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Chapter 9: Problem 1294

The rms. speed of the molecules of a gas in a vessel is $400 \mathrm{~ms}^{-1}$. If half of the gas leaks out, at constant temperature, the r.m.s speed of the remaining molecules will be (A) \(800 \mathrm{~ms}^{-1}\) (B) \(200 \mathrm{~ms}^{-1}\) (C) \(400 \sqrt{2} \mathrm{~ms}^{-1}\) (D) \(400 \mathrm{~ms}^{-1}\)

Short Answer

Expert verified
The new rms speed of the remaining molecules at constant temperature will be: (D) \(400 \, \mathrm{ms}^{-1}\)

Step by step solution

01

Identifying the relevant formula for rms speed

To find the rms speed of the remaining molecules, we need to understand the dependence of rms speed on the mass of gas and the temperature. The formula for rms speed (v_rms) is given by: \( v_{rms} = \sqrt{\frac{3kT}{m}} \) where - \(v_{rms}\) is the root mean square speed of gas molecules - \(k\) is Boltzmann's constant (\(1.38 \times 10^{-23} \, \mathrm{J \, K^{-1}}\)) - \(T\) is the temperature in Kelvin - \(m\) is the mass of a single molecule Since temperature is constant in this case, let's rewrite the formula as: \(v_{rms} = c \sqrt{\frac{1}{m}}\) where \(c = \sqrt{3kT}\) is a constant.
02

Determining the change in the mass of a single molecule

When half of the gas leaks out, the number of molecules in the vessel reduces to half. However, the mass of an individual molecule (\(m\)) remains unchanged. This is important because the rms speed depends on the mass of individual molecules.
03

Calculating the new rms speed

Since the mass of individual molecules remains unchanged and the temperature is constant, we can conclude that the new rms speed will also remain unchanged. From our initial speed formula: \(v_{rms} = c \sqrt{\frac{1}{m}}\) we find that the new rms speed will remain at 400 m/s because the mass and temperature are constant. So the correct option is: (D) \(400 \, \mathrm{ms}^{-1}\)

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Most popular questions from this chapter

The relation between two specific heats of a gas is (A) $\mathrm{C}_{\mathrm{V}}-\mathrm{C}_{\mathrm{P}}=(\mathrm{R} / \mathrm{J})$ (B) \(\mathrm{C}_{\mathrm{P}}-\mathrm{C}_{\mathrm{V}}=\mathrm{J}\) (C) $\mathrm{C}_{\mathrm{P}}-\mathrm{C}_{\mathrm{V}}=(\mathrm{R} / \mathrm{J})$ (D) \(\mathrm{C}_{\mathrm{V}}-\mathrm{C}_{\mathrm{P}}=\mathrm{J}\)

The temperature of a gas at pressure \(\mathrm{P}\) and volume \(\mathrm{V}\) is \(27^{\circ} \mathrm{C}\) Keeping its volume constant if its temperature is raised to \(927^{\circ} \mathrm{C}\), then its pressure will be (A) \(3 \mathrm{P}\) (B) \(2 \mathrm{P}\) (C) \(4 \mathrm{P}\) (D) \(6 \mathrm{P}\)

A gas at \(27^{\circ} \mathrm{C}\) temperature and 30 atmospheric pressure $1 \mathrm{~s}$ allowed to expand to the atmospheric pressure if the volume becomes two times its initial volume, then the final temperature becomes (B) \(-173^{\circ} \mathrm{C}\) (A) \(273^{\circ} \mathrm{C}\) (C) \(173^{\circ} \mathrm{C}\) (D) \(100^{\circ} \mathrm{C}\)

A cylinder contains \(10 \mathrm{~kg}\) of gas at pressure of $10^{\prime} \mathrm{N} / \mathrm{m}^{2}$. The quantity of gas taken out of the cylinder, if final pressure is \(2.5 \times 10^{6} \mathrm{Nm}^{-2}\). will be (temperature of gas is constant) (A) \(5.2 \mathrm{~kg}\) (B) \(3.7 \mathrm{~kg}\) (C) \(7.5 \mathrm{~kg}\) (D) \(1 \mathrm{~kg}\)

If three molecules have velocities \(0.5,1\) and 2 the ratio of rms speed and average speed is (The velocities are in \(\mathrm{km} / \mathrm{s}\) ) (A) \(0.134\) (B) \(1.34\) (C) \(1.134\) (D) \(13.4\)
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