Heat and Thermodynamics 6 Question 39
44. One gram mole of oxygen at $27^{\circ} C$ and one atmospheric pressure is enclosed in a vessel. (a) Assuming the molecules to be moving with $v _{rms}$, find the number of collisions per second which the molecules make with one square metre area of the vessel wall. (b) The vessel is next thermally insulated and moved with a constant speed $v _0$. It is then suddenly stopped. The process results in a rise of the temperature of the gas by $1^{\circ} C$. Calculate the speed $v _0$.
$(1983,8 M)$
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Answer:
Correct Answer: 44. (a) $1.96 \times 10^{27} / s$ (b) $36 m / s$
Solution:
- (a) $v _{\text {rms }}=\sqrt{\frac{3 R T}{M}}=\sqrt{\frac{3 \times 8.31 \times 300}{32 \times 10^{-3}}}=483.4 m / s$
Given, $p _0=1.01 \times 10^{5} N / m^{2}=$ Force per unit area.
Let $n$ molecules of oxygen strike the wall per second per $m^{2}$ and recoil with same speed. Change in momentum is $\left(2 n m v _{rms}\right)$. The change in momentum per unit time is the force.
$ \text { Hence, } \quad p_0 =2 nmv_{\text {rms }} $
$ \therefore \quad n =\frac{p_0}{2mv_{\text {rms }}} $
$=\frac{1.01 \times 10^5}{2 {[\frac{32}{6.02 \times 10^{26}}]} (483.4)} $
$=1.96 \times {10}^{27} / {s} $
(b) $\frac{1}{2}\left(m _{\text {gas }}\right) v _0^{2}=n C _V \Delta T$
$$ \therefore v _0=\sqrt{\frac{2 n C _V \Delta t}{m _{\text {gas }}}}=\sqrt{\frac{(2)(n)\left(\frac{5}{2} \times 8.31\right)(1)}{(n)\left(32 \times 10^{-3}\right)}}=36 m / s $$
