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Chapter 9: Problem 1317

The degrees of freedom for triatomic gas \(1 \mathrm{~s}\) (At room temperature) (A) 8 (B) 6 (C) 4 (D) 2

Short Answer

Expert verified
The correct answer is (B) 6 degrees of freedom. At room temperature, a triatomic gas molecule has 3 translational degrees of freedom and for a nonlinear triatomic molecule, it has 3 rotational degrees of freedom. Vibrational degrees of freedom are not considered, as they require a high amount of energy.

Step by step solution

01

Understand Degrees of Freedom

Degrees of freedom represent the number of independent ways a system can store energy, which can be categorized into three main modes - translational, rotational, and vibrational. For a gas molecule, the total degrees of freedom are the sum of all the possible modes.
02

Translational Degrees of Freedom

In a 3D space, every molecule can move in three independent directions - along the x-axis, y-axis, and z-axis. So, a gas molecule has 3 translational degrees of freedom.
03

Rotational Degrees of Freedom

For a triatomic gas, there are two types of molecules: linear and nonlinear. A linear triatomic molecule has a single axis passing through all three atoms, which restricts its rotation to only two axes. So, it has 2 rotational degrees of freedom. On the other hand, a nonlinear triatomic molecule (such as H2O) has 3 rotational degrees of freedom.
04

Vibrational Degrees of Freedom

At room temperature, vibrational degrees of freedom are usually not activated for most gases, as they require a high amount of energy. Therefore, at room temperature, we will not consider vibrational degrees of freedom.
05

Calculate Total Degrees of freedom

Now we can sum up the degrees of freedom for each mode to find the total degrees of freedom. For a linear triatomic molecule: 3 (translational) + 2 (rotational) + 0 (vibrational) = 5 For a nonlinear triatomic molecule: 3 (translational) + 3 (rotational) + 0 (vibrational) = 6
06

Answer

Comparing our results with the given multiple-choice options, we can see that the most appropriate answer is 6 degrees of freedom, which corresponds to option (B). So the correct answer is (B) 6.

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Most popular questions from this chapter

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) \(6 \mathrm{P}\)

The temperature of an ideal gas is increased from \(27^{\circ} \mathrm{C}\) to \(127^{\circ} \mathrm{C}\), then percentage increase in \(\mathrm{v}_{\mathrm{rms}}\) is (A) \(33 \%\) (B) \(11 \%\) (C) \(15.5 \%\) (D) \(37 \%\)

When temperature of an ideal gas is increased from \(27^{\circ} \mathrm{C}\) to \(227^{\circ} \mathrm{C}\), its rms speed changed from \(400 \mathrm{~ms}^{-1}\) to \(\mathrm{V}_{\mathrm{s}}\). The \(\mathrm{V}_{\mathrm{s}}\) is (A) \(516 \mathrm{~ms}^{-1}\) (B) \(746 \mathrm{~ms}^{-1}\) (C) \(310 \mathrm{~ms}^{-1}\) (D) \(450 \mathrm{~ms}^{-1}\)

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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) \(450 \mathrm{~J}\)
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