For a van der Waals gas, the partial derivative \(\left( {\frac{{\partial U}}{{\partial V}}}\right)\)_T, is

Concept:

• According to the first law of thermodynamics, "the total energy in a system remains constant, although it may be converted from one form to another".
• It can be represented mathematically as,

\(\Delta {\rm{U = }}\Delta {\rm{Q + dW}}\)

where, \(\Delta {\rm{Q }}\) is the amount of heat given or lost

\(\Delta {\rm{U }}\) is the change in internal energy

and \(dW\) is the amount of work done.​

• For an infinitesimal change,

\(d{\rm{U}} = Tds + ( - PdV)\)

\(dU = Tds - pdV\)

\({\left( {{{dU} \over {dV}}} \right)_T} = T{\left( {{{ds} \over {dV}}} \right)_T} - p \)

\({\left( {{{dU} \over {dV}}} \right)_T} = T{\left( {{{dP} \over {dT}}} \right)_V} - p \)

\(\left[ {From\;maxwells\;relation,\;{{\left( {{{ds} \over {dV}}} \right)}_T} = {{\left( {{{dP} \over {dT}}} \right)}_V}} \right] \)

• For one mole of Vander Waals gas,

\(\left( {{\rm{P + }}{{\rm{a}} \over {{{\rm{V}}_{\rm{m}}}^{\rm{2}}}}} \right)\left( {{{\rm{V}}_{\rm{m}}}{\rm{ - b}}} \right){\rm{ = RT}}\)

Explanation:

• The equation for one mole of vander walls gas is given by,

\(\left( {{\rm{P + }}{{\rm{a}} \over {{{\rm{V}}_{\rm{m}}}^{\rm{2}}}}} \right)\left( {{{\rm{V}}_{\rm{m}}}{\rm{ - b}}} \right){\rm{ = RT}}\)

\(P = {{{\rm{RT}}} \over {\left( {{{\rm{V}}_{\rm{m}}}{\rm{ - b}}} \right)}} - {{\rm{a}} \over {{{\rm{V}}_{\rm{m}}}^{\rm{2}}}}\)

Thus from 1st law of thermodynamics, we get,

\({\left( {{{dU} \over {dV}}} \right)_T} = T{\left( {{{dP} \over {dT}}} \right)_V} - p\)

\( = T{\left[ {{d \over {dT}}\left( {{{{\rm{RT}}} \over {\left( {{{\rm{V}}_{\rm{m}}}{\rm{ - b}}} \right)}} - {{\rm{a}} \over {{{\rm{V}}_{\rm{m}}}^{\rm{2}}}}} \right)} \right]_T} - P \)

\( = T{{\rm{R}} \over {\left( {{{\rm{V}}_{\rm{m}}}{\rm{ - b}}} \right)}} - P \)

\( = {{{\rm{RT}}} \over {\left( {{{\rm{V}}_{\rm{m}}}{\rm{ - b}}} \right)}} - P\)

\( = {{{\rm{RT}}} \over {\left( {{{\rm{V}}_{\rm{m}}}{\rm{ - b}}} \right)}} - \left( {{{{\rm{RT}}} \over {\left( {{{\rm{V}}_{\rm{m}}}{\rm{ - b}}} \right)}} - {{\rm{a}} \over {{{\rm{V}}_{\rm{m}}}^{\rm{2}}}}} \right) \)

\( = {{{\rm{RT}}} \over {\left( {{{\rm{V}}_{\rm{m}}}{\rm{ - b}}} \right)}} - {{{\rm{RT}}} \over {\left( {{{\rm{V}}_{\rm{m}}}{\rm{ - b}}} \right)}} + {{\rm{a}} \over {{{\rm{V}}_{\rm{m}}}^{\rm{2}}}} \)

​\(= {{\rm{a}} \over {{{\rm{V}}_{\rm{m}}}^{\rm{2}}}}\)

Conclusion:

Hence, for a van der Waals gas, the partial derivative \(\left( {\frac{{\partial U}}{{\partial V}}}\right)\)T, is \( {{\rm{a}} \over {{{\rm{V}}_{\rm{m}}}^{\rm{2}}}}\)