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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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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Specific Heat Capacities
Specific heat capacities are essential properties that help us understand how a substance reacts to temperature changes. There are two primary specific heats associated with gases: specific heat at constant volume ( C_v ) and specific heat at constant pressure ( C_p ).

Let's delve into what they mean:
  • C_v: This is the amount of heat required to raise the temperature of 1 mole of a gas by 1 degree Celsius when the volume is kept constant. It's directly related to the degrees of freedom ( f ) of the gas molecule.

  • C_p: In contrast, this is the heat needed to raise the temperature of 1 mole of the gas by 1 degree Celsius while maintaining constant pressure.

These specific heats are connected to the internal energy and work done by the gas. Understanding these two helps us to grasp how gases absorb and dissipate energy.
Mayer's Equation
Mayer's equation provides a vital bridge between the specific heats of gases at constant volume and constant pressure. It is given by:

\[C_p - C_v = R\]

where R is the universal gas constant. This equation shows us the difference between C_p and C_v. The difference arises because under constant pressure conditions, a gas needs to do work to expand while gaining heat.

Using Mayer's equation, we can express the specific heats in terms of the degrees of freedom, \(f\).
  • At constant volume: \(C_v = \frac{f}{2} R\)

  • At constant pressure: \(C_p = \left( \frac{f}{2} + 1 \right) R\)

Thus, specific heats are not standalone values but are intricately connected through Mayer's equation, helping us to better analyze any given thermodynamic process.
Universal Gas Constant
The universal gas constant, R, is a fundamental constant pivotal in thermodynamics and gas laws. Its value is approximately 8.314 J/(mol路K), and it appears in various equations involving ideal gases, including Mayer's equation and the ideal gas law.

Some important characteristics include:
  • Role in Equations: R is a unifying constant that links pressure, volume, and temperature in the equations for gases. It helps us find relations between specific heats and degrees of freedom.

  • Connection to Degrees of Freedom: Through the relationship \(C_v = \frac{f}{2} R\), R helps us express specific heats in terms of molecular structure.

Its importance transcends simple calculations, as R provides a backbone for theoretical and practical understanding of gas behavior under different conditions.

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Most popular questions from this chapter

Water flows at the rate of \(0.1500 \mathrm{~kg} / \mathrm{min}\) through a tube and is heated by a heater dissipating \(25.2 \mathrm{~W}\). The inflow and outflow water temperatures are \(15.2^{\circ} \mathrm{C}\) and \(17.4^{\circ} \mathrm{C}\), respectively. When the rate of flow is increased to \(0.2318 \mathrm{~kg} /\) min and the rate of heating to \(37.8 \mathrm{~W}\), the inflow and outflow temperatures are unaltered. Find the rate of loss of heat from the tube

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 \(-40^{\circ} \mathrm{C}\)

A pendulum clock loses 12 s a day if the temperature is \(40^{\circ} \mathrm{C}\) and gains \(4 \mathrm{~s}\) day if the temperature is \(20^{\circ} \mathrm{C}\). The temperature at which the clock will show correct time, and the co-efficient of linear expansion \((\alpha)\) of the metal of the pendulum shaft are respectively. [2016] (A) \(60^{\circ} \mathrm{C} ; \alpha=1.85 \times 10^{-4} /{ }^{\circ} \mathrm{C}\) (B) \(30^{\circ} \mathrm{C} ; \alpha=1.85 \times 10^{-3} /{ }^{\circ} \mathrm{C}\) (C) \(55^{\circ} \mathrm{C} ; \alpha=1.85 \times 10^{-2} /{ }^{\circ} \mathrm{C}\) (D) \(25^{\circ} \mathrm{C} ; \alpha=1.85 \times 10^{-5} /{ }^{\circ} \mathrm{C}\)

A vessel of volume \(V\) contains a mixture of 1 mole of hydrogen and 1 mole oxygen (both considered as ideal). Let \(f_{1}(v) d v\) denote the fraction of molecules with speed between \(v\) and \((v+d v)\) with \(f_{2}(v) d v\), similarly for oxygen. Then, (A) \(f_{1}(v)+f_{2}(v)=f(v)\) obeys the Maxwell's distribution law. (B) \(f_{1}(v), f_{2}(v)\) will obey the Maxwell's distribution law separately. (C) neither \(f_{1}(v)\) nor \(f_{2}(v)\) will obey Maxwell's distribution law. (D) \(f_{2}(v)\) and \(f_{1}(v)\) will be the same.

A copper plate of length \(1 \mathrm{~m}\) is riveted to two steel plates of same length and same cross-section area at \(0^{\circ} \mathrm{C}\). Calculate tension (in kilo newton) generated in copper plate when heated to \(20^{\circ} \mathrm{C}\). \(Y_{\text {copper }}=\frac{1}{2} \times Y_{\text {steel }}=2 \times 10^{11} \mathrm{~N} / \mathrm{m}^{2} Y=\) Young's modules \(\alpha_{\text {copper }}=18 \times 10^{-6} \mathrm{~K}^{-1}\) \(\alpha_{\text {steel }}=11 \times 10^{-6} \mathrm{~K}^{-1} \alpha=\) coefficient of linear expansion \text { Area of each plate }=50 \mathrm{~cm}^{2}
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