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 done on the gas in a reversible isothermal process: $ w_{\mathrm{on}} =-nRT\,\ln!\left(\frac{V_2}{V_1}\right) =nRT\,\ln!\left(\frac{V_1}{V_2}\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 $w_{\mathrm{on}}>0$ denotes work done on the gas, so compression gives positive work. Texts that define work as work done by the gas use the opposite sign.
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 heating 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 $p_1$): \(w_a=-p_1(V_2-V_1)=p_1(V_1-V_2) =nRT\!\left(1-\frac{V_2}{V_1}\right)\) Numerically: $w_a=1.247\times10^5\,\text{J}$.
-
(b) Isothermal: \(w_b=-nRT\,\ln\!\left(\frac{V_2}{V_1}\right) =nRT\,\ln\!\left(\frac{V_1}{V_2}\right) =2.494\times10^5\,\ln 2\) $w_b=1.729\times10^5\,\text{J}$.
-
(c) Isobaric (at $p_2$): \(w_c=-p_2(V_2-V_1)=p_2(V_1-V_2)=2w_a\) $w_c=2.494\times10^5\,\text{J}$.
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 because the gas reaches the same temperature in a smaller volume. Entropy is a state function, so the same value is obtained for all three paths.
- Important remark: Here $\Delta U=0$ because the initial and final temperatures are equal and the gas is ideal. Only path (b) is isothermal throughout; paths (a) and (c) pass through intermediate temperatures.