{"id":2086,"date":"2016-03-28T23:01:03","date_gmt":"2016-03-28T23:01:03","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=2086"},"modified":"2024-08-21T17:01:59","modified_gmt":"2024-08-21T17:01:59","slug":"resolucao-comentada-dos-exercicios-de-vestibulares-sobre-trabalho-de-um-gas-transformacoes-ciclicas","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/fisica-termica\/termodinamica\/trabalho-de-um-gas-transformacao-ciclica\/resolucao-comentada-dos-exercicios-de-vestibulares-sobre-trabalho-de-um-gas-transformacoes-ciclicas\/","title":{"rendered":"Trabalho de um g\u00e1s \u2013 Transforma\u00e7\u00f5es c\u00edclicas – Resolu\u00e7\u00e3o"},"content":{"rendered":"

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

Trabalho de um g\u00e1s \u2013 Transforma\u00e7\u00f5es c\u00edclicas<\/b><\/span><\/span><\/span><\/p>\n

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

02-<\/b><\/span><\/span><\/span>\u00a0W=P.\u0394V\u00a0 —\u00a0 30=P(5 \u2013 2)\u00a0 —\u00a0 P=10N\/m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>(Pa)\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- A<\/b><\/span><\/span><\/p>\n

03-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>1g \u2013 5.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>cal\u00a0 —\u00a0 4.000g \u2013 Q cal\u00a0 —\u00a0 Q=2.10<\/b><\/span><\/span>7<\/b><\/span><\/span><\/sup>cal\u00a0 — 1cal \u2013 4,2J\u00a0 —\u00a0 2.10<\/b><\/span><\/span>7<\/b><\/span><\/span><\/sup>cal \u2013 W J\u00a0 —\u00a0 W=8,4.10<\/b><\/span><\/span>7<\/b><\/span><\/span><\/sup>J\u00a0 —\u00a0 rendimento de 15%\u00a0 —\u00a0 W<\/b><\/span><\/span>\u00fatil<\/b><\/span><\/span><\/sub>=0,15.8,4.10<\/b><\/span><\/span>7<\/b><\/span><\/span><\/sup>=1,26.10<\/b><\/span><\/span>7\u00a0<\/b><\/span><\/span><\/sup>J\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- E\u00a0<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/p>\n

04-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Trata-se de uma transforma\u00e7\u00e3o de calor (gerado pela queima do combust\u00edvel) em trabalho (movimenta\u00e7\u00e3o do ve\u00edculo)\u00a0 —<\/b><\/span><\/span><\/p>\n

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

05-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>a) W<\/b><\/span><\/span>AB<\/b><\/span><\/span><\/sub>=\u00e1rea do tri\u00e2ngulo=b.h\/2=10.10\/2\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0W<\/b><\/span><\/span>AB<\/b><\/span><\/span><\/sub>=50 J<\/b><\/span><\/span><\/p>\n

b) W<\/b><\/span><\/span>BC<\/b><\/span><\/span><\/sub>=\u00e1rea do ret\u00e2ngulo=b.h=(20 \u2013 10).10\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>W<\/b><\/span><\/span>BC<\/b><\/span><\/span><\/sub>=100 J<\/b><\/span><\/span><\/p>\n

c)<\/b><\/span><\/span><\/span>\u00a0W<\/b><\/span><\/span><\/span>CD<\/b><\/span><\/span><\/span><\/sub>= zero<\/b><\/span><\/span><\/span><\/p>\n

d) W<\/b><\/span><\/span><\/span>AD<\/b><\/span><\/span><\/span><\/sub>=50 + 100 + 0\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>W<\/b><\/span><\/span><\/span>AD<\/b><\/span><\/span><\/span><\/sub>=150 J\u00a0<\/b><\/span><\/span><\/span><\/p>\n

06-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Observe que o menor trabalho corresponde \u00e0 menor \u00e1rea, que \u00e9 a hachurada na figura abaixo\u00a0 —\u00a0 <\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

W=\u00e1rea do ret\u00e2ngulo=b.h\u00a0 —\u00a0W=(V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0\u2013 V<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>).P<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0 —<\/b><\/span><\/span>\u00a0 R- B<\/b><\/span><\/span><\/p>\n

