Calculate the work done if $1 \ mole$ of a certain gas is compressed isothermally and reversibly at $300 \ K$ from an initial pressure $x \ bar$ to a final pressure $2x \ bar$ $[R = 8.314 \ J \ K^{-1} \ mol^{-1}]$. (in $kJ$)

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

$\Delta G$ is net energy available to do useful work and is thus a measure of "free energy". Show mathematically that $\Delta G$ is a measure of free energy. Find the unit of $\Delta G$. If a reaction has positive enthalpy change and positive entropy change,under what condition will the reaction be spontaneous?

The standard state Gibbs free energies of formation of $C$ (graphite) and $C$ (diamond) at $T = 298 \ K$ are:
$\Delta_f G^0[C(\text{graphite})] = 0 \ kJ \ mol^{-1}$
$\Delta_f G^0[C(\text{diamond})] = 2.9 \ kJ \ mol^{-1}$
The standard state means that the pressure should be $1 \ bar$,and the substance should be pure at a given temperature. The conversion of graphite [$C$ (graphite)] to diamond [$C$ (diamond)] reduces its volume by $2 \times 10^{-6} \ m^3 \ mol^{-1}$. If $C$ (graphite) is converted to $C$ (diamond) isothermally at $T = 298 \ K$,the pressure at which $C$ (graphite) is in equilibrium with $C$ (diamond) is:
[Useful information: $1 \ J = 1 \ kg \ m^2 \ s^{-2} ; 1 \ Pa = 1 \ kg \ m^{-1} \ s^{-2} ; 1 \ bar = 10^5 \ Pa$ ] (in $bar$)

For a certain reaction,the enthalpy change and entropy change are $40.63 \ kJ \ mol^{-1}$ and $100 \ J \ K^{-1} \ mol^{-1}$ respectively. What will be the value of $\Delta G$ at $27 \ ^oC$?

What is the nature of the reaction at $298 \ K$,if the entropy change and enthalpy change for a chemical reaction are $7.4 \ cal \ K^{-1}$ and $-2.5 \times 10^3 \ cal$,respectively?