SATHEE: Heat and Thermodynamics 4 Question 6

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 $J m o l^{- 1} K^{- 1}$ ) 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

(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

$γ = \frac{C_{p}}{C_{V}} = 1 + \frac{2}{f}$

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, $\frac{1}{2} m v_{rms}^{2} = \frac{n}{2} k_{B} T$

where, $n$ is the degree of freedom.

Since, $H C l$ is a diatomic molecule that has rotational, translational and vibrational motion.

$\begin{array}{ll} So, & n = 7 \Rightarrow \frac{1}{2} m v_{r m s}^{2} = \frac{7}{2} k_{B} T \\ Here, & v_{r m s}^{2} = \bar{v} \Rightarrow T = \frac{m {\bar{v}}^{2}}{7 k_{B}} \end{array}$