\/p>\n

21-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Press\u00e3o na profundidade de 20m=2atm (devido \u00e0 coluna de \u00e1gua) + 1atm (press\u00e3o atmosf\u00e9rica na superf\u00edcie da \u00e1gua)\u00a0 —\u00a0 P=3atm\u00a0 —\u00a0 isot\u00e9rmica\u00a0 —\u00a0 P.V=K\u00a0 —\u00a0\u00a0\u00a0 P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>.V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=P.V\u00a0 —\u00a0 1.V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=3.V\u00a0 —\u00a0 V=V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\/3\u00a0 —\u00a0 V=0,333V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- A<\/b><\/span><\/span><\/p>\n

22-<\/b><\/span><\/span><\/span>\u00a0Como a varia\u00e7\u00e3o de volume dos pneus \u00e9 desprez\u00edvel, trata-se de uma isom\u00e9trica, onde P\/T=constante, sendo P diretamente proporcional a T\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- C<\/b><\/span><\/span><\/p>\n

23-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>\u00a0At\u00e9 a explos\u00e3o o volume do recipiente permanece constante (isom\u00e9trica)\u00a0 —\u00a0 P<\/b><\/span><\/span>i<\/b><\/span><\/span><\/sub>\/T<\/b><\/span><\/span>i<\/b><\/span><\/span><\/sub>=P<\/b><\/span><\/span>f<\/b><\/span><\/span><\/sub>\/T<\/b><\/span><\/span>f<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 2,0.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>\/78=4,0.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>\/T<\/b><\/span><\/span>f<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 T<\/b><\/span><\/span>f<\/b><\/span><\/span><\/sub>=156K\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- B<\/b><\/span><\/span><\/p>\n

24-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Isovolum\u00e9trica\u00a0 —\u00a0 P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\/T<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=P\/T\u00a0 —\u00a0 2,3\/(27 + 273)=2,53\/T\u00a0 —\u00a0 2,3T=759\u00a0 —\u00a0 T=330K=330 \u2013 273\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>T=57<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C<\/b><\/span><\/span><\/p>\n

25-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>a) Se a porta do freezer fica aberta por algum tempo, parte do ar de seu interior \u00e9 substitu\u00eddo pelo ar mais quente do exterior. Depois que voc\u00ea fecha a porta da geladeira, a temperatura interna do ar em seu interior ir\u00e1 diminuir e como se trata de uma isovolum\u00e9trica, a press\u00e3o tamb\u00e9m diminui (P\/T=constante) fazendo com que a press\u00e3o externa fique maior que a interna, dificultando a abertura da porta. Mas, ap\u00f3s alguns instantes, entra ar pelo sistema de veda\u00e7\u00e3o da geladeira, diminuindo a diferen\u00e7a entre as press\u00f5es externa e interna, facilitando, ent\u00e3o, a abertura da porta.<\/b><\/span><\/span><\/p>\n

b) Isovolum\u00e9trica\u00a0 —\u00a0 P<\/b><\/span><\/span>oint<\/b><\/span><\/span><\/sub>\/T<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=P<\/b><\/span><\/span>int<\/b><\/span><\/span><\/sub>\/T\u00a0 —\u00a0 1,0.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>\/(27 + 273)=P<\/b><\/span><\/span>int<\/b><\/span><\/span><\/sub>\/(-3 + 273)\u00a0 —\u00a0 P<\/b><\/span><\/span>int<\/b><\/span><\/span><\/sub>=0,9.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>\u00a0N\/m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0(P<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>)\u00a0 —\u00a0 a diferen\u00e7a entre as press\u00f5es interna e externa sobre a porta da geladeira (\u0394P) \u00e9 igual \u00e0 for\u00e7a resultante externa sobre a \u00e1rea da porta\u00a0 —\u00a0 \u0394P=F\/S\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

(1,0.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>\u00a0\u2013 0,9.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>)=F\/(0,6×1)\u00a0 —\u00a0 0,1.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>=F\/0,6\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0F=6,0.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>N<\/b><\/span><\/span><\/p>\n

