An ideal gas is allowed to expand both reversibly and irreversibly in an isolated system. If $$T_i$$ is the initial temperature and $$T_f$$ is the final temperature, which of the following statements is correct?

An isolated system exchanges neither heat nor matter with the surroundings, therefore for every process occurring in it

$$q = 0 \;\;\Longrightarrow\;\; \Delta U = q + w = w \;-(1)$$

For an ideal gas the internal energy depends only on temperature:

$$\Delta U = nC_v\,(T_f - T_i) \;-(2)$$

Combining $$(1)$$ and $$(2)$$ gives

$$nC_v\,(T_f - T_i) = w$$ $$\Longrightarrow\;\; T_f = T_i + \dfrac{w}{nC_v} \;-(3)$$

Next, compare the work terms for expansion:

• Sign convention used in chemical thermodynamics: $$w$$ is positive when work is done on the system, negative when work is done by the system.
• During any expansion the system does work on the surroundings, so $$w \lt 0$$.

Maximum (most negative) work is obtained in a reversible expansion, while any irreversible expansion does less (is numerically less negative) work:

$$w_{\text{rev}} \lt w_{\text{irrev}} \;-(4)$$

Substitute $$(4)$$ into $$(3)$$:

$$T_{f,\text{rev}} = T_i + \dfrac{w_{\text{rev}}}{nC_v}$$ $$T_{f,\text{irrev}} = T_i + \dfrac{w_{\text{irrev}}}{nC_v}$$

Because the denominators are the same and $$w_{\text{rev}} \lt w_{\text{irrev}}$$, the increment in temperature is more negative for the reversible path. Hence

$$(T_f)_{\text{irrev}} \gt (T_f)_{\text{rev}}$$

Therefore, in an isolated ideal-gas expansion the irreversible process ends at a higher final temperature than the reversible process.

Option A which is: $$(T_f)_{irrev} \;>\; (T_f)_{rev}$$