07-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Trecho CD\u00a0 —\u00a0 W<\/b><\/span><\/span>CD<\/b><\/span><\/span><\/sub>=P.\u0394V=P.(V<\/b><\/span><\/span>D<\/b><\/span><\/span><\/sub>\u00a0\u2013 V<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>)=5,8.(0,8 \u2013 1,6)\u00a0 —\u00a0 W<\/b><\/span><\/span>CD<\/b><\/span><\/span><\/sub>= – 4,64 J\u00a0 —<\/b><\/span><\/span>\u00a0 R- A<\/b><\/span><\/span><\/p>\n

08-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>O trabalho no trecho AB \u00e9 numericamente igual \u00e0 \u00e1rea do trap\u00e9zio, hachurada na figura,\u00a0 e de <\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

valor W=(B + b).h\/2=(30.10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>\u00a0+10.10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>).20.10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>\/2\/2\u00a0 —\u00a0 W=400.10<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>J=40.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>J\u00a0 —\u00a0 W=40kJ\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- C<\/b><\/span><\/span><\/p>\n

09-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Observe as \u00e1reas que fornecem os respectivos trabalhos:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

10-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>a) De A para B\u00a0 —\u00a0 isovolum\u00e9trica\u00a0 —\u00a0 P<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\/T<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=P<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\/T<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 5\/T<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=2\/T<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 T<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=5\/2T<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0(I)\u00a0 —\u00a0 de B para C\u00a0 —\u00a0 isob\u00e1rica\u00a0 —\u00a0 V<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\/T<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=V<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>\/T<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 2\/T<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=5\/T<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 T<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=2\/5T<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>\u00a0(II)\u00a0 —\u00a0 comparando I com II\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>T<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=T<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>\u00a0ou T<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\/T<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>=1<\/b><\/span><\/span><\/p>\n

b) Trabalho pela \u00e1rea\u00a0 —\u00a0 W<\/b><\/span><\/span>AB<\/b><\/span><\/span><\/sub>=0\u00a0 —\u00a0 W<\/b><\/span><\/span>BC<\/b><\/span><\/span><\/sub>=P.(V<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>\u00a0\u2013 V<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>)=2(5 \u2013 2)\u00a0 —\u00a0 W<\/b><\/span><\/span>BC<\/b><\/span><\/span><\/sub>=6J\u00a0 —\u00a0 W<\/b><\/span><\/span>total<\/b><\/span><\/span><\/sub>= 0 + 6\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>W<\/b><\/span><\/span>total<\/b><\/span><\/span><\/sub>=6J<\/b><\/span><\/span><\/p>\n

11<\/b><\/span><\/span><\/span>–<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>W=P.\u0394V\u00a0 —\u00a0 30=P(5 \u2013 2)\u00a0 —\u00a0 P=10N\/m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>(Pa)\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- A<\/b><\/span><\/span><\/p>\n

12-<\/b><\/span><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a) V=S.h=500.20=10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/span><\/sup>\u00a0cm<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/span><\/sup>\u00a0 —\u00a0 V=10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/span><\/sup>.10<\/b><\/span><\/span><\/span>-6<\/b><\/span><\/span><\/span><\/sup>\u00a0 —\u00a0 V=10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/span><\/sup>\u00a0m<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/span><\/sup>\u00a0 —\u00a0 PV=nRT\u00a0 —\u00a0 (10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/span><\/sup>\u00a0+ 10m\/5.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/span><\/sup>).10<\/b><\/span><\/span><\/span>-2 <\/b><\/span><\/span><\/span><\/sup>= 0,5.8.300\u00a0 —\u00a0 10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/span><\/sup>\u00a0+ 2m=1,2.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/span><\/sup>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>m=10<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sup>kg<\/b><\/span><\/span><\/span><\/p>\n

b) 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 S.0,2\/300=S.h\/420 \u00a0—\u00a0 h=0,28m\u00a0 —\u00a0 W=P.\u0394V=1,0.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>.5.10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>.(0,28 \u2013 0,20)\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>W=400J<\/b><\/span><\/span><\/p>\n

