Question

(a) The van der Waals equation for \({\rm{n}}\) moles of a gas is \(\left( {P + \frac{{{n^{\rm{2}}}a}}{{{V^{\rm{2}}}}}} \right)\left( {V - nb} \right) = nRT\) where \(P\)is the pressure,\(V\) is the volume, and\(T\) is the temperature of the gas. The constant\(R\) is the universal gas constant and\(a\)and\(b\)are positive constants that are characteristic of a particular gas. If \(T\) remains constant, use implicit differentiation to find\(\frac{{dV}}{{dP}}\).

(b) Find the rate of change of volume with respect to pressure of \({\rm{1}}\) mole of carbon dioxide at a volume of \(V = {\rm{10}}L\) and a pressure of \(P = {\rm{2}}{\rm{.5atm}}\). Use \({\rm{a}} = {\rm{3}}{\rm{.592}}{{\rm{L}}^{\rm{2}}}{\rm{ - atm}}/{\rm{mol}}{{\rm{e}}^{\rm{2}}}\)and \(b = {\rm{0}}{\rm{.04267}}L/mole\).

Short answer

From the van der Waals equation \(\left( {P + \frac{{{n^{\rm{2}}}a}}{{{V^{\rm{2}}}}}} \right)\left( {V - nb} \right) = nRT\) by implicit differentiation the value \(\frac{{dV}}{{dP}} = \frac{{\left( {nb - V} \right){V^{\rm{2}}}}}{{P{V^{\rm{2}}} + {n^{\rm{2}}}a\left( {{\rm{1}} + nbV - {V^{\rm{2}}}} \right)}}\), and rate of change of volume with respect to pressure is \({\rm{9}}{\rm{.5674}}L/atm\).

Step-by-step solution

Given Information

(a) The given van der Waals equation for \({\rm{n}}\) moles of a gas is\(\left( {P + \frac{{{n^{\rm{2}}}a}}{{{V^{\rm{2}}}}}} \right)\left( {V - nb} \right) = nRT\). (b) The values to find rate of change of volume with respect to pressure are \(n = {\rm{1}}\), \(V = {\rm{10L}}\), \(P = {\rm{2}}{\rm{.5atm}}\), \({\rm{a}} = {\rm{3}}{\rm{.592}}{{\rm{L}}^{\rm{2}}}{\rm{ - atm}}/{\rm{mol}}{{\rm{e}}^{\rm{2}}}\) and \(b = {\rm{0}}{\rm{.04267}}L/mole\).

Definition of Derivative

The derivative of a real-valued function measures the sensitivity of the function's value (output value) to changes in its argument in mathematics (input value).

Implicit differentiation

Differentiate the van der Waals equation for \({\rm{n}}\) moles, with respect to \(P\), such that \(T\) remains constant.

\(\begin{array}{c}\left( {P + \frac{{{n^{\rm{2}}}a}}{{{V^{\rm{2}}}}}} \right)\left( {V - nb} \right) = nRT\\\left( {{\rm{1}} - \frac{{{n^{\rm{2}}}a}}{V}\frac{{dV}}{{dP}}} \right)\left( {V - nb} \right) + \left( {P + \frac{{{n^{\rm{2}}}a}}{{{V^{\rm{2}}}}}} \right)\left( {\frac{{dV}}{{dP}} - {\rm{0}}} \right) = {\rm{0}}\\\left( { - {n^{\rm{2}}}a + \frac{{{n^{\rm{3}}}ab}}{V} + P + \frac{{{n^{\rm{2}}}a}}{{{V^{\rm{2}}}}}} \right)\frac{{dV}}{{dP}} = nb - V\\\frac{{dV}}{{dP}} = \frac{{\left( {nb - V} \right){V^{\rm{2}}}}}{{P{V^{\rm{2}}} + {n^{\rm{2}}}a\left( {{\rm{1}} + nbV - {V^{\rm{2}}}} \right)}}\end{array}\)

Rate of change of volume with respect to pressure

Substitute the values \(n = {\rm{1}}\), \(V = {\rm{10L}}\), \(P = {\rm{2}}{\rm{.5atm}}\), \({\rm{a}} = {\rm{3}}{\rm{.592}}{{\rm{L}}^{\rm{2}}}{\rm{ - atm}}/{\rm{mol}}{{\rm{e}}^{\rm{2}}}\) and \(b = {\rm{0}}{\rm{.04267}}L/mole\) in \(\frac{{dV}}{{dP}}\) and find the rate of change of volume with respect to pressure.

\(\begin{array}{l}\frac{{dV}}{{dP}} = \frac{{\left( {nb - V} \right){V^{\rm{2}}}}}{{P{V^{\rm{2}}} + {n^{\rm{2}}}a\left( {{\rm{1}} + nbV - {V^{\rm{2}}}} \right)}}\\\frac{{dV}}{{dP}} = \frac{{\left( {\left( {\rm{1}} \right)\left( {{\rm{0}}{\rm{.04267}}} \right) - \left( {{\rm{10}}} \right)} \right){{\left( {{\rm{10}}} \right)}^{\rm{2}}}}}{{\left( {{\rm{2}}{\rm{.5}}} \right){{\left( {{\rm{10}}} \right)}^{\rm{2}}} + {{\left( {\rm{1}} \right)}^{\rm{2}}}\left( {{\rm{3}}{\rm{.592}}} \right)\left( {{\rm{1}} + \left( {\rm{1}} \right)\left( {{\rm{0}}{\rm{.04267}}} \right)\left( {{\rm{10}}} \right) - {{\left( {{\rm{10}}} \right)}^{\rm{2}}}} \right)}}\\\frac{{dV}}{{dP}} = \frac{{ - {\rm{995}}{\rm{.733}}}}{{{\rm{250}} + \left( {{\rm{3}}{\rm{.592}}} \right)\left( {{\rm{1}} + {\rm{0}}{\rm{.4267}} - {\rm{100}}} \right)}}\\\frac{{dV}}{{dP}} = {\rm{9}}{\rm{.5674}}L/atm\end{array}\)

Therefore, from the van der Waals equation \(\left( {P + \frac{{{n^{\rm{2}}}a}}{{{V^{\rm{2}}}}}} \right)\left( {V - nb} \right) = nRT\) by implicit differentiation the value \(\frac{{dV}}{{dP}} = \frac{{\left( {nb - V} \right){V^{\rm{2}}}}}{{P{V^{\rm{2}}} + {n^{\rm{2}}}a\left( {{\rm{1}} + nbV - {V^{\rm{2}}}} \right)}}\), and rate of change of volume with respect to pressure for \(n = {\rm{1}}\), \(V = {\rm{10L}}\), \(P = {\rm{2}}{\rm{.5atm}}\), \({\rm{a}} = {\rm{3}}{\rm{.592}}{{\rm{L}}^{\rm{2}}}{\rm{ - atm}}/{\rm{mol}}{{\rm{e}}^{\rm{2}}}\) and \(b = {\rm{0}}{\rm{.04267}}L/mole\) is \({\rm{9}}{\rm{.5674}}L/atm\).