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Question: If a gas has \('n'\) degrees of freedom, the ratio of the specific heats of the gas \(\gamma \) is ...

If a gas has n'n' degrees of freedom, the ratio of the specific heats of the gas γ\gamma is
A) 1+n2A)\text{ }\dfrac{1+n}{2}
B) 1+n2B)\text{ 1+}\dfrac{n}{2}
C) 1+1nC)\text{ 1+}\dfrac{1}{n}
D) 1+2nD)\text{ 1+}\dfrac{2}{n}

Explanation

Solution

We can find the specific heats by finding the internal energy of one mole of gas and relating it to the specific heat capacity at constant volume. From this we can get the specific heat at constant pressure by relating it with the universal gas constant. Therefore, we get the two specific heats and can find their ratio.

Formula used:
E=12nRTE=\dfrac{1}{2}nRT
dEdT=CV\dfrac{dE}{dT}={{C}_{V}}
CPCV=R{{C}_{P}}-{{C}_{V}}=R

Complete step-by-step answer:
First, we will try to find the specific heat capacity at constant volume for the gas by finding the internal energy of one mole of the gas.
The internal energy of one mole of a gas with nn degrees of freedom is given by
E=12nRTE=\dfrac{1}{2}nRT --(1)
Where TT is the temperature of the gas and R=8.314J.mol1K1R=8.314J.mo{{l}^{-1}}{{K}^{-1}} is the universal gas constant.
Now, the specific heat at constant volume CV{{C}_{V}} of a gas is related to its internal energy EE by
CV=dEdT{{C}_{V}}=\dfrac{dE}{dT} --(2)
Putting (1) in (2), we get,
CV=d(12nRT)dT=12nR{{C}_{V}}=\dfrac{d\left( \dfrac{1}{2}nRT \right)}{dT}=\dfrac{1}{2}nR --(3)
Now, the specific heat at constant volume CV{{C}_{V}} is related to the specific heat at constant pressure CP{{C}_{P}} by
CP=CV+R{{C}_{P}}={{C}_{V}}+R
Using (3), we get,
CP=12nR+R=R(12n+1){{C}_{P}}=\dfrac{1}{2}nR+R=R\left( \dfrac{1}{2}n+1 \right) --(4)
Now, ratio of the specific heats γ\gamma of a gas is the ratio of the specific heat at constant pressure CP{{C}_{P}} to the specific heat at constant volume CV{{C}_{V}}. Therefore,
γ=CPCV\gamma =\dfrac{{{C}_{P}}}{{{C}_{V}}}
Using (3) and (4), we get,
γ=R(12n+1)12nR=12nR(1+112n)12nR=(1+112n)=1+2n\gamma =\dfrac{R\left( \dfrac{1}{2}n+1 \right)}{\dfrac{1}{2}nR}=\dfrac{\dfrac{1}{2}nR\left( 1+\dfrac{1}{\dfrac{1}{2}n} \right)}{\dfrac{1}{2}nR}=\left( 1+\dfrac{1}{\dfrac{1}{2}n} \right)=1+\dfrac{2}{n}
Hence, the required value of γ\gamma is 1+2n1+\dfrac{2}{n}.
Therefore, the correct option is D) 1+2nD)\text{ 1+}\dfrac{2}{n}.

Note: Students get confused while writing the ratio of the specific heats and tend to write the ratio with the specific heat capacity at constant volume in the numerator. A good way to remember the ratio is to keep in mind that the ratio is always greater than one and the specific heat at constant pressure is always greater than the specific heat at constant volume and therefore, if the ratio has to be greater than tone, the specific heat at constant pressure should be in the numerator.