13-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Volume da caixa\u00a0 —\u00a0 V=\u2113<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>=(4.10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>)<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 V=64.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>m<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 \u00e1rea de cada parede\u00a0 —\u00a0 S=\u2113<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=(4.10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 S=16.10<\/b><\/span><\/span>-4<\/b><\/span><\/span><\/sup>m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 a 300K\u00a0 —\u00a0 press\u00e3o nas paredes\u00a0 —\u00a0 P=1atm=10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>N\/m<\/b><\/span><\/span>2\u00a0<\/b><\/span><\/span><\/sup>\u00a0—\u00a0 P=F<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\/S\u00a0 —\u00a0 10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>=F<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0\/16.10<\/b><\/span><\/span>-4<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 F<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=160N\u00a0 —\u00a0 a 330K\u00a0 —\u00a0 c\u00e1lculo da press\u00e3o no final desta transforma\u00e7\u00e3o que \u00e9 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 10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>\/300=P\/330\u00a0 —\u00a0 P=1,1.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>N\/m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 P=F<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\/S\u00a0 —\u00a0 1,1.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>=F<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\/16.10<\/b><\/span><\/span>-4<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 F<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=176N\u00a0 —\u00a0 \u0394F=F<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0\u2013 F<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=176 \u2013 160\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>\u0394F=16N<\/b><\/span><\/span>\"\"<\/p>\n

14-<\/b><\/span><\/span><\/span>\u00a0Antes\u00a0 —\u00a0 P<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>=P<\/b><\/span><\/span>atm<\/b><\/span><\/span><\/sub>=1atm=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<\/b><\/span><\/span><\/sub>=S.h=20.10<\/b><\/span><\/span>-4<\/b><\/span><\/span><\/sup>.1\u00a0 —\u00a0 V<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>=2.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>m<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 depois\u00a0 —\u00a0 P<\/b><\/span><\/span>d<\/b><\/span><\/span><\/sub>=P<\/b><\/span><\/span>atm<\/b><\/span><\/span><\/sub>\u00a0+ P=P + 10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 V<\/b><\/span><\/span>d<\/b><\/span><\/span><\/sub>=S.h\u00a0 —\u00a0 V<\/b><\/span><\/span>d<\/b><\/span><\/span><\/sub>=20.10<\/b><\/span><\/span>-4.<\/b><\/span><\/span><\/sup>.0,8\u00a0 —\u00a0 V<\/b><\/span><\/span>d<\/b><\/span><\/span><\/sub>=1,6.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>m<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 isot\u00e9rmica\u00a0 —\u00a0 P<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>V<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>=P<\/b><\/span><\/span>d<\/b><\/span><\/span><\/sub>.V<\/b><\/span><\/span>d<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>.2.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>=(P + 10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>).1,6.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 P=25.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>N\/m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 na situa\u00e7\u00e3o depois, a press\u00e3o com que o \u00eambolo comprime o g\u00e1s vale\u00a0 —\u00a0 press\u00e3o=for\u00e7a\/\u00e1rea=peso\/\u00e1rea\u00a0 —\u00a0 25.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>=peso\/20.10<\/b><\/span><\/span>-4<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

peso=50N\u00a0 —\u00a0 peso=m.g\u00a0 —\u00a0 50=m.10\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>m=5,0kg<\/b><\/span><\/span><\/p>\n

15<\/b><\/span><\/span><\/span>–<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>R- B<\/b><\/span><\/span>\u00a0 —\u00a0 veja teoria e lembre-se de que nesse caso o trabalho \u00e9 negativo (sentido anti-hor\u00e1rio).<\/b><\/span><\/span><\/p>\n

16-<\/b><\/span><\/span><\/span>\u00a0Como a transforma\u00e7\u00e3o c\u00edclica \u00e9 no sentido hor\u00e1rio, o trabalho realizado \u00e9 positivo (predomina a expans\u00e3o \u2013 do g\u00e1s sobre o ambiente) e fornecido pela \u00e1rea do ciclo\u00a0 —\u00a0 W=base x altura=((0,3 \u2013 0,1).(30 \u2013 10)\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>W= + 4J\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- B<\/b><\/span><\/span><\/p>\n

17-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>W positivo \u2013 sentido hor\u00e1rio\u00a0 —\u00a0 W=(5 \u2013 2).(800 \u2013 300)\u00a0 —\u00a0 W= + 1.500J\u00a0 —<\/b><\/span><\/span>\u00a0 R- A<\/b><\/span><\/span><\/p>\n

