Ideal-Gas Processes — Work, $\Delta U$ and $\Delta S$
Theoretical Background
- Equation of state: $pV=nRT$
- Internal energy of an ideal gas: $U=U(T)$, so $\Delta U$ depends only on the temperature change.
- Isothermal process: $\Delta U=0$
- Work in a reversible isothermal: \(w=nRT\,\ln\!\left(\frac{V_2}{V_1}\right)\)
- Entropy change (reversible): \(\Delta S=nR\,\ln\!\left(\frac{V_2}{V_1}\right)\)
- State vs path functions: $\Delta U$ and $\Delta S$ depend only on initial and final states, while $w$ and $q$ depend on the path.
Sign convention reminder. Here $w>0$ denotes work on the gas. Many textbooks adopt the opposite convention ($w<0$ for work on the system): always check the definition in use.
Validity note. The results assume an ideal gas and reversible transformations. For real gases or irreversible processes, the formulas must be corrected.
Exercise
A sample of $n=100\,\text{mol}$ of ideal hydrogen at $T=300\,\text{K}$ is compressed from $V_1=4.0\,\text{m}^3$ to $V_2=2.0\,\text{m}^3$.
Calculate the work on the gas along three different reversible paths:
- (a) Isobaric compression (at $p_1$) followed by isochoric cooling to the final state.
- (b) Direct isothermal compression from $V_1$ to $V_2$.
- (c) Isochoric heating to $p_2$, followed by isobaric compression at $p_2$.
Then evaluate $\Delta U$ and $\Delta S$, and compare results.
Step-by-Step Solution
Step 1. Calculate initial and final pressures
From $pV=nRT$:
\(p_1=\frac{nRT}{V_1},\qquad p_2=\frac{nRT}{V_2}.\)
With $nRT=2.494\times10^5\,\text{J}$, we have $p_2=2p_1$.
Step 2. Work for each path
-
(a) Isobaric (at $p_1$): \(w_a=p_1(V_2-V_1)=nRT\!\left(\frac{V_2}{V_1}-1\right)\) Numerically: $w_a=1.247\times10^5\,\text{J}$.
-
(b) Isothermal: \(w_b=nRT\,\ln\!\left(\frac{V_2}{V_1}\right) =2.494\times10^5\,\ln(0.5)\) $w_b=1.729\times10^5\,\text{J}$.
-
(c) Isobaric (at $p_2$): \(w_c=p_2(V_2-V_1)=2p_1(V_2-V_1)=2w_a\) $w_c=2.494\times10^5\,\text{J}$.
Step 3. Internal energy change
Since $T$ is the same at initial and final state:
\(\Delta U=0.\)
Step 4. Entropy change
Use the isothermal reference (reversible):
\(\Delta S=nR\,\ln\!\left(\frac{V_2}{V_1}\right).\)
Numerically:
\(\Delta S=100(8.314)\,\ln(0.5)=-5.76\times10^2\,\text{J K}^{-1}.\)
Notes and Discussion
- The work depends on the path: $w_a<w_b<w_c$. This illustrates that work is path-dependent.
- The internal energy change is zero for all cases because $U$ of an ideal gas depends only on $T$.
- The entropy decreases, consistent with the system becoming more ordered upon compression. This is a state function, so the same value emerges regardless of path.
- Important remark: Always check whether $\Delta U=0$ holds: it is true here because the process is isothermal and the gas is ideal. For real gases this statement is not exact.