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I am supposed to calculate the change in energy upon changing

both the temperature from $T_1$ to $T_2$ and the volume from $V_1$ to $V_2$.

Now I was wondering whether this solution is correct:

We can treat both transition independent from each other, as the energy is not path-dependent and therefore we could have $$\Delta(E) = \frac{3}{2} k_BN (T_2-T_1) -N k_B T ln(\frac{V_2}{V_1})$$.

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  • $\begingroup$ What is $T$ on the right hand side? Also, where did your second term on the right hand side come from? $\endgroup$
    – Brian Moths
    Commented Dec 8, 2014 at 23:54
  • $\begingroup$ @NowIGetToLearnWhatAHeadIs yes, actually I don't really know. The second term comes from $\int_{V_1}^{V_2} \frac{N k_B T}{V} dV = N k_B T ln( \frac{V_2}{V_1})$ $\endgroup$
    – Xin Wang
    Commented Dec 9, 2014 at 0:04

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Your idea is correct but the calculation is not. For $N$ moles of an ideal gas you have $U = cNRT$, look that if $T$ doesn't change, $U$ also doesn't change. So you start at the state $(T_1,V_1)$ and want to go to the state $(T_2,V_2)$. You then use two process corresponding to the following sequence of states:

$$(T_1,V_1)\to (T_2,V_1)\to (T_2,V_2)$$

On the first, you indeed have $\Delta U_1 = cNRT_2 - cNRT_1 = cNR(T_2-T_1)$ but the second one however has temperature equal to $T_2$ so that $\Delta U_2 = 0$. This implies that the total change in energy is $\Delta U = cNR(T_2-T_1)$ which is just the first term you wrote.

Edit: About your explanation to the second term, the quantity $Nk_B T/V$ is indeed pressure, since $PV = Nk_B T$. Now, because of that your integral gives in truth the negative of the amount of work on the second process. Recall that $\delta W = -PdV$ and so

$$W = -\int_\Gamma PdV$$

But by the first law of Thermodynamics, $\Delta U = W + Q$, so that change of energy is not made up just of work, but of heat also. By the formula for energy you could then see that for this process $Q = -W$.

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  • $\begingroup$ why does a change in volume not result in a change of energy? I mean there is the integral $\Delta E = \int p dV = \int \frac{N k_B T}{V} dV = N k_B T ln(V)$ $\endgroup$
    – Xin Wang
    Commented Dec 9, 2014 at 0:08
  • $\begingroup$ well, actually this answer surprises me. I was first suppose to calculate the energy by changing the temperature of an ideal gas from $T_1$ to $T_2$. THen I was supposed to do it for a volume change $V_1$ to $V_2$, after that for both couples processes(which is essentially my question). Finally, I am now supposed to calculate the necessary volume change to maintain the energy if we raise the temperature from $T_1$ to $T_2$. If I understand you correctly, then it is impossible to accomplish the last situation. $\endgroup$
    – Xin Wang
    Commented Dec 9, 2014 at 0:23
  • $\begingroup$ I didn't say to calculate the volume change to maintain the energy. I said that since the energy of an ideal gas depends just on the temperature, on the second process the change will be zero because both initial and final temperature will be $T_2$. In that case the total change in energy will just come from the first process. What I said on the edit, is that the integral you speak of gives you work, not total change in energy, which accounts for heat as well. Now, if you use the formula $U = cNRT$ for the energy, you see that on the second process $\Delta U_2 = 0$. $\endgroup$
    – Gold
    Commented Dec 9, 2014 at 0:27
  • $\begingroup$ yes, but this is a homework question and there I am also supposed to say how much I need to change my volume in order to maintain the energy if the temperature is raised. If I understand you correctly, this is impossible. $\endgroup$
    – Xin Wang
    Commented Dec 9, 2014 at 0:31
  • $\begingroup$ Again, if temperature is held fixed, you cannot change the internal energy of the gas, this would violate the equation of state $U = cNRT$. I believe the problem was built with the intention of making you see that the internal energy of an ideal gas depends just on its temperature. So it is really impossible to do it. To be able to change the energy back you would need to change the temperature back anyway. $\endgroup$
    – Gold
    Commented Dec 9, 2014 at 0:45
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No, your solution is not correct. The energy difference is by definition the first term in your formula, you should drop the second term. The work done during the process is path dependent, so you don't have enough information to calculate the work done, neither the transferred heat (which is the sum of the energy difference and the work).

One possible way to perform that process is to first heat the gas and keep the volume fixed (for this the work is zero and the heat is given by your formula), then (assuming $V_2>V_1$) you instanteniously change the volume (move the walls out). This does not change the velocity of any particles, so no energy and temperature change happens, and no heat is transferred. This is of course an irreversible process, you can perform the change in the volume in reversible ways too, giving some nonzero work and nonzero heat if you keep T constant.

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  • $\begingroup$ well, actually this answer surprises me. I was first suppose to calculate the energy by changing the temperature of an ideal gas from $T_1$ to $T_2$. THen I was supposed to do it for a volume change $V_1$ to $V_2$, after that for both couples processes(which is essentially my question). Finally, I am now supposed to calculate the necessary volume change to maintain the energy if we raise the temperature from $T_1$ to $T_2$. If I understand you correctly, then it is impossible to accomplish the last situation. $\endgroup$
    – Xin Wang
    Commented Dec 9, 2014 at 0:23
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As long as the temperature of an ideal gas does not change its internal energy will remain the same. Change in volume without changing the temperature as in an isothermal process will not bring about any change in the internal energy of the gas. However a change in temperature of the gas by any means would result in change in internal energy of the gas as given by the first term on the right of your expression for energy change.

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