Question

Calculate the work done in joules when $3$ moles of an ideal gas at ${27}^{\circ }C$ expands isothermally and reversibly from $10$ atm to $1$ atm ($1atm=1.013\times {10}^{5}N{m}^{-2}$). What will be the work done if the expansion is against the constant pressure of $1$ atm?

Answer 1

$n=3$ moles, $T={27}^{o}C=300K$

$P_1=10atm$ $P_2=1atm$

Work done in isothermal reversible process,

$W=-2.303nRT\log \left(P_1/P_2\right)$

$=-2.303\times 3\times 8.314\times 300\log \left(\frac{10}{1}\right)$

$=5744.14J$

Work done against constant pressure of $1atm$

$=P_{opp}(V_f-V_i)$

$=-1atm(\frac{nRT}{P_2}-\frac{nRT}{P_1})$

$=-1\times 3\times 8.314\times 300(\frac{1}{1}-\frac{1}{10})$

$=6734.34J$ .