Question

Calculate $q$ for the isothermal reversible expansion of one mole of an ideal gas from an initial volume of $12L$ to a final volume of $120L$ at constant temperature of $273K$.

• A. $-5227J$
• B. $+5227J$
• C. $3120J$
• D. $Noneoftheabove$

Answer 1

The correct option is B $+5227J$

For an isothermal reversible expansion:
$w=-2.303\text{nRT log}\frac{V_2}{V_1}$
$=-2.303\times 1\times 8.314\times 273\text{log}\frac{120}{12}=-5227.2J$
From the first law:
$\Delta U=q+w$
As $\Delta U=0$
$\therefore q=-w=5227.2J$