18-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>O trabalho \u00e9 positivo \u2013 sentido hor\u00e1rio\u00a0 —\u00a0 \u00e1rea do paralelogramo=base x altura\u00a0 —\u00a0 W=(6V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0\u2013 2V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>).(P<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0\u2013 P<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>)\u00a0 —\u00a0 <\/b><\/span><\/span>W=4V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>.(P<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0\u2013 P<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>)<\/b><\/span><\/span><\/p>\n

19-<\/b><\/span><\/span><\/span>\u00a0A temperatura mais alta corresponde ao ponto cuja isoterma se encontra mais afastada dos <\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

eixos (ponto B) conforme a figura e o trabalho \u00e9 positivo (sentido hor\u00e1rio) e de valor\u00a0 —\u00a0 W=(4.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>\u00a0\u2013 1,5.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>).(4.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>\u00a0\u2013 2.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>)=2,5.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>.2.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 W= +5.10<\/b><\/span><\/span>-1<\/b><\/span><\/span><\/sup>J\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

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

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

\"\"<\/p>\n

W<\/b><\/span><\/span>total<\/b><\/span><\/span><\/sub>=W<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0+ W<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=b<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>.h<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0+ b<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>.h<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=(4 \u2013 1).(1) + (7 \u2013 4).(2 \u2013 1)=3 + 3\u00a0 —\u00a0 W<\/b><\/span><\/span>total<\/b><\/span><\/span><\/sub>= 6J\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- D<\/b><\/span><\/span><\/p>\n

21<\/b><\/span><\/span><\/span>–<\/b><\/span><\/span><\/span> W=b.h\/2=(6.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>-2.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>).(3.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>\u00a0\u2013 1.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>)\/2\u00a0 —\u00a0 W=4,0.10<\/b><\/span><\/span>-1<\/b><\/span><\/span><\/sup>J\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- B<\/b><\/span><\/span><\/p>\n

22<\/b><\/span><\/span><\/span>–\u00a0<\/b><\/span><\/span>Em toda transforma\u00e7\u00e3o c\u00edclica o trabalho realizado \u00e9 numericamente igual \u00e0 \u00e1rea interna do ciclo\u00a0 —\u00a0 observe na figura que a menor \u00e1rea \u00e9 a do ciclo K (menor que quatro quadr\u00edculos)\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- C<\/b><\/span><\/span><\/p>\n

23-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>a) d=m\/V\u00a0 —\u00a0 V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=60,1,2\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=50m<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>
\n<\/b><\/span><\/span>b) Equa\u00e7\u00e3o geral dos gases perfeitos\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>\u00a0= P.V\/T\u00a0 —\u00a0 50\/300 = V\/500\u00a0 —\u00a0 V=250\/3\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>V=83,3 m<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0(V>V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0o g\u00e1s sofreu expans\u00e3o)<\/b><\/span><\/span><\/p>\n

c) Expans\u00e3o isob\u00e1rica\u00a0 —\u00a0 W=P.(V \u2013 V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>)=10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>.(83,3 \u2013 50)\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>W=33,3.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>=3,3.10<\/b><\/span><\/span>6<\/b><\/span><\/span><\/sup>\u00a0J<\/b><\/span><\/span><\/p>\n

24<\/b><\/span><\/span><\/span>–<\/b><\/span><\/span><\/span> Em um ciclo fechado o trabalho \u00e9 numericamente igual \u00e0 \u00e1rea da figura. Seu valor \u00e9 negativo devido ao sentido anti-hor\u00e1rio.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

 <\/p>\n

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

I. Verdadeira\u00a0 —\u00a0 observe que o processo de 1 para 2 ocorre \u00e0 volume constante e que o de 3 para 1 ocorre \u00e0 press\u00e3o constante.<\/b><\/span><\/span><\/p>\n

II. Falsa\u00a0 —\u00a0 observe que nesse processo o volume n\u00e3o varia (\u2206V=0) e, como W=P.\u2206V, W=0.<\/b><\/span><\/span><\/p>\n

III. Verdadeira\u00a0 —\u00a0 nesse processo o trabalho \u00e9 negativo, pois o volume est\u00e1 diminuindo e o meio exterior exerce W sobre o g\u00e1s.<\/b><\/span><\/span><\/p>\n

