Q.Calculate the work involved in expansion and compression process.

Step 1. For a single-step (irreversible) expansion or compression against a CONSTANT external pressure $P_{ext}$, starting from $w=-F\Delta x$ (7.2) and $F=P_{ext}A$ (7.3): substituting gives $w=-P_{ext}A\Delta x$, and since $A\Delta x$ is the volume change, $w=-P_{ext}(V_f-V_i)$ (7.4), i.e. $w=P_{ext}\Delta V$ for a compression written with $-\Delta V$ positive (7.5).

Step 2. For a REVERSIBLE process, where the external pressure tracks the internal pressure at every infinitesimal step ($P_{ext}=P_{int}\pm dP$), the work is instead an integral: $w_{rev}=-\int_{V_i}^{V_f}P_{int}\,dV$.

Step 3. For $n$ moles of ideal gas ($P_{int}=nRT/V$), this integral evaluates to $w_{rev}=-nRT\ln(V_f/V_i)=-2.303\,nRT\log(V_f/V_i)$ (7.6). …

Derive both the single-step formula (from $w=-F\Delta x$) and the reversible integral formula (from $w_{rev}=-\i …

• Giving only the single-step formula $w=-P\Delta V$ without also covering the reversible (integral) case the question is really asking about …