SATHEE: Some Basic Concepts of Chemistry - Result Question 33

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

Show Answer

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, $F = 3$

$∴ γ = \frac{C_{p}}{C_{v}} = 1 + \frac{2}{3} = \frac{5}{3} = 1.67$

For diatomic gas, $F = 5$

$∴ γ = \frac{C_{p}}{C_{v}} = 1 + \frac{2}{5} = \frac{7}{5} = 1.40$

For triatomic gas, $F = 6$ or 7 (depending upon the nature of the gas)

$\begin{matrix} ∴ γ = \frac{C_{p}}{C_{v}} = 1 + \frac{2}{6} = 1.33 \\ γ = \frac{C_{p}}{C_{v}} = 1 + \frac{2}{7} = 1.29 \end{matrix}$

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