A sample of gas enclosed in a vessel has a density of 0.90\text{ kg/m}^3 and exerts a pres… — JAMB UTME

A sample of gas enclosed in a vessel has a density of $0.90\text{ kg/m}^3$ and exerts a pressure of $3.0 \times 10^5\text{ N/m}^2$ on the walls of the vessel. Based on the kinetic theory of gases, what is the root-mean-square (r.m.s.) speed of the gas molecules in $\text{m/s}$ ?

Answer: 1000 m/s

Answer

The root-mean-square speed of the gas molecules is

1000\text{ m/s}

By applying the kinetic theory formula

P = \frac{1}{3} \rho v_{\text{rms}}^2

, rearranging gives

v_{\text{rms}} = \sqrt{\frac{3P}{\rho}}

. Substituting

P = 3.0 \times 10^5\text{ N/m}^2

and

\rho = 0.90\text{ kg/m}^3

results in

v_{\text{rms}} = \sqrt{\frac{9.0 \times 10^5}{0.90}} = \sqrt{1.0 \times 10^6} = 1000\text{ m/s}

Step-by-Step Solution

Identify the kinetic theory equation relating gas pressure, density, and microscopic molecular speed.

P = \frac{1}{3} \rho v_{\text{rms}}^2

According to the kinetic theory of gases, the macroscopic pressure exerted by gas molecules colliding with container walls is proportional to the gas density and the square of their r.m.s. speed.

Make

v_{\text{rms}}

the subject of the formula.

v_{\text{rms}} = \sqrt{\frac{3P}{\rho}}

Multiplying both sides by

3

and dividing by density

\rho

isolates

v_{\text{rms}}^2

, taking the square root yields

v_{\text{rms}}

Substitute the given numerical values into the equation.

v_{\text{rms}} = \sqrt{\frac{3 \times 3.0 \times 10^5}{0.90}} = \sqrt{1,000,000} = 1000\text{ m/s}

Performing the division yields

1.0 \times 10^6\text{ m}^2/\text{s}^2

, whose square root gives the speed in

\text{m/s}

Key Concept

Kinetic Theory Pressure Equation relating macroscopic pressure and density to microscopic root-mean-square velocity (

P = \frac{1}{3}\rho v_{\text{rms}}^2

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