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Chapter 9: Problem 1318

If the degrees of freedom of a gas are \(\mathrm{f}\), then the ratio of two specific heats $\left(\mathrm{C}_{\mathrm{p}} / \mathrm{C}_{\mathrm{v}}\right)$ is given by (A) \(1-(2 / \mathrm{f})\) (B) \(1+(1 / f)\) (C) \((2 / \mathrm{f})+1\) (D) \(1+(3 / \mathrm{f})\)

Short Answer

Expert verified
The correct answer is (C) \((2 / \mathrm{f})+1\), as it matches with the simplified expression we found for the ratio of specific heats, which is \(\frac{f + 2}{f}\).

Step by step solution

01

Recall the relation between specific heats and degrees of freedom

The relation between the specific heats (Cp and Cv) and the degrees of freedom (f) can be expressed as: \(C_p = C_v + R\) Where R is the gas constant, Cp is the heat capacity at constant pressure, and Cv is the heat capacity at constant volume. We also know that: \(C_v = \frac{f}{2} R\)
02

Solve for Cp

We have the relations: \(C_p = C_v + R\) and \(C_v = \frac{f}{2} R\) We can solve for Cp by plugging the expression for Cv in the first equation: \(C_p = \frac{f}{2} R + R\)
03

Find the ratio of specific heats

We are interested in finding the ratio of specific heats: \(\frac{C_p}{C_v}\) Using the expressions for Cp and Cv, we can rewrite the ratio as: \(\frac{C_p}{C_v} = \frac{\frac{f}{2} R + R}{\frac{f}{2} R}\) Divide both the numerator and denominator by R to simplify the ratio: \(\frac{C_p}{C_v} = \frac{\frac{f}{2} + 1}{\frac{f}{2}}\) Now, multiply the numerator and denominator by 2 to remove the fractions: \(\frac{C_p}{C_v} = \frac{f + 2}{f}\)
04

Compare the result with the given options

We have found that the ratio of specific heats is \(\frac{f + 2}{f}\). Comparing this expression with the given options, we can see that it matches with option (C), which is: (C) \((2 / \mathrm{f})+1\) So, the correct answer is (C) \((2 / \mathrm{f})+1\).

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Most popular questions from this chapter

At room temperature \(\left(27^{\circ} \mathrm{C}\right)\), the rms speed of the molecules of certain diatomic gas is found to be $1930 \mathrm{~m} / \mathrm{s}$. The gas is (A) \(\mathrm{O}_{2}\) (B) \(\mathrm{Cl}_{2}\) (C) \(\mathrm{H}_{2}\) (D) \(\mathrm{F}_{2}\)

\(2 \mathrm{~g}\) of \(\mathrm{O}_{2}\) gas is taken at \(27^{\circ} \mathrm{C}\) and pressure \(76 \mathrm{~mm} \mathrm{Hg}\). Find out volume of gas (ln liter) (A) \(3.08\) (B) \(44.2\) (C) \(2.05\) (D) \(2.44\)

At a given temperature the root mean square velocities of Oxygen and hydrogen molecules are in the ratio (A) \(1: 4\) (B) \(1: 16\) (C) \(16: 1\) (D) \(4: 1\)

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 product of the pressure and volume of an ideal gas \(1 \mathrm{~s}\) (A) A constant (B) Directly proportional to its temperature. (C) Inversely proportional to its temperature. (D) Approx. equal to the universal gas constant.
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