IV. Verdadeira\u00a0 —\u00a0 do enunciado\u00a0 —\u00a0 P<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=5P<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 todos os pontos da mesma curva da isoterma possuem a mesma temperatura\u00a0 —\u00a0 T<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=T<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 o processo de 1 para 2 \u00e9 uma isovolum\u00e9trica e seu volume n\u00e3o varia (V<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=V)\u00a0 —\u00a0 P<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>.V\/T<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0= P<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>.V\/T<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 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 P<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\/T<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=5P<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\/T<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 T<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=5T<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 como T<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=T<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 T<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>=5T<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>.<\/b><\/span><\/span><\/p>\n

V. Verdadeira\u00a0 —\u00a0 em todo ciclo a temperatura final e inicial s\u00e3o iguais e, consequentemente n\u00e3o ocorre varia\u00e7\u00e3o da energia interna do sistema.<\/b><\/span><\/span><\/p>\n

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

<\/a> 26-<\/b><\/span><\/span><\/span> Neste ciclo voc\u00ea tem\u00a0 — Trecho AB\u00a0 —\u00a0 isot\u00e9rmica de expans\u00e3o\u00a0 —\u00a0 trecho BC\u00a0 —\u00a0 isoc\u00f3rica ou <\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

isovolum\u00e9trica\u00a0 —\u00a0 trecho CD\u00a0 —\u00a0 isot\u00e9rmica de compress\u00e3o\u00a0 — Trecho DA\u00a0 —\u00a0 isoc\u00f3rica ou isovolum\u00e9trica.<\/b><\/span><\/span><\/p>\n

As respostas devem ser em fun\u00e7\u00e3o dos dados fornecidos, ou seja, em fun\u00e7\u00e3o de P<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>, V<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>, P<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>\u00a0e V<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>.<\/b><\/span><\/span><\/p>\n

Primeira linha<\/b><\/span><\/span>\u00a0 —\u00a0 no trecho AB trata-se de uma isot\u00e9rmica (mesma temperatura T<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=T<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=T)\u00a0 —\u00a0 P<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>.V<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\/T=P<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>V<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\/T ou P<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>.V<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=P<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>.V<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0= constante\u00a0 —\u00a0 equa\u00e7\u00e3o dos gases perfeitos\u00a0 —\u00a0 P<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>.V<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=nRT<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 T<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=T<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=T=P<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>.V<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\/nR (I)\u00a0 —\u00a0 P=P<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>(dado); V=V<\/b><\/span><\/span>A\u00a0<\/b><\/span><\/span><\/sub>(dado) e T=P<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>.V<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\/nR\u00a0 —<\/b><\/span><\/span>segunda linha<\/b><\/span><\/span>\u00a0 —\u00a0 no trecho BC <\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

trata-se de uma isoc\u00f3rica ou isovolum\u00e9trica (volume constante V<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=V<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>=V)\u00a0 —\u00a0 do trecho AB\u00a0 — P<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>.V<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=P<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>.V<\/b><\/span><\/span>B\u00a0<\/b><\/span><\/span><\/sub>(II)\u00a0 —\u00a0 do trecho BC (III) \u00a0—\u00a0 V<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=V<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 (III) em (II)\u00a0 — P<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>.V<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=P<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>.V<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 P<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=P<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>.V<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\/V<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 P= P<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>.V<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\/V<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>; V=V<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>\u00a0e T= P<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>.V<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\/nR (de I)\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0terceira linha\u00a0<\/b><\/span><\/span>\u00a0—\u00a0 isot\u00e9rmica\u00a0 —\u00a0 T<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>=T<\/b><\/span><\/span>D<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 equa\u00e7\u00e3o dos gases ideais\u00a0 —\u00a0 P<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>.V<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>=nRT<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 T<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>=P<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>.V<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>\/nR\u00a0 —\u00a0 P=P<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>; V=V<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>\u00a0e T= P<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>.V<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>\/nR\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>quarta linha<\/b><\/span><\/span>\u00a0 —\u00a0 analogamente\u00a0 —\u00a0 P=P<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>.V<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>\/V<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>, V=V<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0e T= P<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>.V<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>\/nR.<\/b><\/span><\/span><\/p>\n

