Work in an isobaric process

Problem

One mole of ideal gas at $T_1 = 300\ \text{K}$ and $p = 1\ \text{atm}$ is heated at constant pressure up to $T_2 = 500\ \text{K}$. Calculate the work done by the gas. Check the formula $L = nR\,\Delta T$.

Solution

Work at constant pressure. In an isobaric process the gas expands against a fixed pressure, so

$L = p\,\Delta V$

Using the equation of state. From $pV = nRT$ at constant pressure it follows that $p\,\Delta V = nR\,\Delta T$, so

$L = nR\,\Delta T = 8{,}31\cdot(500-300) = 8{,}31\cdot 200 = 1662\ \text{J}$

Conclusion. The area of the rectangle in the $p\text{-}V$ plane (base $\Delta V$, height $p$) represents exactly this work:

$\ev{L\approx 1{,}66\cdot 10^3\ \text{J}}$

Links

Topics: Thermodynamics Concepts: Thermodynamic processes Objects: Ideal gas