An ideal gas is expanded irreversibly from 5 L to 10 L against a constant external pressure of 1 bar — Thermodynamics and Thermochemistry Chemistry Question

💡 Solution & Explanation

For an ideal gas, an isenthalpic process ($\Delta H = 0$) must also be isothermal ($\Delta T = 0$), so $\Delta U = 0$. By the first law, $q = -W$. The work done by expansion is $W = -P_{\text{ext}}\Delta V = -1\text{ bar} \times (10 - 5)\text{ L} = -5\text{ bar}\cdot\text{L} = -500\text{ J}$. Therefore, heat absorbed is $q = +500\text{ J}$.