# Energy as a state function

## Homework Statement:

(Figure 1) shows two processes, A and B, for moving 3.45 × 10^22 particles of a monatomic ideal gas from state i to state f. Which process requires a smaller magnitude of the energy Q transferred thermally to the gas?

## Relevant Equations:

$$\Delta E=W+Q$$
$$PV=Nk_BT$$
$$Q=\tfrac{d}{2}Nk_BT \text{ (isochoric process)}$$
$$Q=NC_p\Delta T \text{ (isobaric process) }$$
Here is the figure: The answer is $$Q_A<Q_B$$ which I can show by calculation using the above equations. What's confusing to me is I thought that the change in internal energy was a state function. Which would mean since the initial and final points are the same, $$\Delta E_A=\Delta E_B$$ or by the 1st law of thermodynamics $$W_A+Q_A=W_B+Q_B$$ Since W is the area under the curve (and positive in this case since the gas is being compressed), $$W_A<W_B$$ So the work done on the gas in process A is less than in process B, and the energy transferred thermally to the gas in process A is less than in process B. Which, since all values are positive, seems like it contradicts the original assumption that the change in internal energy was a state function, since it seems like the change in internal energy has to be less in process A. What am I missing here?

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mfb
Mentor
The answer is ##Q_A<Q_B## which I can show by calculation using the above equations.
Can you show that calculation? There is heat involved in both parts of both transitions.

Can you show that calculation? There is heat involved in both parts of both transitions.
Sure, energy transferred thermally in the isobaric processes is
$$Q=NC_p\Delta T$$
$$Q=N\left(\tfrac{d}{2}+1\right)k_b\Delta T$$
And ##\Delta T = \frac{P\Delta V}{Nk_B}## by the ideal gas law so
$$Q=P\Delta V\left(\tfrac{d}{2}+1\right)$$
And then for the isochoric processes
$$Q=\tfrac{d}{2}Nk_B\Delta T$$
And ##\Delta T = \frac{\Delta P V}{Nk_B}## so
$$Q = \tfrac{d}{2}(\Delta P)V$$
So the overall energy transferred thermally in each process is
$$P\Delta V\left(\tfrac{d}{2}+1\right)+\tfrac{d}{2}(\Delta P)V$$
Taking d to be 3 since this is a monatomic gas, the energy transferred thermally in process A is then
$$9680\cdot1.3(\tfrac{3}{2}+1)+\tfrac{3}{2}\cdot12720\cdot0.45=40046J$$
And then the energy transferred thermally in process B is
$$22400\cdot1.3(\tfrac{3}{2}+1)+\tfrac{3}{2}\cdot12720\cdot1.75=106190J$$
So ##Q_A<Q_B##, which my homework grader says is correct.

mfb
Mentor
What is the sign of ##\Delta V##?

• Chestermiller
What is the sign of ##\Delta V##?
Oh, that was it! So ##|Q_A|<|Q_B|## like the question asked, but ##Q_A>Q_B##, and so there is no contradiction. Thanks!

• mfb and Chestermiller