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Chapter 9: Problem 1319

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) \(1.67\)

Short Answer

Expert verified
A diatomic gas molecule has a total of 6 degrees of freedom: 3 translational, 2 rotational, and 1 vibrational. Using the equipartition theorem, we can calculate the specific heat capacities at constant volume (Cv) and constant pressure (Cp) as \(C_v = 3R\) and \(C_p = 4R\), respectively. Then, we can find the ratio \(\left(\dfrac{C_p}{C_v}\right) = \dfrac{4R}{3R}\), which simplifies to \(\dfrac{4}{3} = 1.33\). Thus, the correct answer is (B) 1.33.

Step by step solution

01

Determine degrees of freedom

A diatomic molecule has 3 translational, 2 rotational, and 1 vibrational degree of freedom. So in total, there are 3+2+1= 6 degrees of freedom.
02

Calculate specific heat capacities

For a diatomic gas molecule, use equipartition theorem to find the specific heat capacities. The specific heat capacity at constant volume, Cv, is given by: \[C_v = \dfrac{f}{2} R\] Where f is the total degrees of freedom, and R is the gas constant. Since we have calculated 6 degrees of freedom, we get: \[C_v = \dfrac{6}{2} R = 3 R\] The specific heat capacity at constant pressure, Cp, is given by: \[C_p = C_v + R\] \[C_p = 3R + R = 4R\]
03

Calculate the ratio Cp/Cv

Now that we have Cp and Cv, we can calculate the ratio: \[\left(\dfrac{C_p}{C_v}\right) = \dfrac{4R}{3R}\]
04

Simplify the ratio

Simplify the ratio by canceling the common factor R: \[\left(\dfrac{C_p}{C_v}\right) = \dfrac{4}{3}\]
05

Convert ratio to decimal form

Convert 4/3 to decimal form: \[\left(\dfrac{C_p}{C_v}\right) = 1.33\] Since the ratio is 1.33, the correct answer is (B) 1.33.

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Most popular questions from this chapter

Under constant temperature, graph between \(\mathrm{p}\) and \((1 / \mathrm{V})\) is (A) Hyperbola (B) Circle (C) Parabola (D) Straight line

\- decrease the volume of gas by \(\supset \%\) at constant temperature the pressure should be (A) Increased by \(5.26 \%\) (B) Decreased by \(5.26 \%\) (C) Decreased by \(11 \%\) (D) Increased by \(11 \%\)

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 volume of a gas at pressure \(21 \times 10^{4} \mathrm{Nm}^{-2}\) and temperature \(27^{\circ} \mathrm{C}\) is 83 Liters. If $\mathrm{R}=8.3 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$. Then the quantity of gas in \(\mathrm{g}\) -mole will be (A) 42 (B) 7 (C) 14 (D) 15

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) \(37 \%\)
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