When $\Delta H_{vap} = 30 \ kJ \ mol^{-1}$ and $\Delta S_{vap} = 75 \ J \ mol^{-1} \ K^{-1}$,then the temperature of vapour,at one atmosphere is . . . . $K$.

The relation between $\Delta G$ and $\Delta H$ is

Calculate the work done during the reversible isothermal expansion of $2 \ mol$ of an ideal gas at $298 \ K$ from $10 \ L$ to $20 \ L$ in $J$. (in $.9$)

For the reaction $A \longrightarrow B$,$\Delta H^{\circ} = 7.5 \, kJ \, mol^{-1}$ and $\Delta S^{\circ} = 25 \, J \, K^{-1} \, mol^{-1}$. The value of $\Delta G^{\circ}$ and the temperature at which the reaction reaches equilibrium are,respectively:

The densities of graphite and diamond at $298 \, K$ are $2.25 \, g \, cm^{-3}$ and $3.31 \, g \, cm^{-3}$ respectively. If the standard free energy difference $(\Delta G^o)$ is $1895 \, J \, mol^{-1}$, the pressure at which graphite will be transformed into diamond at $298 \, K$ is:

$2 \, mol$ of zinc is dissolved in $HCl$ at $25 \, ^\circ C$. The work done in an open vessel is: (in $, kJ$)