Now the friends try a homework problem. Two identical cylinders with a movable p
ID: 1584548 • Letter: N
Question
Now the friends try a homework problem. Two identical cylinders with a movable piston contain 0.4 mol of helium gas at a temperature of 300 K. The temperature of the gas in the first cylinder is increased to 465 K at constant volume by doing work W1 and transferring energy Q1 by heat. The temperature of the gas in the second cylinder is increased to 465 K at constant pressure by doing work W2 while transferring energy Q2 by heat.
Find
(delta) E internal, Q1 , and W1 at constant volume
Find
(delta)Einternal, Q2 and W2 at constant pressure
Explanation / Answer
Einternal = n*Cv*dT = n*(3/2)*R*dT
Einternal = 0.4*(3/2)*8.314*(465-300) = 823.1 J
Q1 = n*CV*dT = 823.1 J
from I law of thermodynamics
W1 = Q1 - Einternal = 0
============================================
for cylinder 2
Einternal = n*Cv*dT = n*(3/2)*R*dT
Einternal = 0.4*(3/2)*8.314*(465-300) = 823.1 J
Q2 = n*Cp*dT = n*(5/2)*R*dT = 0.4*(5/2)*8.314*(465-300) = 1372 J
from I law of thermodynamics
W2 = Q2 - Einternal = 1372 - 823.1 = 548.9 J
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