Answer
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Hint: 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:
$E=\dfrac{1}{2}nRT$
$\dfrac{dE}{dT}={{C}_{V}}$
${{C}_{P}}-{{C}_{V}}=R$
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 $n$ degrees of freedom is given by
$E=\dfrac{1}{2}nRT$ --(1)
Where $T$ is the temperature of the gas and $R=8.314J.mo{{l}^{-1}}{{K}^{-1}}$ is the universal gas constant.
Now, the specific heat at constant volume ${{C}_{V}}$ of a gas is related to its internal energy $E$ by
${{C}_{V}}=\dfrac{dE}{dT}$ --(2)
Putting (1) in (2), we get,
${{C}_{V}}=\dfrac{d\left( \dfrac{1}{2}nRT \right)}{dT}=\dfrac{1}{2}nR$ --(3)
Now, the specific heat at constant volume ${{C}_{V}}$ is related to the specific heat at constant pressure ${{C}_{P}}$ by
${{C}_{P}}={{C}_{V}}+R$
Using (3), we get,
${{C}_{P}}=\dfrac{1}{2}nR+R=R\left( \dfrac{1}{2}n+1 \right)$ --(4)
Now, ratio of the specific heats $\gamma $ of a gas is the ratio of the specific heat at constant pressure ${{C}_{P}}$ to the specific heat at constant volume ${{C}_{V}}$. Therefore,
$\gamma =\dfrac{{{C}_{P}}}{{{C}_{V}}}$
Using (3) and (4), we get,
$\gamma =\dfrac{R\left( \dfrac{1}{2}n+1 \right)}{\dfrac{1}{2}nR}=\dfrac{\dfrac{1}{2}nR\left( 1+\dfrac{1}{\dfrac{1}{2}n} \right)}{\dfrac{1}{2}nR}=\left( 1+\dfrac{1}{\dfrac{1}{2}n} \right)=1+\dfrac{2}{n}$
Hence, the required value of $\gamma $ is $1+\dfrac{2}{n}$.
Therefore, the correct option is $D)\text{ 1+}\dfrac{2}{n}$.
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.
Formula used:
$E=\dfrac{1}{2}nRT$
$\dfrac{dE}{dT}={{C}_{V}}$
${{C}_{P}}-{{C}_{V}}=R$
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 $n$ degrees of freedom is given by
$E=\dfrac{1}{2}nRT$ --(1)
Where $T$ is the temperature of the gas and $R=8.314J.mo{{l}^{-1}}{{K}^{-1}}$ is the universal gas constant.
Now, the specific heat at constant volume ${{C}_{V}}$ of a gas is related to its internal energy $E$ by
${{C}_{V}}=\dfrac{dE}{dT}$ --(2)
Putting (1) in (2), we get,
${{C}_{V}}=\dfrac{d\left( \dfrac{1}{2}nRT \right)}{dT}=\dfrac{1}{2}nR$ --(3)
Now, the specific heat at constant volume ${{C}_{V}}$ is related to the specific heat at constant pressure ${{C}_{P}}$ by
${{C}_{P}}={{C}_{V}}+R$
Using (3), we get,
${{C}_{P}}=\dfrac{1}{2}nR+R=R\left( \dfrac{1}{2}n+1 \right)$ --(4)
Now, ratio of the specific heats $\gamma $ of a gas is the ratio of the specific heat at constant pressure ${{C}_{P}}$ to the specific heat at constant volume ${{C}_{V}}$. Therefore,
$\gamma =\dfrac{{{C}_{P}}}{{{C}_{V}}}$
Using (3) and (4), we get,
$\gamma =\dfrac{R\left( \dfrac{1}{2}n+1 \right)}{\dfrac{1}{2}nR}=\dfrac{\dfrac{1}{2}nR\left( 1+\dfrac{1}{\dfrac{1}{2}n} \right)}{\dfrac{1}{2}nR}=\left( 1+\dfrac{1}{\dfrac{1}{2}n} \right)=1+\dfrac{2}{n}$
Hence, the required value of $\gamma $ is $1+\dfrac{2}{n}$.
Therefore, the correct option is $D)\text{ 1+}\dfrac{2}{n}$.
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.
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