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

\"\"<\/p>\n

27-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Press\u00e3o \u00e0 profundidade de 15m\u00a0 —\u00a0 P=P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0+ P<\/b><\/span><\/span>\u00e1gua a 15m<\/b><\/span><\/span><\/sub>=1,0 + 1,5\u00a0 —\u00a0 P=2,5 atm\u00a0 —\u00a0 trata-se de uma isot\u00e9rmica (temperatura permanece constante dentro da bolha)\u00a0 —\u00a0 P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>.V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=P<\/b><\/span><\/span>i<\/b><\/span><\/span><\/sub>.V<\/b><\/span><\/span>i<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 1.V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=2,5V<\/b><\/span><\/span>i<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=2,5V<\/b><\/span><\/span>i<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- D<\/b><\/span><\/span><\/p>\n

 <\/p>\n

28-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>a) Se a porta do freezer fica aberta por algum tempo, parte do ar de seu interior \u00e9 substitu\u00eddo pelo ar mais quente do exterior. Depois que voc\u00ea fecha a porta da geladeira, a temperatura interna do ar em seu interior ir\u00e1 diminuir e como se trata de uma isovolum\u00e9trica, a press\u00e3o tamb\u00e9m diminui (P\/T=constante) fazendo com que a press\u00e3o externa fique maior que a interna,<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0dificultando a abertura da porta. Mas, ap\u00f3s alguns instantes, entra ar pelo sistema de veda\u00e7\u00e3o da geladeira, diminuindo a diferen\u00e7a entre as press\u00f5es externa e interna, facilitando, ent\u00e3o, a abertura da porta.<\/b><\/span><\/span><\/p>\n

b) Se a press\u00e3o externa \u00e9 maior que a interna, como Press\u00e3o=For\u00e7a\/\u00c1rea da porta, a for\u00e7a externa ter\u00e1 intensidade maior que a for\u00e7a interna e a intensidade da for\u00e7a resultante ser\u00e1 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=F<\/b><\/span><\/span>ext<\/b><\/span><\/span><\/sub>\u00a0\u2013 F<\/b><\/span><\/span>int<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 a transforma\u00e7\u00e3o \u00e9 isovolum\u00e9trica (volume constante)\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

P<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\/T<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=P<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\/T<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 1,0.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>x1.000=P<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\/(25 + 273)\u00a0 —\u00a0 P<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=P<\/b><\/span><\/span>int<\/b><\/span><\/span><\/sub>=0,9.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>N\/m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0(P<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>)\u00a0 —\u00a0<\/b><\/span><\/span>\u0394<\/b><\/span><\/span><\/span>P=F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>\/S\u00a0 —\u00a0 (1,0.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>\u00a0\u2013 0,9.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>)=F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>\/2\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=0,2.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>N\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=2,0.10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>N<\/b><\/span><\/span><\/p>\n

29-<\/b><\/span><\/span><\/span> R- A<\/b><\/span><\/span>\u00a0 —\u00a0 veja teoria<\/b><\/span><\/span><\/p>\n

30-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Isob\u00e1rica\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>.<\/b><\/span><\/span><\/sub>V<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub>\/T=constante\u00a0 —\u00a0 V \u00e9 diretamente proporcional a T\u00a0 —<\/b><\/span><\/span>\u00a0 R- B<\/b><\/span><\/span><\/p>\n

31-<\/b><\/span><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Isovolum\u00e9trica\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/span><\/sub>\/T<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/span><\/sub>=P\/T\u00a0 —\u00a0 30\/(27 + 273)=32\/T\u00a0 —\u00a0 30\/300=T\/32\u00a0 —\u00a0 T=320K\u00a0 —\u00a0 T=320 \u2013 273\u00a0 —\u00a0 T=47<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/span><\/sup>C\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span><\/p>\n

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

32-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Trata-se de uma transforma\u00e7\u00e3o isob\u00e1rica, pois a press\u00e3o exercida pelo ar no interior do cilindro \u00e9 igual \u00e0 press\u00e3o externa (atmosf\u00e9rica) exercida sobre a superf\u00edcie livre da \u00e1gua\u00a0 —\u00a0 observe que o n\u00edvel da \u00e1gua externa e internamente continuar\u00e1 sendo<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0de 2m, pois o sistema est\u00e1 em equil\u00edbrio e o empuxo \u00e9 o mesmo (o peso do ar n\u00e3o varia)\u00a0 —\u00a0 volume inicial do ar\u00a0 —\u00a0 V<\/b><\/span><\/span>i<\/b><\/span><\/span><\/sub>=S.h=S.(2 + 9)\u00a0 —\u00a0 V<\/b><\/span><\/span>i<\/b><\/span><\/span><\/sub>=11.S (S \u00e9 a \u00e1rea de se\u00e7\u00e3o transversal do cilindro)\u00a0 —\u00a0 volume final do ar\u00a0 —\u00a0 V=S.(H + 2)\u00a0 —\u00a0 isob\u00e1rica\u00a0 —\u00a0 V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\/T<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=V\/T\u00a0 —\u00a0 11.S\/300=(H + 2).S\/360\u00a0 —\u00a0 H=11,2m\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- D<\/b><\/span><\/span><\/p>\n

