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2 moles of an ideal gas are in a closed system with a temperature of $25\ \mathrm{^\circ C}$.

The gas is cooled under constant pressure and afterwards heated gas under constant volume.

During the cooling process the gas has lost $1000\ \mathrm J$ of heat to the environment.

It is known that the molar heat capacity of the gas is $C=25\ \mathrm{J/(^\circ C\ mol)}$

What is the work done by the system during the cooling and heating combined?

Official answer: $333\ \mathrm J$.

The problem: I understand how to get to this solution, however I don't understand why an alternative method yields an incorrect answer. Here's what I tried:

First, during the heating process there is no work done because the volume is constant.

$Q=-1000\ \mathrm J$

We know that:

$\Delta E=\frac32\cdot n\cdot R\cdot\Delta T$

Now, $Q=n\cdot C\cdot \Delta T$, therefore:

$-1000=25\times2\times\Delta T$ and therefore $\Delta T=-20$

So $\Delta E=\frac32\cdot n\cdot R\cdot\Delta T=\frac32\times2\times8.314\times(-20)=-498.84\ \mathrm J$

By the first law of thermodynamics: $\Delta E=Q+W$, therefore $-498.84=-1000+W$ and $W=501.16\ \mathrm J$

Could anyone please tell me where is my mistake?

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    $\begingroup$ When a problem requires calculations using values, always write the values with the correct units and carry the units through the calculation. Do not omit the units while performing intermediate steps and do not just reintroduce units at the end of the calculation. $\endgroup$ – user7951 Oct 2 '20 at 18:49
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The question didn't specify which process the given molar heat capacity was specified for. So we will assume that it is molar heat capacity for the first process. The first process is isobaric so $$q=nC_p\,\Delta T$$

You calculated $\Delta T$ correctly. The only mistake you did was assume that the gas was monoatomic. The expression for $\Delta U$ you wrote was $\Delta U=\frac32nR\,\Delta T$. This expression would be correct if this was a monoatomic gas. But we don't know that do we?

The correct expression to use here is $$\Delta U=nC_V\,\Delta T$$ It is given that molar heat capacity for first process (constant pressure) is $C_p=25\ \mathrm{J/(^\circ C\ mol)}$ and we know that $$C_p=C_V+R$$ So you can calculate the value of $C_V$ and use it to get the value of $\Delta U$ which I calculated to be $$\Delta U=667.2\ \mathrm J$$ Plug this into first law of thermodynamics and you will get the correct answer.

Edit: I used the sign $\Delta U$ for Change in Internal Energy intead of $\Delta E$

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Firstly, I do not know what coefficient $3/2$ of your $\Delta E$ means, it makes none sense.

From data of the problem above, we can consider this process as an irreversible one. Additionally, because the latter process is isochoric process, the work done by that also equals $0$. So the work done by all process is the work done by isobaric process.

In the other hand, similarly to your calculation of $\Delta T$, now we can calculate the first work by equation $\Delta W=-p\,\mathrm dV=-n\times R\times \Delta T=- 2\ \mathrm{mol}\times8.314\ \mathrm{J\ K^{-1}\ mol^{-1}}\times 20\ \mathrm{^\circ C}\approx-333\ \mathrm J$. So, the work done is $333\ \mathrm J$.

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    $\begingroup$ What is it about the data of the problem that leads you to conclude that this process is irreversible??? $\endgroup$ – Chet Miller Oct 2 '20 at 16:56
  • $\begingroup$ I managed to get the answer using this method but can't figure out what's wrong with the other method. For an ideal gas we were taught that the energy is 3/2*nRT $\endgroup$ – EL_9 Oct 2 '20 at 16:57
  • $\begingroup$ That's only for a monoatomic ideal gas. $\endgroup$ – Chet Miller Oct 2 '20 at 17:11
  • $\begingroup$ The Equation you used is the internal energy involving in kinetic molecular energy, not the change in the internal energy. $\endgroup$ – Phan Đức Oct 2 '20 at 17:29
  • $\begingroup$ @ChetMiller This was never mentioned in my class and was used extensively even when it wasn't known that the gas was monatomic. $\endgroup$ – EL_9 Oct 2 '20 at 17:34

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