Two moles of an ideal gas (cv=R) was compressed adiabetically against constay pressure of 3 atm. which was initially at350k and 1 atm. pressure. what is the work involved in the pressure?
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Answer:
It is given that,
Cv=R,
No of moles,n=2
External pressure,pext=3atm
initial temperature, T1=350K
p2=1atm
Now,
nCv(T2-T1)=-Pext×nR(T2/P2-T1/P1)
2R(T2-T1)=-3×2R(T2/1-T1/3)
T2-T1=-3(3T2-T1/3)
T2-T1=T1-3T2
T2=T1/2
so, Work done,w=nCv∆T
w=2×R×(T2-T1)
W=2R(T1/2-T1
w=2R(T1-2T1/2)
W=R(-T1)
w=-350R
Explanation:
Hope it helps you frnd........
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