Consider the expansion given in problem $6.2$,for $1 \, mol$ of an ideal gas conducted reversibly.

(N/A) For a reversible isothermal expansion of an ideal gas,the work done $w$ is given by $w = -2.303 \, nRT \, \log \frac{V_f}{V_i}$.
Since the process is reversible and isothermal,the heat exchanged $q = -w = 2.303 \, nRT \, \log \frac{V_f}{V_i}$.
Given $n = 1 \, mol$,$R = 0.08206 \, L \, atm \, K^{-1} \, mol^{-1}$,$T = 298 \, K$,$V_i = 2 \, L$,and $V_f = 10 \, L$.
Substituting the values: $q = 2.303 \times 1 \times 0.08206 \times 298 \times \log \frac{10}{2}$.
$q = 2.303 \times 0.08206 \times 298 \times \log 5$.
$q = 2.303 \times 0.08206 \times 298 \times 0.6990 \approx 393.66 \, L \, atm$.