A gas for which $\gamma$ is $3/2$ is heated at constant pressure. The percentage of heat supplied for external work is nearly

Heat supplied at constant pressure is given by $\Delta Q = n C_p \Delta T$

Work done by the gas is given by $\Delta W = P \Delta V = n R \Delta T$

The fraction of heat supplied used for external work is $\dfrac{\Delta W}{\Delta Q} = \dfrac{n R \Delta T}{n C_p \Delta T} = \dfrac{R}{C_p}$

Using Mayer's relation $C_p - C_v = R$ and $\gamma = \dfrac{C_p}{C_v}$, we get $C_p - \dfrac{C_p}{\gamma} = R$

$\dfrac{R}{C_p} = 1 - \dfrac{1}{\gamma}$

Given $\gamma = \dfrac{3}{2}$, substituting this value:

$\dfrac{\Delta W}{\Delta Q} = 1 - \dfrac{2}{3} = \dfrac{1}{3}$

Percentage of heat used for external work $= \dfrac{1}{3} \times 100\% = 33.33\%$

This is nearly $33\%$.

Answer: $33\ \%$