Question

\(a)\)The gas law for a fixed mass m for an ideal gas at absolute temperature T, pressure P and volume V is PV=mRT, where R is the gas constant. Show that:

\(\frac{{\partial P}}{{\partial V}}\frac{{\partial V}}{{\partial T}}\frac{{\partial T}}{{\partial P}} = 1\)

\(b)\)For the ideal gas at part \((a)\) show that:

\(T\frac{{\partial P}}{{\partial T}}\frac{{\partial V}}{{\partial T}} = mR\)

Short answer

The general equation, commonly known as the ideal gas law, is the state equation of a hypothetical ideal gas. Although it has significant drawbacks, it is the good approximation of the behavior of various gases under many conditions.

Step-by-step solution

Given data

Given: \(PV = mRT\)

\(P = \frac{{mRT}}{V}\)

\(\frac{{\partial P}}{{\partial V}} = - \frac{{ - mRT}}{{{V^2}}}\)

Where T is constant.

To find: \(\frac{{\partial V}}{{\partial T}}\)

\(PV = mRT\)

\(V = \frac{{mRT}}{P}\)

\(\frac{{\partial V}}{{\partial T}} = \frac{{mR}}{P}\)

Where P is constant.

To find: \(\frac{{\partial T}}{{\partial P}}\)

\(PV = mRT\)

\(T = \frac{{PV}}{{mR}}\)

\(\frac{{\partial T}}{{\partial P}} = \frac{V}{{mR}}\)

Where V is constant.

Substituting the values

Plug these values into \(\frac{{\partial P}}{{\partial V}}\frac{{\partial V}}{{\partial T}}\frac{{\partial T}}{{\partial P}}\)

\(\frac{{\partial P}}{{\partial V}}\frac{{\partial V}}{{\partial T}}\frac{{\partial T}}{{\partial P}}\)\( = - \frac{{mRT}}{{{V^2}}} \times \frac{{mR}}{P} \times \frac{V}{{mR}}\)

\( = - \frac{{mRT}}{P}\)

But \(PV = mRT\), it follows that \(\frac{{mRT}}{{PV}} = 1\)

Thus:

\(\frac{{\partial P}}{{\partial V}}\frac{{\partial V}}{{\partial T}}\frac{{\partial T}}{{\partial P}} = - \frac{{mRT}}{{PV}}\)

\( = - 1\)

Hence, \(\frac{{\partial P}}{{\partial V}}\frac{{\partial V}}{{\partial T}}\frac{{\partial T}}{{\partial P}} = - 1\).