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Chapter 10: Problem 41

If the degrees of freedom of a gas molecule be \(f\), then the ratio of two specific heats \(C_{p} / C_{v}\) is given by (A) \(\frac{2}{f}+1\) (B) \(1-\frac{2}{f}\) (C) \(1+\frac{1}{f}\) (D) \(1-\frac{1}{f}\)

Short Answer

Expert verified
The relationship between the ratio of two specific heats, \(C_p / C_v\), and the degrees of freedom of a gas molecule (\(f\)) is given by: \[\frac{C_p}{C_v} = \frac{2}{f} + 1\].

Step by step solution

01

Relationship between the specific heats and the degrees of freedom of a gas molecule

To relate the specific heats and the degrees of freedom, we can refer to the Mayer's equation, which states: \[C_p - C_v = R\] where, \(C_p\) is the molar specific heat at constant pressure, \(C_v\) is the molar specific heat at constant volume, and \(R\) is the universal gas constant.
02

Relating the degrees of freedom with molar specific heats

Now, let's recall the expressions for molar specific heats in terms of degrees of freedom. For a gas molecule, the molar specific heat at constant volume, \(C_v\), is given by: \[C_v = \frac{f}{2} R\] and the molar specific heat at constant pressure, \(C_p\), is derived from Mayer's equation: \[C_p = C_v + R = \left(\frac{f}{2} + 1\right) R\].
03

Calculating the ratio of specific heats

Now that we have expressions for \(C_p\) and \(C_v\) in terms of the degrees of freedom, we can calculate their ratio: \[\frac{C_p}{C_v} = \frac{\left(\frac{f}{2} + 1\right) R}{\frac{f}{2}R}\].
04

Simplifying the expression

We can simplify the above expression further by canceling the universal gas constant, \(R\), in both the numerator and the denominator: \[\frac{C_p}{C_v} = \frac{\frac{f}{2} + 1}{\frac{f}{2}}\].
05

Identifying the correct answer

Now, let's compare the expression that we obtained with the given answer choices: (A) \(\frac{2}{f} + 1\) (B) \(1 - \frac{2}{f}\) (C) \(1 + \frac{1}{f}\) (D) \(1 - \frac{1}{f}\) None of the exact given choices match our derived expression. However, with a small amount of manipulation, we can make our expression look like option (A). Multiplying the numerator and denominator of our expression by 2, we get: \[\frac{C_p}{C_v} = \frac{f + 2}{f} = \frac{2}{f} + 1\]. This matches option (A), so the correct answer is: (A) \(\frac{2}{f} + 1\)

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Most popular questions from this chapter

The root mean square velocity of the gas molecules is \(300 \mathrm{~m} / \mathrm{s}\). What will be the root mean square speed of the molecules if the atomic weight is double and absolute temperature is halved? (A) \(300 \mathrm{~m} / \mathrm{s}\) (B) \(150 \mathrm{~m} / \mathrm{s}\) (C) \(600 \mathrm{~m} / \mathrm{s}\) (D) \(75 \mathrm{~m} / \mathrm{s}\)

A bimetallic strip is formed by two identical strips, one of copper and the other of brass. The coefficients of linear expansion of the two metals are \(\alpha_{C}\) and \(\alpha_{B}\). On heating, the temperature of the strip goes up by \(\Delta T\) and the strip bends to form an arc of radius of curvature \(R\). Then \(R\) is (A) proportional to \(\Delta T\). (B) inversely proportional to \(\Delta T\). (C) proportional to \(\left|\alpha_{B}-\alpha_{C}\right|\). (D) inversely proportional to \(\left|\alpha_{B}-\alpha_{C}\right|\).

On the Celsius scale, the absolute zero of temperature is at (A) \(0^{\circ} \mathrm{C}\) (B) \(-32^{\circ} \mathrm{C}\) (C) \(100^{\circ} \mathrm{C}\) (D) \(-273.15^{\circ} \mathrm{C}\)

At what temperature, the Fahrenheit and the Celsius scales will give numerically equal (but opposite in sign) values? (A) \(-40^{\circ} \mathrm{F}\) and \(40^{\circ} \mathrm{C}\) (B) \(11.43^{\circ} \mathrm{F}\) and \(-11.43^{\circ} \mathrm{C}\) (C) \(-11.43^{\circ} \mathrm{F}\) and \(+11.43^{\circ} \mathrm{C}\) (D) \(+40^{\circ} \mathrm{F}\) and \(-40^{\circ} \mathrm{C}\)

At what temperature will the resistance of a copper wire become three times its value at \(0^{\circ} \mathrm{C}\) (Temperature coefficient of resistance for copper \(=4 \times 10^{-3} /{ }^{\circ} \mathrm{C}\) ) (A) \(400^{\circ} \mathrm{C}\) (B) \(450^{\circ} \mathrm{C}\) (C) \(500^{\circ} \mathrm{C}\) (D) \(550^{\circ} \mathrm{C}\)
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