Fraction of heat energy used to do expansion work of a gas?

AI Thread Summary
The discussion centers on calculating the fraction of heat energy used for expansion work in a monatomic ideal gas under constant pressure. The initial attempt incorrectly calculated the ratio of internal energy change to heat added, resulting in a fraction of 3/5 instead of the correct 2/5. The correct approach involves finding the ratio of work done (W) to heat added (Q), leading to W/Q = 2/5. Clarification is provided that the relationship (nRΔT) = PΔV is derived from the ideal gas law, where pressure remains constant. This highlights the importance of correctly interpreting the question to arrive at the right answer.
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Homework Statement


Heat Q flows into a monatomic ideal gas, and the volume increases while the pressure is kept constant. What fraction of the heat energy is used to do the expansion work of the gas?

Homework Equations


ΔU = Q - W
Q = nC_pΔT
W = PΔV
C_p = C_v + R => C_p = 3R/2 + R => C_p = 5R/2

The Attempt at a Solution


So we sub in the Q and W in the internal energy formula and we get:
ΔU = nC_pΔT - PΔV
ΔU = (5/2)(nRΔT) - PΔV
ΔU = (5/2)(PΔV) - PΔV
ΔU = (3/2)(PΔV)

Therefore we can take ΔU/Q to find the fraction:
(3/2)(PΔV)/(5/2)(PΔV)

Which gives me 3/5. However the answer is 2/5. What am I doing wrong?
Personally, I think the answer key is wrong.

Thanks.
 
Last edited:
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I know this is an old question, but I thought I'd reply for future students that need help with this question.

All of the work shown is right, except for the very end. The question wants the ratio of W/Q, while the original poster found the ratio of ΔU/Q.

W/Q =
PΔV / (5/2)(PΔV) = 2/5
 
Could someone please explain to me how (nRΔT) = PΔV in the third step of this derivation?
 
BabyPhys said:
Could someone please explain to me how (nRΔT) = PΔV in the third step of this derivation?
Ideal gas law. PV=nRT. If P is constant, then PΔV=nRΔT

Chet
 
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