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Chapter 9: Problem 1266

At what temperature is the kinetic energy of a gas molecule double that of its value at \(27^{\circ} \mathrm{C}\) (A) \(54^{\circ} \mathrm{C}\) (B) \(108^{\circ} \mathrm{C}\) (C) \(327^{\circ} \mathrm{C}\) (D) \(300^{\circ} \mathrm{C}\)

Short Answer

Expert verified
The temperature at which the kinetic energy of a gas molecule is double that of its value at 27°C is approximately 327°C. The correct answer is (C) \(327^{\circ} \mathrm{C}\).

Step by step solution

01

Understand the relationship between kinetic energy and temperature

A gas molecule's kinetic energy (represented by K) is directly proportional to its temperature (represented by T), which is given by the following formula: \[ K = \frac{3}{2} kT \] where k is the Boltzmann constant.
02

Set up an equation for the doubled kinetic energy

We are asked to find the temperature when the kinetic energy is double that at 27°C. Let's denote the initial temperature (27°C) as \(T_1\) and the final temperature (which we need to find) as \(T_2\). First, we must convert the given temperature to Kelvin (K) by adding 273.15: \[T_1 = 27^{\circ} C + 273.15 \Rightarrow T_1 = 300.15 K\] Since we know that the kinetic energy at the final temperature, let's call it \(K_2\), is double the kinetic energy at the initial temperature, \(K_1\), we can write that as: \[K_2 = 2 \times K_1\] Now substituting the formula of kinetic energy from step 1: \[\frac{3}{2} kT_2 = 2 \times \frac{3}{2} kT_1\]
03

Solve for the final temperature

Now, we can cancel out the constants in the equation, which leaves us with: \[T_2 = 2 \times T_1\] Substitute the value of \(T_1\) we found earlier: \[T_2 = 2 \times 300.15\] \[T_2 = 600.3 K\] To convert the temperature back to Celsius, subtract 273.15: \[T_2 = 600.3 - 273.15 = 327.15^{\circ} C\] Thus, the temperature at which the kinetic energy of a gas molecule is double that of its value at 27°C is approximately 327°C. So, the correct answer is: (C) \(327^{\circ} \mathrm{C}\)

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Most popular questions from this chapter

A diatomic gas molecule has translational, rotational and vibrational degrees of freedom. The $\left(\mathrm{C}_{\mathrm{p}} / \mathrm{C}_{\mathrm{V}}\right)$ is (A) \(1.29\) (B) \(1.33\) C) \(1.4\) (D) \(1.67\)

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 gas at \(27^{\circ} \mathrm{C}\) has a volume \(\mathrm{V}\) and pressure \(\mathrm{P}\). On heating its pressure is doubled and volume becomes three times. The resulting temperature of the gas will be (A) \(1527^{\circ} \mathrm{C}\) (B) \(600^{\circ} \mathrm{C}\) (C) \(162^{\circ} \mathrm{C}\) (D) \(1800^{\circ} \mathrm{C}\)

The mean kinetic energy of a gas at \(300 \mathrm{~K}\) is \(100 \mathrm{~J}\). mean energy of the gas at \(450 \mathrm{~K}\) is equal to (A) \(100 \mathrm{~J}\) (B) \(150 \mathrm{~J}\) (C) \(3000 \mathrm{~J}\) (D) \(450 \mathrm{~J}\)

A sample of gas is at \(0^{\circ} \mathrm{C}\). To what temperature it must be raised in order to double the rms speed of molecule. (A) \(270^{\circ} \mathrm{C}\) (B) \(819^{\circ} \mathrm{C}\) (C) \(100^{\circ} \mathrm{C}\) (D) \(1090^{\circ} \mathrm{C}\)
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