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Chapter 8: Problem 1094

Each molecule of a gas has \(\mathrm{f}\) degrees of freedom. The radio \(\left(\mathrm{C}_{\mathrm{p}} / \mathrm{C}_{\mathrm{v}}\right)=\gamma\) for the gas is (A) \(1+(\mathrm{f} / 2)\) (B) \(1+(1 / \mathrm{f})\) (C) \(1+(2 / \mathrm{f})\) (D) \(1+\\{(\mathrm{f}-1) / 3\\}\)

Short Answer

Expert verified
The correct option for the ratio \(\left(\mathrm{C}_{\mathrm{p}} / \mathrm{C}_{\mathrm{v}}\right)=\gamma\) for the gas is (C) \(1+(2 / \mathrm{f})\).

Step by step solution

01

Relationship between specific heat capacities and degrees of freedom

To find the relationship between the specific heat capacities and the degrees of freedom (\(f\)), we can recall that: \(C_p=C_v+nR\) Where \(C_p\) = specific heat capacity at constant pressure, \(C_v\) = specific heat capacity at constant volume, \(n\) = number of moles, and \(R\) = ideal gas constant.
02

Equation of specific heat capacity at constant volume

For a gas with \(f\) degrees of freedom, the specific heat capacity at constant volume (\(C_v\)) is given by: \(C_v=\frac{f}{2}nR\) We are going to use this equation to express \(\frac{C_p}{C_v}\) in terms of \(f\).
03

Derive the expression for gamma (\(\gamma\))

To find the ratio \(\gamma\), we can substitute the expression of \(C_v\) (from step 2) into the equation from step 1: \(C_p - C_v = nR\) Then divide both sides by \(C_v\): \(\frac{C_p}{C_v}-1 = \frac{2}{f}\) Add 1 on both sides of the equation to get the ratio: \(\frac{C_p}{C_v} = \gamma = 1 + \frac{2}{f}\) Comparing our derived expression with the given options, we find that it matches with option (C).
04

Answer

The correct option for the ratio \(\left(\mathrm{C}_{\mathrm{p}} / \mathrm{C}_{\mathrm{v}}\right)=\gamma\) for the gas is (C) \(1+(2 / \mathrm{f})\)

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Most popular questions from this chapter

What is an adiabatic Bulk modulus of hydrogen gas at NTP? (A) \(1.4\left(\mathrm{~N} / \mathrm{M}^{2}\right)\) (B) \(1.4 \times 10^{5}\left(\mathrm{~N} / \mathrm{M}^{2}\right)\) (C) \(1 \times 10^{-8}\left(\mathrm{~N} / \mathrm{M}^{2}\right)\) (D) \(1 \times 10^{5}\left(\mathrm{~N} / \mathrm{M}^{2}\right)\)

Carnot engine working between \(300 \mathrm{~K}\) and \(600 \mathrm{~K}\) has work output of \(800 \mathrm{~J}\) per cycle. What is amount of heat energy supplied to the engine from source per cycle (A) \(1600\\{\mathrm{~J} /\) (cycle)\\} (B) \(2000\\{\mathrm{~J} /(\mathrm{cycle})\\}\) (C) \(1000\\{\mathrm{~J} /(\) cycle \()\\}\) (D) \(1800\\{\mathrm{~J} /(\) cycle \()\\}\)

A Carnot engine Converts one sixth of the heat input into work. When the temperature of the sink is reduced by \(62^{\circ} \mathrm{C}\) the efficiency of the engine is doubled. The temperature of the source and sink are (A) \(80^{\circ} \mathrm{C}, 37^{\circ} \mathrm{C}\) (B) \(95^{\circ} \mathrm{C}, 28^{\circ} \mathrm{C}\) (C) \(90^{\circ} \mathrm{C}, 37^{\circ} \mathrm{C}\) (D) \(99^{\circ} \mathrm{C}, 37^{\circ} \mathrm{C}\)

An ideal gas at \(27 \mathrm{C}\) is Compressed adiabatically, to \(\\{8 / 27\\}\) of its original Volume. If \(\mathrm{v}=(5 / 3)\), then the rise in temperature is (A) \(225 \mathrm{k}\) (B) \(450 \mathrm{~K}\) (C) \(375 \mathrm{~K}\) (D) \(405 \mathrm{~K}\)

Starting with the same initial Conditions, an ideal gas expands from Volume \(\mathrm{V}_{1}\) to \(\mathrm{V}_{2}\) in three different ways. The Work done by the gas is \(\mathrm{W}_{1}\) if the process is purely isothermal, \(\mathrm{W}_{2}\) if purely isobaric and \(\mathrm{W}_{3}\) if purely adiabatic Then (A) \(\mathrm{W}_{2}>\mathrm{W}_{1}>\mathrm{W}_{3}\) (B) \(\mathrm{W}_{2}>\mathrm{W}_{3}>\mathrm{W}_{1}\) (C) \(\mathrm{W}_{1}>\mathrm{W}_{2}>\mathrm{W}_{3}\) (D) \(\mathrm{W}_{1}>\mathrm{W}_{3}>\mathrm{W}_{2}\)
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