Question

Calculate $q,w,\Delta U$ for the isothermal reversible expansion of $1\text{mole}$ of an ideal gas from an initial pressure of $1.0\text{bar}$ to a final pressure of $0.1\text{bar}$ at a constant temperature of $273\text{K}$.

• A. $w=q=\Delta U=0$
• B. $w=q=5.22\text{kJ}\text{and}\Delta U=0$
• C. $q=-w=5.22\text{kJ}\text{and}\Delta U=0$
• D. $q=-w=6.22\text{kJ}\text{and}\Delta U=0$

Answer 1

The correct option is C $q=-w=5.22\text{kJ}\text{and}\Delta U=0$

In an isothermal process, temperature remains constant so $\Delta U$ is zero
According to the first law of thermodynamics,
$\Delta U=w+q0=w+qq=-w\text{For a reversible isothermal process,}w=-2.303\text{nRT}{\log }_{10}\frac{P_1}{P_2}=-2.303\times 1\times 8.314\times 273{\log }_{10}\frac{1}{0.1}=-5227.1\text{J}=-5.22\text{kJ}$
Thus, $q=-w=5.227\text{kJ}$ and $\Delta U=0$