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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 pressure is exerted by the gas on the walls of the container because (A) It sticks with the walls (B) It is accelerated towards the walls (C) It loses kinetic energy (D) On collision with the walls there is a change in momentum

The pressure and temperature of an ideal gas in a closed vessel are $720 \mathrm{kPa}\( and \)40^{\circ} \mathrm{C}\( respectively. If \)(1 / 4)^{\text {th }}$ of the gas is released from the vessel and the temperature of the remaining gas is raised to \(353^{\circ} \mathrm{C}\), final pressure of the gas is (A) \(1440 \mathrm{kPa}\) (B) \(540 \mathrm{kPa}\) (C) \(1080 \mathrm{kPa}\) (D) \(720 \mathrm{kPa}\)

The molecules of a given mass of a gas have a rms velocity of $200 \mathrm{~m} / \mathrm{s}\( at \)27^{\circ} \mathrm{C}\( and \)1.0 \times 10^{5} \mathrm{Nm}^{-2}\( pressure when the temperature is \)127^{\circ} \mathrm{C}$ and pressure is \(0.5 \times 10^{5} \mathrm{Nm}^{-2}\), the rms velocity in \(\mathrm{m} / \mathrm{s}\) will be (A) \(100 \sqrt{2}\) (B) \([(100 \sqrt{2}) / 3]\) (C) \([(400) / \sqrt{3}]\) (D) None of these

For a gas, the rms speed at \(800 \mathrm{~K}\) is (A) Four times the value at \(200 \mathrm{~K}\) (B) Twice the value at \(200 \mathrm{~K}\) (C) Half the value at \(200 \mathrm{~K}\) (D) same as at \(200 \mathrm{~K}\)

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) \(\mathrm{C}_{\mathrm{V}}-\mathrm{C}_{\mathrm{P}}=\mathrm{J}\)
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