Question

Compute the density (in units of m3

) of an ideal gas under the following

conditions:

(a) At 0 C and 760 Torr pressure (1 Torr ¼ 1 mmHg). This is called the

Loschmidt number.

(b) In a vacuum of 103

Torr at room temperature (20 C). This number is a

useful one for the experimentalist to know by heart (103

Torr ¼ 1 μ).

Answer 1

According to ideal gas law:

$$n=\frac{p}{k_BT}.$$

(a) In this scenario, at 760 torr and 0°C:

$$n=\frac{760·133.322}{(1.38·10^{-23})273.15}=2.687·10^{25}\text{ m}^{-3}.$$

(b) At room temperature:

$$n=\frac{(10^{-3})·133.322}{(1.38·10^{-23})(273.15+20)}=3.294·10^{19}\text{ m}^{-3}.$$