Heat and Thermodynamics 4 Question 6
6. The specific heats, $C _p$ and $C _V$ of a gas of diatomic molecules, $A$ are given (in units of $J mol^{-1} K^{-1}$ ) by 29 and 22 , respectively. Another gas of diatomic molecules $B$, has the corresponding values 30 and 21 . If they are treated as ideal gases, then
(2019 Main, 9 April II)
(a) $A$ has a vibrational mode but $B$ has none
(b) Both $A$ and $B$ have a vibrational mode each
(c) $A$ has one vibrational mode and $B$ has two
(d) $A$ is rigid but $B$ has a vibrational mode
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Solution:
- Key Idea A diatomic gas molecule has 5 degrees of freedom, i.e. 3 translational and 2 rotational, at low temperature ranges ( $250 K$ to $750 K$ ). At temperatures above $750 K$, molecular vibrations occurs and this causes two extra degrees of freedom.
Now, in given case,
For gas $A, C _p=29, C _V=22$
For gas $B, C _p=30, C _V=21$
By using
$$ \gamma=\frac{C _p}{C _V}=1+\frac{2}{f} $$
We have,
For gas $A, 1+\frac{2}{f}=\frac{29}{22} \approx 1.3 \Rightarrow f=6.67 \approx 7$
So, gas $A$ has vibrational mode of degree of freedom.
For gas $B$,
$$ 1+\frac{2}{f}=\frac{30}{21} \approx 1.4 \Rightarrow f=5 $$
Hence, gas $B$ does not have any vibrational mode of degree of freedom.
Key Idea According to the law of equipartition of energy, $\frac{1}{2} m v _{\text {rms }}^{2}=\frac{n}{2} k _B T$
where, $n$ is the degree of freedom.
Since, $HCl$ is a diatomic molecule that has rotational, translational and vibrational motion.
$$ \begin{array}{ll} \text { So, } & n=7 \Rightarrow \frac{1}{2} m v _{rms}^{2}=\frac{7}{2} k _B T \\ \text { Here, } & v _{rms}^{2}=\bar{v} \Rightarrow T=\frac{m \bar{v}^{2}}{7 k _B} \end{array} $$
