Question

O cantitate niu=1 mol de gaz monoatomic suferă transformare ciclică 1 --> 2 --> 3 --> 4 --> 1 reprezentată în figură.
t1=27°C
Se cere:
a) calculați Q cedat;
b) L total= ?;
c) deltaU= ?;
d) eta= ?

Answer 1

T₁=27°C=300 K

Q₁₂=υCv(T₂-T₁)

p₁/T₁=p₂/T₂=>T₂=T₁p₂/p₁=300*3p/p=900 K

Q₁₂=υCv(T₂-T₁)=3υR(T₂-T₁)/2=3*1*8,314(900-300)/2=12,47*600=7482 J ->pe transformarea 1-2 gazul primeste caldura.

Q₂₃=υCp(T₃-T₂)

V₂/T₂=V₃/T₃=>T₃=T₂V₃/V₂=900*3V/V=2700 K

Q₂₃=υCp(T₃-T₂)=5υR(T₃-T₂)/2=5*1*8,314(2700-900)/2=20,78*1800=37413 J->pe transformarea 2-3 gazul primeste caldura

Qtp=Q₁₂+Q₂₃=7482+37413=44895 J

Q₃₄=υCv(T₄-T₃)

p₃/T₃=p₄/T₄=>T₄=T₃p₄/p₃=2700*3p/p=8100 K

Q₃₄=υCv(T₄-T₃)=3υR(T₄-T₃)/2=3*1*8,314(8100-2700)/2=12,47*5400=67338 J->pe transformarea 3-4 gazul cedeaza caldura

Q₄₁=υCp(T₁-T₄)=5υR(T₁-T₄)/2=5*1*8,314(300-8100)/2=20,78*(-7800)=-162084 J->pe transformarea 4-1 gazul cedeaza caldura

Qtc=Q₃₄+Q₄₁=67338-162084=-94746 J

Qt=Qtc+Qtp=44895-94746=-49851 J

Deoarece Qt<0=>gazul cedeaza mai multa caldura decat primeste.

Lt=(p₂-p₁)(V₄-V₁)=(3p-p)(3V-V)=2p*2V=4pV

pV=υRT=1*8,31*300=2493

Lt=4pV=4*2493=9972 J

ΔU=Qt-Lt=-49851 -9972=-59823 J

η=Lt/Qtp=9972/44895=0,22=>η=22%