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Chapter 9: Problem 1320

The value of \(\mathrm{C}_{\mathrm{v}}\) for one mole of neon gas is (A) \((3 / 2) \mathrm{R}\) (B) \((7 / 2) \mathrm{R}\) (C) \((1 / 2) \mathrm{R}\) (D) \((5 / 2) \mathrm{R}\)

Short Answer

Expert verified
The value of \(C_v\) for one mole of neon gas is (A) \( (3 / 2) \cdot R \).

Step by step solution

01

Determine the Degrees of Freedom for a Monatomic Gas

Neon is a monatomic gas, meaning it is composed of single atoms with no molecular bonds. The degrees of freedom for a monatomic gas are related to the number of independent ways in which an atom can store its energy. For a monatomic gas, there are only three translational degrees of freedom, corresponding to the motion in the x, y, and z directions in Cartesian space.
02

Calculate the Molar Heat Capacity (Cv) Using Equipartition Theorem

The equipartition theorem is used to calculate the heat capacity at constant volume (Cv) for different types of gases. For a monatomic gas like neon with only three translational degrees of freedom, the equipartition theorem expression takes the form: Cv = \( \frac{f}{2} \cdot R \) Where: Cv = Molar heat capacity at constant volume f = degrees of freedom (3 for neon) R = Gas constant (8.314 J/mol·K)
03

Plug in the Values and Calculate Cv

Now, we can plug the values into the equation to calculate the molar heat capacity (Cv) for one mole of neon gas: Cv = \( \frac{3}{2} \cdot R \) The answer choice corresponding to this value for Cv is (A), therefore: The value of Cv for one mole of neon gas is (A) \( (3 / 2) \cdot R \).

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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) \(1.67\)

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 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) \(37 \%\)

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) \(1090^{\circ} \mathrm{C}\)

At \(0 \mathrm{~K}\) which of the following properties of a gas will be zero (A) Kinetic energy (B) Density (C) Potential energy (D) Vibrational energy
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