Answer to Question #125049 in Molecular Physics | Thermodynamics for NAna

Question

The density of a gas of relative molecular mass 28 at a certain temperature is 0.90 kgm-3.

The root mean square speed of the gas molecules at that temperature is 602 ms-1. Assuming that the rate of diffusion of a gas is inversely proportional to the square root of its density, calculate the density of the gas standard temperature and pressure if its root mean square speed at is 490 ms-1.

Answer 1

Given:

$M = 28\frac{g}{mol}\\ \overline{v_0} = 602\frac{m}{s}\\ \rho_0 = 0.9\frac{kg}{m^3}\\ \overline{v_1} = 490\frac{m}{s}\\ T_1 = 273K\\ p_1 = 10^5Pa\\ r \propto \frac{1}{\sqrt\rho}$

Find: $\rho_1$

Solution:

Let's use the kinetic gas equation:

$p=\dfrac{1}{3}m_0n\overline{v^2}=\dfrac{1}{3}\dfrac{m_0N_A\nu}{V}\overline{v^2}=\dfrac{1}{3}\dfrac{M\nu}{V}\overline{v^2}=\dfrac{1}{3}\dfrac{m}{V}\overline{v^2}=\dfrac{1}{3}\rho\overline{v^2}\\ \rho=\dfrac{\ 3p\ }{\overline{v^2}} \\ \rho_1 = \dfrac{\ 3p_1\ }{\overline{v_1^2}} = \dfrac{3\times10^5Pa}{(490\frac{m}{s})^2} \approx 1.25\frac{kg}{m^3}$

Answer: $\rho=1.25\frac{kg}{m^3}$