Heat and Thermodynamics 4 Question 9
10. An ideal gas is enclosed in a cylinder at pressure of $2 atm$ and temperature, $300 K$. The mean time between two successive collisions is $6 \times 10^{-8} s$. If the pressure is doubled and temperature is increased to $500 K$, the mean time between two successive collisions will be close to (2019 Main, 12 Jan II)
(a) $4 \times 10^{-8} s$
(b) $3 \times 10^{-6} s$
(c) $2 \times 10^{-7} s$
(d) $0.5 \times 10^{-8} s$
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Answer:
Correct Answer: 10. (a)
Solution:
- Mean time elapsed between two successive, collisions is $t=\frac{\lambda}{v}$
where, $\lambda=$ mean free path length and
$$ v=\text { mean speed of gas molecule } $$
$\therefore \quad t=\frac{\left(\frac{k _B t}{\sqrt{2} \pi d^{2} p}\right)}{\sqrt{\frac{8}{\pi} \cdot \frac{k _B T}{M}}} \Rightarrow t=\frac{C \sqrt{T}}{p}$
where, $C=\frac{1}{4 d^{2}} \sqrt{\frac{k _B M}{\pi}}=$ a constant for a gas.
So, $\quad \frac{t _2}{t _1}=\frac{\sqrt{T _2}}{\sqrt{T _1}} \cdot\left(\frac{p _1}{p _2}\right)\quad \cdots(i)$
Here given, $\frac{p _1}{p _2}=\frac{1}{2}, \sqrt{\frac{T _2}{T _1}}=\sqrt{\frac{500}{300}}=\sqrt{\frac{5}{3}}$
and
$$ t _1=6 \times 10^{-8} s $$
Substituting there values in (i), we get
$$ t _2=6 \times 10^{-8} \times \sqrt{\frac{5}{3}} \times \frac{1}{2}=3.86 \times 10^{-8} s \approx 4 \times 10^{-8} s $$
