# Clarification on Clausius Inequality

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!

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.

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 $\Delta S>Q/T-Q/T_0>0$

• 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 – Siddharth Prakash May 22 '15 at 14:48
• 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 – facenian May 22 '15 at 16:50