Kinetic Theory

For a gas, the fraction of molecular volume to total volume is proportional to the density compared with liquid water. From the chapter’s Example 12.1, for water vapour at $100^\circ\mathrm{C}$ and 1 atm this fraction was estimated as about $6\times10^{-4}$. Using the same method for oxygen at STP, the order is the same: the molecules occupy only a very small fraction of the total volume.

So the estimated fraction is **$6\times10^{-4}$**.

Use the ideal gas equation:

$$PV=\mu RT$$

For 1 mole at STP:
- $P=1\,\text{atm}=1.01\times10^5\,\text{Pa}$
- $T=273\,\text{K}$
- $\mu=1$ mol
- $R=8.31\,\text{J mol}^{-1}\text{K}^{-1}$

So,

$$V=\frac{RT}{P}$$

$$V=\frac{8.31\times273}{1.01\times10^5}\,\text{m}^3$$

$$V\approx\frac{2268}{1.01\times10^5}\,\text{m}^3\approx2.24\times10^{-2}\,\text{m}^3$$

Now convert to litres:

$$2.24\times10^{-2}\,\text{m}^3=22.4\,\text{L}$$

Hence, the molar volume at STP is 22.4 litres.

The plot of $PV/T$ versus $P$ for oxygen at two temperatures shows how a real gas departs from ideal-gas behaviour. At low pressure and high temperature, the curve approaches a constant value, because the gas behaves more nearly like an ideal gas. The two different curves correspond to two different temperatures, and the lower-pressure part is closer to the ideal-gas value.

So the figure is meant to show the departure of a real gas from ideal-gas behaviour and its approach to the ideal-gas line at low pressure.

The dotted plot represents the ideal-gas behaviour for comparison. It shows the value expected if the gas obeyed $PV=\mu RT$ exactly, i.e. if $PV/T$ remained constant.

So it signifies the ideal-gas curve.

For a fixed amount of gas, in the ideal region,

$$\frac{PV}{T}=\mu R$$

is a constant. In the plot, the curve at the higher temperature lies closer to ideal-gas behaviour and has the higher value of $PV/T$ at the same pressure region shown.

Thus the temperature corresponding to the upper curve is the larger one, so

$$\boxed{T_1>T_2}$$

At the point where the curves meet the **$y$-axis**, pressure is very small, so the gas behaves ideally. Hence

$$\frac{PV}{T}=\mu R$$

For $1.00\times10^{-3}\,\text{kg}$ of oxygen:
- molar mass of $O_2 = 32\,\text{g mol}^{-1}=0.032\,\text{kg mol}^{-1}$
- number of moles

$$\mu=\frac{m}{M_0}=\frac{1.00\times10^{-3}}{0.032}=3.125\times10^{-2}\,\text{mol}$$

Now

$$\frac{PV}{T}=\mu R=(3.125\times10^{-2})(8.31)$$

$$\frac{PV}{T}\approx2.60\times10^{-1}\,\text{J K}^{-1}$$

So the value at the intercept is **$2.6\times10^{-1}\,\text{J K}^{-1}$**.

At low pressure/high temperature,

$$\frac{PV}{T}=\mu R$$

So the value at the $y$-axis intercept depends on the number of moles, not directly on the gas.

For oxygen:

$$\mu_{O_2}=\frac{1.0\times10^{-3}}{0.032}=3.125\times10^{-2}\,\text{mol}$$

For hydrogen to give the same $PV/T$, it must have the same number of moles:

$$\mu_{H_2}=3.125\times10^{-2}\,\text{mol}$$

Mass of hydrogen needed:

$$m=\mu M_0=(3.125\times10^{-2})(2.02\times10^{-3})\,\text{kg}$$

$$m\approx6.31\times10^{-5}\,\text{kg}$$

So the required mass of hydrogen is about **$6.3\times10^{-5}\,\text{kg}$**. Since $1.0\times10^{-3}$ kg of hydrogen has a different number of moles, it would not give the same intercept value.