33-\u00a0<\/b><\/span><\/span>Utilizando a equa\u00e7\u00e3o de estado dos gases ideais, temos: p<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>V<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0= n<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>RT<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0e p<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0= n<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>RT<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>e, portanto,\u00a0\u00a0\"\". Como a parede \u00e9 diat\u00e9rmica e os gases est\u00e3o em equil\u00edbrio t\u00e9rmico, T<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0= T<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>; como a parede pode mover-se sem atrito na horizontal e est\u00e1 em repouso, p<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0= p<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>. Portanto,\u00a0\u00a0\"\". Mas\u00a0\u00a0\"\"\u00a0, logo, V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0= 2 V<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>. Substituindo esse resultado em V<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0+ V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0= V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>, obtemos 3V<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0= V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>, ou seja, V<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0= V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\/3 e, consequentemente,<\/b><\/span><\/span>V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0= 2V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\/3. \u00a0<\/b><\/span><\/span><\/p>\n

34-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Isot\u00e9rmica\u00a0 —\u00a0 P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>.V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=P.V\u00a0 —\u00a0 200.9=1.V\u00a0 —\u00a0 V=1.800 \u2113\u00a0\u00a0 —\u00a0 40\u2113 – 1min\u00a0 —\u00a0 1.800 \u2113 – t min\u00a0 —\u00a0 t=1.800\/40=45min\u00a0 —<\/b><\/span><\/span><\/p>\n

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

35-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Como n\u00e3o h\u00e1 trocas de calor com o meio externo, a temperatura continua sendo T\u00a0 —\u00a0 press\u00e3o final\u00a0 —\u00a0 P<\/b><\/span><\/span>i<\/b><\/span><\/span><\/sub>.V<\/b><\/span><\/span>i<\/b><\/span><\/span><\/sub>=P<\/b><\/span><\/span>f<\/b><\/span><\/span><\/sub>.V<\/b><\/span><\/span>f<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 P.(1\/3)V=P<\/b><\/span><\/span>f<\/b><\/span><\/span><\/sub>.V\u00a0 —\u00a0 P<\/b><\/span><\/span>f<\/b><\/span><\/span><\/sub>=P\/3\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- B<\/b><\/span><\/span><\/p>\n

36-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>antes de comprimir a bomba voc\u00ea tem n<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0mol de ar no interior da bomba e n<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0no interior do pneu\u00a0 —\u00a0 depois que o \u00eambolo desceu voc\u00ea ter\u00e1 no interior do pneu n mol tal que n=n<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0+ n<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 PV\/RT=P<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>V<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\/RT<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0+ P<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\/RT<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 o processo \u00e9 isot\u00e9rmico\u00a0 —\u00a0 PV=P<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>V<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0+ P<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 P<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=1 atm\u00a0 —\u00a0 P<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=3atm\u00a0 —\u00a0 V<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0<\/b><\/span><\/span>(interior da bomba)<\/b><\/span><\/span><\/sub>=S.h=24cm<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>.30cm\u00a0 —\u00a0 V<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=720cm<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 V<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=0,72L\u00a0 —\u00a0 P.2,4=1.0,72 + 3.2,4\u00a0 —\u00a0 P=3,3 atm\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- C<\/b><\/span><\/span><\/p>\n

37-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\/T<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=P.V\/T\u00a0 —\u00a0 T=T<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0(do enunciado) \u00a0—\u00a0 1,6.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>.3=P.4\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0P=1,2.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>\u00a0Nm<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/p>\n

