Question

One gram-mole of oxygen at ${27}^{\circ }C$ and $1atm$ pressure is enclosed in a vessel. Assuming the molecules to be moving with $v_{rms}$, the number of collisions per second which the molecules make with $1m^2$ area of the vessel wall is approximately $N=x\times {10}^{27}{m}^{-2}$. Find $x$. (Round off to closest integer)

Answer 1

From gas law, we have

$PV=\frac{m}{M}RT$

or $PV=\frac{m^{\prime }N}{M^{\prime }{N}_A}RT$ ($m^{\prime }$, mass of one mole; $N$, total number of molecules)

or $PV=NkT$ $(as\frac{R}{N_{AV}}=k)$

or $P=n_{0}kT$ (i)

Here $n_0=\frac{N}{V}$ the number of molecules per unit volume of gas. Students should note that equation (i) relates the gas pressure, temperature and molecular number density. This expression can be used as a standard equation in several numerical cases as an alternative form of gas law.

Thus, here molecular number density is given as

$n_0=\frac{P}{kT}$

$=\frac{\left(1atm\right)}{(1.38\times {10}^{-23})\times 300}$

$=\frac{{10}^5}{(1.38\times {10}^{-23})\times 300}=2.44\times {10}^{25}{m}^{-3}$

The rms speed of gas molecules is given as

$v_{rms}=\sqrt{\frac{3RT}{M}}$

$=\sqrt{\frac{3\times 8.314\times 300}{32\times {10}^{-3}}}=483.4m/s$

Now the number of collision per square metre of vessel wall is given by

$N=\frac{n_0{v}_{rms}}{6}=\frac{2.44\times {10}^{25}\times 483.4}{6}=1.97\times {10}^{27}{m}^{-2}$