Question

The molecules of a given mass of a gas have r.m.s velocity of $200 \, ms^{-1}$ at $27^{\circ} C$ and $1.0 \times 10^5 \, Nm^{-2}$ pressure. When the temperature and pressure of the gas are respectively, $127^{\circ} C$ and $0.05 \times 10^5 \, Nm^{-2}$, the r.m.s. velocity of its molecules in $ms^{-1}$ is

• A. $\frac{400}{\sqrt{3}}$
• B. $\frac{100 \sqrt{2}}{3}$
• C. $\frac{100}{3}$
• D. $100 \sqrt{2}$

Answer 1

The Correct Option is A

 To find the root mean square (r.m.s) velocity of the gas molecules at the new temperature and pressure conditions, we can use the formula for r.m.s velocity of a gas, which is given by:

\(v_{\text{rms}} = \sqrt{\frac{3kT}{m}}\)

where:

• \(v_{\text{rms}}\) is the r.m.s velocity.
• \(k\) is the Boltzmann constant.
• \(T\) is the temperature in Kelvin.
• \(m\) is the mass of the gas molecule.

 

However, for the purposes of this problem, it's more convenient to use the relationship with pressure and temperature given by:

\(v_{\text{rms}} \propto \sqrt{\frac{P}{T}}\)

where:

• \(P\) is the pressure of the gas.
• \(T\) is the absolute temperature in Kelvin.

 

Given:

• Initial r.m.s velocity \((v_{\text{rms1}}) = 200 \, \text{ms}^{-1}\)
• Initial temperature \(= 27^{\circ} C = 300 \, \text{K}\)
• Initial pressure \(= 1.0 \times 10^5 \, \text{Nm}^{-2}\)
• Final temperature \(= 127^{\circ} C = 400 \, \text{K}\)
• Final pressure \(= 0.05 \times 10^5 \, \text{Nm}^{-2}\)

 

First, calculate the ratio of the r.m.s velocities:

\(\frac{v_{\text{rms2}}}{v_{\text{rms1}}} = \sqrt{\frac{P_2}{P_1} \cdot \frac{T_1}{T_2}}\)

Substitute the given values:

\(\frac{v_{\text{rms2}}}{200} = \sqrt{\frac{0.05 \times 10^5}{1.0 \times 10^5} \cdot \frac{300}{400}}\)

This simplifies to:

\(\frac{v_{\text{rms2}}}{200} = \sqrt{\frac{1}{20} \cdot \frac{3}{4}}\)

\(\frac{v_{\text{rms2}}}{200} = \sqrt{\frac{3}{80}}\)

Calculate the final r.m.s velocity:

\(v_{\text{rms2}} = 200 \times \sqrt{\frac{3}{80}}\)

\(v_{\text{rms2}} = 200 \times \frac{\sqrt{3}}{\sqrt{80}}\)

\(v_{\text{rms2}} = 200 \times \frac{\sqrt{3}}{2\sqrt{20}}\)

\(v_{\text{rms2}} = 200 \times \frac{\sqrt{3}}{10\sqrt{2}}\)

\(v_{\text{rms2}} = \frac{200\sqrt{3}}{20\sqrt{2}}\)

\(v_{\text{rms2}} = \frac{20\sqrt{3}}{2\sqrt{2}}\)

\(v_{\text{rms2}} = \frac{400}{\sqrt{3}}\)

Therefore, the r.m.s velocity of the molecules at the new temperature and pressure is \(\frac{400}{\sqrt{3}} \, \text{ms}^{-1}\), which corresponds to the correct answer given.