Heat in an isothermal process

Problem

One mole of ideal gas at $T = 350\ \text{K}$ expands isothermally from $V_1 = 5\ \text{L}$ to $V_2 = 15\ \text{L}$. Calculate the work done and the heat absorbed. Check that $\Delta U = 0$.

Solution

Internal energy. In an isothermal process the temperature does not change, and for an ideal gas $U$ depends only on $T$: hence $\Delta U = 0$.

First law. From $\Delta U = Q - L = 0$ it follows that $Q = L$: all the heat absorbed is converted into work.

Calculating the work. The volume ratio is $V_2/V_1 = 15/5 = 3$, so

$L = nRT\ln\frac{V_2}{V_1} = 8{,}31\cdot 350\cdot \ln 3 = 2908{,}5\cdot 1{,}0986 \approx 3196\ \text{J}$

Conclusion. The heat absorbed equals the work, and the internal energy remains unchanged:

$\ev{L=Q\approx 3{,}20\cdot 10^3\ \text{J},\ \Delta U=0}$

Links

Topics: Thermodynamics Concepts: Thermodynamic processes Objects: Ideal gas