38-\u00a0<\/b><\/span><\/span>P<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>V<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\/T<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0+ P<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\/T<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>= P<\/b><\/span><\/span>f<\/b><\/span><\/span><\/sub>V<\/b><\/span><\/span>f<\/b><\/span><\/span><\/sub>\/T<\/b><\/span><\/span>f<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 P<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>V<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0+ P<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>= P<\/b><\/span><\/span>f<\/b><\/span><\/span><\/sub>V<\/b><\/span><\/span>f<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 P<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>V<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0+ P<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>= P<\/b><\/span><\/span>f<\/b><\/span><\/span><\/sub>(V<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0+ V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>)\u00a0 —\u00a0 P<\/b><\/span><\/span>f<\/b><\/span><\/span><\/sub>=(P<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0+ P<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>)\/2\u00a0 —\u00a0 (5,0 + 18,0)\/2\u00a0 —\u00a0 P<\/b><\/span><\/span>f<\/b><\/span><\/span><\/sub>=11,5 atm\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- B<\/b><\/span><\/span><\/p>\n

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

\"\"<\/p>\n

40-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>a) Situa\u00e7\u00e3o antes – cilindro fechado\u00a0 —\u00a0 situa\u00e7\u00e3o depois \u2013 \u00eambolo j\u00e1 deslocado<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Antes\u00a0 —\u00a0 P<\/b><\/span><\/span>A\u2019<\/b><\/span><\/span><\/sub>=P<\/b><\/span><\/span>atm<\/b><\/span><\/span><\/sub>=1,0.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>N\/m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 V<\/b><\/span><\/span>A\u2019\u00ad<\/b><\/span><\/span><\/sub>=4,5 unidades\u00a0 —\u00a0 T<\/b><\/span><\/span>A\u2019<\/b><\/span><\/span><\/sub>=27<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C=300K<\/b><\/span><\/span><\/p>\n

Depois\u00a0 —\u00a0 P<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=?\u00a0 —\u00a0 V<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=3 unidades\u00a0 —\u00a0 T<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=27<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C=300K<\/b><\/span><\/span><\/p>\n

Isot\u00e9rmica\u00a0 —\u00a0 P<\/b><\/span><\/span>A\u2019<\/b><\/span><\/span><\/sub>.V<\/b><\/span><\/span>AP<\/b><\/span><\/span><\/sub>A<\/b><\/span><\/span>.V<\/b><\/span><\/span><\/sub>A\u00a0 —\u00a0 1.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>.4,5=P<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>.3\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0P<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=1,5.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>N\/m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/p>\n

b) Parte inferior do cilindro<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

P<\/b><\/span><\/span>B\u2019<\/b><\/span><\/span><\/sub>.V<\/b><\/span><\/span>B\u2019<\/b><\/span><\/span><\/sub>\/T<\/b><\/span><\/span>B\u2019<\/b><\/span><\/span><\/sub>=P<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>.V<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\/T<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 1.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>.7,5\/300=1,5.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>.9\/T<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>T<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=540K ou 267<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C<\/b><\/span><\/span><\/p>\n

41-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>a)<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0= F\/S + P<\/b><\/span><\/span>atmosf\u00e9rica<\/b><\/span><\/span><\/sub>\u00a0= m.g\/S + 10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>\u00a0= 160×10\/0,16 + 10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>\u00a0 = 10.000 + 10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>\u00a0= 10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>+ 10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>\u00a0= 0,1.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>\u00a0+ 10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>\u00a0\u00a0\u00a0—\u00a0\u00a0<\/b><\/span><\/span>P<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>= 1,1.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>\u00a0Pa<\/b><\/span><\/span><\/p>\n

b) Pela lei geral dos gases\u00a0 —\u00a0 (p.V\/T) = constante\u00a0 —\u00a0 (P<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>.V<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\/T<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>) = (P<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>.V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\/T<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>) e como V = H.S pode-se ainda escrever\u00a0 —\u00a0 \u00a0(P<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>.V<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\/T<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>) = (P<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>.V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\/T<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>)\u00a0 — \u00a0(P<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>.S.H<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\/T<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>) = (P<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>.S.H<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\/T<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>)\u00a0 — \u00a0(P<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>.H<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\/T<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>) = (P<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>.H<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\/T<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>), mas o processo ocorre sob press\u00e3o constante e logo P<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0= P<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0ent\u00e3o (H<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\/T<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>) = (H<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\/T<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>)\u00a0 —\u00a0 a partir dos dados dispon\u00edveis \u00a0— \u00a0(H<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\/T<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>) = (H<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\/T<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>) \u00a0— (4\/300) = (6\/T<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>) \u00a0— T<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0= 1800\/4\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>T<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0= 450 K<\/b><\/span><\/span><\/p>\n

