I'm having trouble understanding why reversible adiabatic processes are isentropic.

I understand that in a reversible adiabatic process there is no heat exchange and so $dQ = TdS = 0.$ However, if I have a thermally isolated ideal gas in a piston, and I reversibly compress the gas, even though there is no heat exchange (the process is reversible and adiabatic), I have increased the temperature of the gas! Doesn't that mean I've increased it's entropy?

It occurs to me that the decrease in volume probably exactly matches the increase in temperature in terms of entropy changes so that the total entropy change is $0.$ But still, if I try and do some calculations, I'm stuck. Say I change the volume from $V_1$ to $V_2$ by reversibly compressing the piston. Am not I changing the pressure and temperature simultaneously as I do that? How do I integrate the ideal gas law here?

  • $\begingroup$ Also, dont forget your original statement (no entropy change) applies to all gases, not just ideal ones. $\endgroup$
    – lalala
    Commented Jun 4, 2021 at 11:01

1 Answer 1


The short answer is that you're right - the change in entropy due to the decrease in volume cancels out the change in entropy due to the increase in temperature. But you're also right that this involves a simultaneous change in $p$ and $T$, and it would be nice to know by how much each of these changes. Below I've done the calculation. It was a bit more involved than I was expecting, and I hope I didn't make any mistakes!

To do this calculation we first need to note that the ideal gas law, $$pV=nRT,\tag{i}$$ tells us how pressure, volume and temperature relate to one another, but if you want to do any real calculations, you also need to know how the internal energy $U$ behaves. This is given by $$ U =n\; c_V T,\tag{ii} $$ where $c_V$ is the dimensionless heat capacity at constant volume. (Some people will define an ideal gas in such a way that $c_V$ is allowed to be a function of $U$, but here I'll assume it's constant. To a good approximation, $c_V=\frac{3}{2} R$ for a monatomic gas, or $\frac{5}{2} R$ for a diatomic one.) Equation $(\mathrm{ii})$ can't be derived from Equation $(\mathrm{i})$, so both of them are needed in order to define the properties of an ideal gas.

With this in mind, let's start with the fundamental equation of thermodynamics (for systems without chemical reactions): $$ dU = TdS - pdV. $$ Because we're considering an adiabatic process we know that $TdS = 0$, so $$ dU = -pdV = - \frac{nRT}{V} dV, $$ where the second equality is obtained by substituting the ideal gas law $(\mathrm{i})$. But we also know from the heat capacity equation $(\mathrm{ii})$ that $nT = U/c_V$. This gives us $$ dU = - \frac{U\;R}{c_VV} dV, $$ or $$ c_V\frac{1}{U}dU = - R\frac{1}{V} dV. $$ Now we can integrate both sides: $$ c_V \int_{U_1}^{U_2} \frac{1}{U}dU = -R\int_{V_1}^{V_2} \frac{1}{V}dV, $$ or $$ c_V\left( \ln U_2 - \ln U_1 \right) =R \left( \ln V_1 - \ln V_2 \right). $$ A quick note about interpretation is in order here. We're integrating both sides over different ranges ($U_1$ to $U_2$ and $V_1$ to $V_2$) and then setting them equal. This is because we want to know how much the internal energy will change if we reversibly change the volume by a certain amount, so we're looking for $U_2$ and $U_1$ as a function of $V_1$ and $V_2$. Anyway, now we can use some logarithm identities to get $$ c_V \ln \frac{U_2}{U_1} = R\ln \frac{V_1}{V_2} $$ or $$ \ln \left(\frac{U_2}{U_1}\right)^{c_V} = \ln \left(\frac{V_1}{V_2}\right)^R, $$ so $$ \left(\frac{U_2}{U_1}\right)^{c_V} = \left(\frac{V_1}{V_2}\right)^{R}, $$ or $$ V_1^{R} U_1^{c_V} = V_2 ^{R} U_2^{c_V}. $$ This means that, for a reversible process, the quantity $VU^{c_V}$ must remain constant. Now, finally, we can substitute $U$ from Equation $(\mathrm{ii})$ to get $$ V^R(c_VnT)^{c_V} = \text{constant for an adiabatic process.} $$ There's a factor of $(c_Vn)^{c_V}$ that we can ignore by incorporating it into the constant on the right hand side, so $$ V^RT^{c_V} = \text{constant for an adiabatic process,} $$ or $$ V_1^R T_1^{c_V} = V_2^R T_2^{c_V}. $$

Given any values for $V_1$, $V_2$ and $T_1$, you can use this to work out $T_2$. By substituting into $(\mathrm{i})$ you can also work out the change in pressure.

  • 2
    $\begingroup$ Hi, sir. I've edited your answer. Firstly, I removed the infinitesimal $dU$ & wrote the correct one $dV.$ I've seen you used $U= nC_v R T$ but my book didn't write any $R$ in it. $C_v$ itself has $R$ in its definition; for monoatomic gases, $C_v= \frac{3}{2} R.$ So, I removed $R$ & modified the subsequent calculations. Yours were also correct as you wrote $R$ explicitly. However, if you feel I've done wrong, then rollback my edit; it's your answer, after-all:) Thanks for the answer & sorry for my edit if it annoyed you. $\endgroup$
    – user36790
    Commented Nov 11, 2015 at 4:04

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