Some Basic Concepts of Chemistry - Result Question 33
####31. Ratio of $C_p$ and $C_v$ of a gas ’ $X$ ’ is 1.4. The number of atoms of the gas ’ $X$ ’ present in 11.2 litres of it at NTP will be
(a) $6.02 \times 10^{23}$
(b) $1.2 \times 10^{23}$
(c) $3.01 \times 10^{23}$
(d) $2.01 \times 10^{23}$
[1989]
[1988]
(a) $\frac{1.8}{224} \times 10^{22}$ atoms
(b) $\frac{6.02}{22400} \times 10^{23}$ molecules
(c) $\frac{1.32}{224} \times 10^{23}$ electrons
(d) all the above
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Solution:
- (a) $C_p / C_v=1.4$ shows that the gas is diatomic. 22.4 litre at NTP $\equiv 6.02 \times 10^{23}$ molecules
$11.2 L$ at NTP $=3.01 \times 10^{23}$ molecules
No. of atoms in gas $=3.01 \times 10^{23} \times 2$ atoms $=6.02 \times 10^{23}$ atoms
$r=\frac{C_p}{C_v}=1+\frac{2}{F}$, where $F=$ degree of freedom of the gas molecules For mono atomic gas, $F=3$
$\therefore \gamma=\frac{C_p}{C_v}=1+\frac{2}{3}=\frac{5}{3}=1.67$
For diatomic gas, $F=5$
$\therefore \gamma=\frac{C_p}{C_v}=1+\frac{2}{5}=\frac{7}{5}=1.40$
For triatomic gas, $F=6$ or 7 (depending upon the nature of the gas)
$ \begin{matrix} \therefore \quad \gamma=\frac{C_p}{C_v}=1+\frac{2}{6}=1.33 \\ \quad \gamma=\frac{C_p}{C_v}=1+\frac{2}{7}=1.29 \end{matrix} $