c) Para o c\u00e1lculo da efici\u00eancia do processo R ser\u00e1 necess\u00e1rio determinar a varia\u00e7\u00e3o de energia potencial gravitacional da plataforma, DE<\/b><\/span><\/span>P<\/b><\/span><\/span><\/sub>\u00a0= m.g.DH e a quantidade de calor no processo Q que \u00e9 dada por Q = m.c.DT\u00a0 —\u00a0 DE<\/b><\/span><\/span>P<\/b><\/span><\/span><\/sub>\u00a0= m.g.DH = 160.10.(6 \u2013 4) = 3200 J\u00a0 —\u00a0 para o c\u00e1lculo da quantidade de calor \u00e9 necess\u00e1rio conhecer a massa de ar no cilindro. —\u00a0\u00a0 como a densidade do ar a 300 K foi fornecida pode-se fazer \u00a0— \u00a0d = m\/V \u00a0—\u00a0 m = d.V = 1,1.0,16.4\u00a0 —\u00a0 d= 0,704 kg\u00a0 —\u00a0 Q = m.c.DT = 0,704.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>.(450 \u2013 300) — \u00a0Q=105.600 J\u00a0 —\u00a0\u00a0 efici\u00eancia\u00a0 —\u00a0 \u00a0R = 3.200\/105.600\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R= 0,03 = 3%<\/b><\/span><\/span><\/p>\n

42-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>A lei geral dos gases afirma que P.V\/T \u00e9 constante\u00a0 —\u00a0 entre o estado A e B o volume \u00e9 constante e logo P\/T = constante\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

(P\/T)<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0= (P\/T)<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 P\/300 = 2P\/T\u00a0 —\u00a0 1\/300 = 2\/T\u00a0 —\u00a0 T = 600 K\u00a0 —\u00a0 entre o estado B e C a temperatura \u00e9 constante e logo P.V = constante\u00a0 —\u00a0 (P.V)<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0= (P.V)<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>\u00a0\u00a0— \u00a02P.V = P<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>.(3V\/2) \u00a0— 4P\/3 = P<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 R- D<\/b><\/span><\/span><\/p>\n

43-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>a) Falsa- \u00e9 isovolum\u00e9trica\u00a0 —\u00a0 b) Correta- todos os pontos da curva AB tem a mesma temperatura\u00a0 —\u00a0 c) Falsa \u2013 o volume \u00e9 constante\u00a0 —\u00a0 d) Falsa\u00a0 —\u00a0 mais afastada, T<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>>T<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- B<\/b><\/span><\/span><\/p>\n

44-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>a) p.V\/T = constante — 150.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>.V\/300 = 160.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>.1,05.V\/T — (1\/2)= 160.1,05.\/T\u00a0\"\"\u00a0T = 168.2 = 336 K\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0T = 63\u00b0C<\/b><\/span><\/span><\/p>\n

b) p.V\/T = constante — 150.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>.V\/300 = p.1,08.V\/360 — (10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\/2)= 1,08.p\/360 — \u00a0p = 167.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0Pa\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0p = 1,67.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>\u00a0Pa<\/b><\/span><\/span><\/p>\n

45-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Isovolum\u00e9trica\u00a0 —\u00a0 P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\/T<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=P\/T\u00a0 —\u00a0 P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\/(27 + 273)=P\/(273 \u2013 18)\u00a0 —\u00a0 P=(255\/300)P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 P=0,85P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 P=85% da press\u00e3o inicial\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

R- D\u00a0<\/b><\/span><\/span><\/p>\n

46-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Isot\u00e9rmica\u00a0 —\u00a0 P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>.V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=P.V\u00a0 —\u00a0 1.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>.100=P.140\u00a0 —\u00a0 P=0,71.10<\/b><\/span><\/span>5\u00a0<\/b><\/span><\/span><\/sup>Pa\u00a0 —\u00a0 P=71.000Pa\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- C<\/b><\/span><\/span><\/p>\n

