Consider a heat reservoir which gains heat $Q$ irreversibly at temperature $T$ from the surroundings which is at temperature $T_0$. The entropy change of reservoir is then given by $\frac{Q}{T}$, while that of the surroundings is $-\frac{Q}{T_0}$.

My question is, how is this possible? According to the Clausius inequality, the entropy change of a irreversible process is greater than that due to heat transfer. Please help, thank you!


2 Answers 2


The equation:

$$ dS = \frac{dQ}{T} $$

only applies to reversible processes. For an irreversible process $dS \gt dQ/T$.

To see this start with the expression for the change in internal energy:

$$ dU = dQ - dW $$

The internal energy is a state function, so this equation always applies whether the process is reversible or irreversible. So for a reversible process we have:

$$ dU = TdS - dW_{rev} $$

Suppose we make the same change in $U$ with an irreversible process then we have:

$$ dU = dQ_{irrev} - dW_{irrev} $$

And because $dU$ is the same in both cases we equate the two expressions to get:

$$ TdS - dW_{rev} = dQ_{irrev} - dW_{irrev} $$

which rearranges to:

$$ dS = \frac{dQ_{irrev}}{T} + \frac{dW_{rev} - dW_{irrev}}{T} $$

But we know that the work from a reversible process is always greater than the work from an irreversible process i.e. $dW_{rev} - dW_{irrev} > 0$, and this means:

$$ dS = \frac{dQ_{irrev}}{T} + \Delta $$

for some positive number $\Delta$ that depends on the details of the irreversible process.

  • $\begingroup$ This answer is the clearest way I've seen to demonstrate the Clausius Inequality. My question would be, is it too simple? Is there something missing in this formulation? Other arguments seem unnecessarily complex in comparison. $\endgroup$
    – michael b
    Commented Apr 25, 2021 at 0:41

The change of entropy is Q/T only if the heat transfer is reversible if the process is irreversible you can't obtain the change of entropy through the formula Q/T

After the heat transfer the change of entropy o the whole system is $\Delta S>Q/T-Q/T_0>0$

  • $\begingroup$ I wouldn't be asking this if i didn't come across one such condition..if you can check my question on" heat transfer through a finite temp difference" i have explained the condition briefly $\endgroup$ Commented May 22, 2015 at 14:48
  • $\begingroup$ Sorry but I don't understand you because you say in your question that the entropy change of the reservoir is Q/T and that is not correct because se process is irreversible $\endgroup$
    – facenian
    Commented May 22, 2015 at 16:50

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