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Chapter 9: Problem 1264

A sealed container with negligible co-efficient of volumetric expansion contains helium (a monatomic gas) when it is heated from \(200 \mathrm{~K}\) to \(400 \mathrm{~K}\), the average \(\mathrm{K} . \mathrm{E}\). of helium atom is (A) Halved (B) Doubled (C) Unchanged (D) Increased by factor \(\sqrt{2}\)

Short Answer

Expert verified
The average kinetic energy of a monatomic gas is given by the formula \(KE = \frac{3}{2}kT\). When the temperature is raised from 200K to 400K, the ratio of the final average kinetic energy to the initial average kinetic energy is \(\frac{KE_2}{KE_1} = \frac{400}{200} = 2\). This means that the average kinetic energy of the helium atoms has doubled during the process, corresponding to option (B).

Step by step solution

01

Recall the formula for the average kinetic energy of a monatomic gas

The average kinetic energy (KE) of a monatomic gas is given by the formula: \[KE = \frac{3}{2}kT\] where \(k\) is the Boltzmann constant and \(T\) is the temperature in Kelvin.
02

Calculate the initial average kinetic energy

At the initial temperature of 200K, the average kinetic energy of the helium atoms is: \[KE_1 = \frac{3}{2}k(200)\]
03

Calculate the final average kinetic energy

When the temperature is raised to 400K, the average kinetic energy of the helium atoms becomes: \[KE_2 = \frac{3}{2}k(400)\]
04

Determine the change in average kinetic energy

Now, let's divide the final average kinetic energy by the initial average kinetic energy to find the change in average kinetic energy: \[\frac{KE_2}{KE_1} = \frac{\frac{3}{2}k(400)}{\frac{3}{2}k(200)}\] The \( \frac{3}{2}k \) factors cancel out, leaving: \[\frac{KE_2}{KE_1} = \frac{400}{200} = 2\]
05

Analyze the results

The ratio of the final average kinetic energy to the initial average kinetic energy is 2. This means that the average kinetic energy of the helium atoms has doubled during the process, which corresponds to option (B).

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Most popular questions from this chapter

Air is pumped into an automobile tube upto a pressure of \(200 \mathrm{kPa}\) in the morning when the air temperature is \(22^{\circ} \mathrm{C}\). During the day, temperature rises to \(42^{\circ} \mathrm{C}\) and the tube expands by $2 \%$ The pressure of the air in the tube at this temperature will be approximately. (A) \(209 \mathrm{kPa}\) (B) \(206 \mathrm{kPa}\) (C) \(200 \mathrm{kPa}\) (D) \(212 \mathrm{kPa}\)

Air is filled in a bottle at atmospheric pressure and it is corked at \(35^{\circ} \mathrm{C}\), If the cork can come out at 3 atmospheric pressure then upto what temperature should the bottle be heated in order to remove the cork. (A) \(325.5^{\circ} \mathrm{C}\) (B) \(651^{\circ} \mathrm{C}\) (C) \(851^{\circ} \mathrm{C}\) (D) None of these

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The root mean square velocity of the molecules in a sample of helium is $(5 / 7)^{\text {th }}$ that of the molecules in a sample of hydrogen. If the temperature of hydrogen sample is \(0^{\circ} \mathrm{C}\), then the temperature of the helium sample is about (A) \(273^{\circ} \mathrm{C}\) (B) \(0^{\circ} \mathrm{C}\) (C) \(0^{\circ} \mathrm{K}\) (D) \(100^{\circ} \mathrm{C}\)

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