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

48-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Estado inicial\u00a0 —\u00a0 \u00a0p = p<\/b><\/span><\/span>0<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 \u00a0V = V<\/b><\/span><\/span>0<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 \u00a0T = T<\/b><\/span><\/span>0<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 1\u00aa transforma\u00e7\u00e3o\u00a0 —\u00a0\u00a0 isot\u00e9rmica\u00a0 —\u00a0 \u00a0T<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0= T<\/b><\/span><\/span>0<\/b><\/span><\/span><\/sub>\u00a0e V<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0= V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\/2\u00a0 —\u00a0 P<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>V<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\/T<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\/T<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 P<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>.(V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\/2)\/T<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\/T<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 P<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\/2=P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 P<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=2P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 2\u00aa transforma\u00e7\u00e3o\u00a0 —\u00a0 Isob\u00e1rica: p<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0= p<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 \u00a0V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0= 2<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub>V<\/b><\/span><\/span>0<\/b><\/span><\/span><\/sub>\u00a0 — P<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>V<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\/T<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=P<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\/T<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 (P<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>.2V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>)\/T<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=(P<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>.V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\/2)\/2T<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 2\/T<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=1\/2T<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 T<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=4T<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- E<\/b><\/span><\/span><\/p>\n

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

\"\"<\/p>\n

50-<\/b><\/span><\/span><\/span>\u00a0Dados\u00a0 —\u00a0 D = 20 cm\u00a0 — \u00a0R = 10 cm = 1 dm\u00a0 —\u00a0 p = 1 atm\u00a0 —\u00a0 T = 27 \u00b0C = 300 K\u00a0 —\u00a0 M = 30 g\/mol\u00a0 —\u00a0 R = 0,08 atm\u00b7L\/mol\u00b7K\u00a0 —\u00a0 \u03c0 = 3\u00a0 —\u00a0 volume da bola\u00a0 —\u00a0 V=(4\u03c0R<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>)\/3=(4.3.1<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>)\/3\u00a0 —\u00a0 V=4dm<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 V=4L\u00a0 —\u00a0 equa\u00e7\u00e3o de Clapeyron\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

pV=(m\/M).R.T\u00a0 —\u00a0 m=(M.p.V)\/RT=30.1.4\/0,08.300=120\/24\u00a0 —\u00a0 m=5g\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- B<\/b><\/span><\/span><\/p>\n

 <\/p>\n

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

Como o enunciado<\/b><\/span><\/span> afirma que, durante o aquecimento n\u00e3o houve deforma\u00e7\u00e3o da lata \u2013 Vi=Vf=V\u00a0 —\u00a0 dados: Pi=1atm\u00a0 — \u00a0\u00a0Ti=25oC=298K\u00a0 —\u00a0 Vi=V\u00a0 —\u00a0 Pf=?\u00a0 —\u00a0 Tf=621oC=894K\u00a0 —\u00a0 Vf=V\u00a0 —\u00a0 trata-se de uma transforma\u00e7\u00e3o isovolum\u00e9trica (isoc\u00f3rica, isom\u00e9trica)\u00a0 —\u00a0 Pi.V\/Ti\u00a0= Pf.V\/Tf\u00a0 —\u00a0 1.V\/298 = Pf.V\/894\u00a0 —\u00a0 Pf=894\/298\u00a0 —\u00a0 Pf=3 atm\u00a0 —\u00a0\u00a0R- C<\/b><\/span><\/span><\/p>\n

 <\/p>\n

Voltar para os exerc\u00edcios<\/span><\/a><\/h3>\n","protected":false},"excerpt":{"rendered":"

Resolu\u00e7\u00e3o comentada dos exerc\u00edcios de vestibulares sobre Tipos de transforma\u00e7\u00f5es gasosas 01–\u00a0R- D\u00a0\u00a0—\u00a0 veja teoria 02-\u00a0Isot\u00e9rmica\u00a0 —\u00a0 P.V=constante\u00a0 —\u00a0 P \u00e9 inversa a V\u00a0 —\u00a0\u00a0R- D 03- R- B\u00a0 —\u00a0 veja teoria 04- R- A\u00a0\u00a0—\u00a0 veja teoria 05-\u00a0Observe que a press\u00e3o sobre o \u00eambolo \u00e9 constante, press\u00e3o atmosf\u00e9rica que \u00e9 constante\u00a0 mais a press\u00e3o devido ao peso, que tamb\u00e9m \u00e9<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":2095,"menu_order":0,"comment_status":"open","ping_status":"open","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-2099","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2099","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=2099"}],"version-history":[{"count":2,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2099\/revisions"}],"predecessor-version":[{"id":2101,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2099\/revisions\/2101"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2095"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=2099"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}