Reading a pV against kT Graph

Paper 4

Physics9702

The molecular gas equation becomes a straight-line graph. Its shape tests ideal behaviour, and its gradient counts the molecules.

Swipe left or right to see the whole diagram →

For a fixed amount of gas, pV against kT is a straight line through the origin. That straight line through the origin is what tells you the gas is ideal, and the gradient of the line is the number N of molecules.

Compare the molecular equation of state with the equation of a straight line.
$pV=N(kT)\qquad y=mx$
Here the vertical quantity is $pV$ , the horizontal quantity is $kT$ , and the gradient is $N$ .
A straight line through the origin tells you that the gas is ideal. That is the whole one-mark answer (9702/41/O/N/25 Q4(b)(i)).
The gradient of that line is $N$ , the number of molecules.

Worked example

Exam version: get N and n from the graph

Use the point on the graph shown below. Find the number of molecules $N$ , then the amount of gas $n$ . (9702/41/O/N/25 Q4(b))

pV=270\ \mathrm{J}

kT=8.0\times10^{-21}\ \mathrm{J}

Use the gradient, or divide the vertical value by the horizontal value.
$N=\dfrac{pV}{kT}$
$N=\dfrac{270}{8.0\times10^{-21}}$
$N=3.4\times10^{22}$
Convert the molecule count to moles.
$n=\dfrac{N}{N_A}$
$n=\dfrac{3.4\times10^{22}}{6.02\times10^{23}}$
$n=0.056\ \mathrm{mol}$

Answer

N=3.4\times10^{22}

Answer

n=0.056\ \mathrm{mol}

Reading a gradient from this graph is the same calculation as using the molecular equation of state. The graph gives the pair of values.

Match the axes to the molecular gas equation. A straight line through the origin supports ideal behaviour, and its gradient is the molecule count $N$ .

Reading a pV against kT Graph

The molecular gas equation becomes a straight-line graph. Its shape tests ideal behaviour, and its gradient counts the molecules.