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What will be the molar specific heat at constant volume of an ideal gas consisting of rigid diatomic molecules?
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Verified Answer
The correct answer is:
$\frac{5}{2} R$
For a gas at temperature $T$, the internal energy,
$$
U=\frac{f}{2} \mu R T
$$
where, $f=$ degree of freedom
$\Rightarrow$ Change in energy, $\Delta U=\frac{f}{2} \mu R \Delta T$
Also, as we know for any gas heat supplied at constant volume
$$
\left(\Delta Q_{V}\right)=\mu C_{V} \Delta T=\Delta U
$$
where, $C_{V}=$ molar specific heat at constant volume From Eqs. (i) and (ii), we get
$$
C_{V}=\frac{1}{2} f R
$$
For diatomic gas, degree of freedom, $f=5$
$$
C_{V}=\frac{5}{2} R
$$
$$
U=\frac{f}{2} \mu R T
$$
where, $f=$ degree of freedom
$\Rightarrow$ Change in energy, $\Delta U=\frac{f}{2} \mu R \Delta T$
Also, as we know for any gas heat supplied at constant volume
$$
\left(\Delta Q_{V}\right)=\mu C_{V} \Delta T=\Delta U
$$
where, $C_{V}=$ molar specific heat at constant volume From Eqs. (i) and (ii), we get
$$
C_{V}=\frac{1}{2} f R
$$
For diatomic gas, degree of freedom, $f=5$
$$
C_{V}=\frac{5}{2} R
$$
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