b) \u00a0* Transforma\u00e7\u00e3o isot\u00e9rmica\u00a0 —\u00a0 como a temperatura (T) e a energia interna (U) de um sistema est\u00e3o associadas \u00e0 energia cin\u00e9tica das mol\u00e9culas, se a temperatura for constante, a energia interna (U) do sistema tamb\u00e9m ser\u00e1 constante. Assim, a varia\u00e7\u00e3o<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

de energia interna (\u0394U) do sistema ser\u00e1 nula\u00a0 —\u00a0 \u0394U=0\u00a0 —\u00a0 \u0394U= Q \u2013 W\u00a0 —\u00a0 0=Q \u2013 W\u00a0 —<\/b><\/span><\/span>\u00a0\u00a0Q=W<\/b><\/span><\/span>\u00a0—<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>portanto em toda transforma\u00e7\u00e3o isot\u00e9rmica todo calor Q recebido pelo sistema ou cedido por ele \u00e9 transformado em trabalho W.<\/b><\/span><\/span><\/p>\n

* Transforma\u00e7\u00e3o isoc\u00f3rica, isom\u00e9trica ou isovolum\u00e9trica\u00a0 —\u00a0 ocorre a volume constante\u00a0 —\u00a0 V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=V\u00a0 —\u00a0 <\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u0394V=0\u00a0 —\u00a0 W=P.\u0394V\u00a0 —\u00a0W=P.0\u00a0 —\u00a0 W=0\u00a0 — \u0394U = Q \u2013 W\u00a0 —\u00a0 \u0394U=Q \u2013 0\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0\u0394U=Q<\/b><\/span><\/span>\u00a0 —\u00a0 assim, todo o calor (Q) recebido pelo sistema \u00e9 igual \u00e0 sua varia\u00e7\u00e3o de energia interna (\u0394U)<\/b><\/span><\/span><\/p>\n

Primeira linha\u00a0 —\u00a0 isot\u00e9rmica\u00a0 —\u00a0 \u0394U=0\u00a0 —\u00a0 Q=W\u00a0 —\u00a0 Q=300J\u00a0\u00a0 —\u00a0 W=Q=300J\u00a0 —\u00a0 segunda linha\u00a0 —\u00a0 <\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

isoc\u00f3rica\u00a0 —\u00a0 W=0\u00a0 —\u00a0\u0394U=Q\u00a0 —\u0394U= -750J\u00a0 —\u00a0 Q= – 750J\u00a0 —\u00a0 terceira linha\u00a0 —\u00a0 isot\u00e9rmica\u00a0 —\u00a0 \u2206U=0\u00a0 —\u00a0 Q=W\u00a0 —\u00a0 W = -150J\u00a0 —\u00a0 Q= – 150J\u00a0 —\u00a0 quarta linha\u00a0 —\u00a0 isoc\u00f3rica\u00a0 —\u00a0 W=0\u00a0 —\u00a0 \u2206U=Q\u00a0 —\u00a0 Q=750J\u00a0 —\u00a0 \u2206U=750J.<\/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 Trabalho de um g\u00e1s \u2013 Transforma\u00e7\u00f5es c\u00edclicas 01-\u00a0R- C\u00a0\u00a0—\u00a0 veja teoria 02-\u00a0W=P.\u0394V\u00a0 —\u00a0 30=P(5 \u2013 2)\u00a0 —\u00a0 P=10N\/m2(Pa)\u00a0 —\u00a0\u00a0R- A 03-\u00a01g \u2013 5.103cal\u00a0 —\u00a0 4.000g \u2013 Q cal\u00a0 —\u00a0 Q=2.107cal\u00a0 — 1cal \u2013 4,2J\u00a0 —\u00a0 2.107cal \u2013 W J\u00a0 —\u00a0 W=8,4.107J\u00a0 —\u00a0 rendimento de 15%\u00a0 —\u00a0 W\u00fatil=0,15.8,4.107=1,26.107\u00a0J\u00a0 —\u00a0\u00a0R- E\u00a0\u00a0 04-\u00a0Trata-se de uma transforma\u00e7\u00e3o<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":2082,"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-2086","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2086","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=2086"}],"version-history":[{"count":4,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2086\/revisions"}],"predecessor-version":[{"id":10751,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2086\/revisions\/10751"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2082"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=2086"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}