Question

One gram-mole of oxygen at $27^\circ {\text{C}}$ and $1\,{\text{atm}}$ 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 $1\,{{\text{m}}^2}$ area of vessel wall is approximately $N = x \times {10^{27}}\,{{\text{m}}^{ - 2}}$. Find x. (Round off to closest integer)

Answer 1

Use the formula for the ideal gas law in terms of the molecular density of the gas. Also use the formula for the root mean square velocity of the gas molecules in terms of Avogadro’s constant and Boltzmann constant. Use the values of molecular density and root mean square velocity to obtain the number of collisions on the wall of the vessel.

Formula used:
The ideal gas equation in terms of molecular density is
$P = nkT$ …… (1)
Here, $P$ is the pressure, $n$ is the molecular density (number of molecules per unit volume), $k$ is the Boltzmann constant and $T$ is the temperature in kelvin.
The root mean square velocity ${v_{rms}}$ of a gas is given by
${v_{rms}} = \sqrt {\dfrac{{3{N_A}kT}}{m}}$ …… (2)
Here, ${N_A}$ is the Avogadro’s number, $k$ is the Boltzmann constant, $T$ is the temperature in Kelvin and $m$ is the mass of a gas molecule in kilogram.

Complete step by step answer:
Determine the molecular density of the oxygen gas.

Convert the temperature $T$ of the gas in kelvin.
$T = \left( {27^\circ {\text{C}}} \right) + 273$
$\Rightarrow T = 300\,{\text{K}}$

Hence, the temperature of the oxygen gas is $300\,{\text{K}}$.

Rearrange equation (1) for the molecular density $n$ of oxygen gas.
$n = \dfrac{P}{{kT}}$

Substitute $1.01 \times {10^5}\,{\text{N/}}{{\text{m}}^2}$ for $P$, $1.38 \times {10^{ - 23}}\,{\text{J}} \cdot {{\text{K}}^{ - 1`}}$ for $k$ and $300\,{\text{K}}$ for $T$ in the above equation.
$n = \dfrac{{1.01 \times {{10}^5}\,{\text{N/}}{{\text{m}}^2}}}{{\left( {1.38 \times {{10}^{ - 23}}\,{\text{J}} \cdot {{\text{K}}^{ - 1`}}} \right)\left( {300\,{\text{K}}} \right)}}$
$\Rightarrow n = 2.44 \times {10^{25}}\,{{\text{m}}^{ - 3}}$

Hence, the molecular density of the oxygen gas is $2.44 \times {10^{25}}\,{{\text{m}}^{ - 3}}$.

Now determine the root mean square velocity of the oxygen molecules.

Substitute $6.02 \times {10^{23}}{\text{ molecules/mol}}$ for ${N_A}$, $1.38 \times {10^{ - 23}}\,{\text{J}} \cdot {{\text{K}}^{ - 1`}}$ for $k$, $300\,{\text{K}}$ for $T$and $32 \times {10^{ - 3}}\,{\text{kg}}$ for $m$ in equation (2).
${v_{rms}} = \sqrt {\dfrac{{3\left( {6.02 \times {{10}^{23}}{\text{ molecules/mol}}} \right)\left( {1.38 \times {{10}^{ - 23}}\,{\text{J}} \cdot {{\text{K}}^{ - 1`}}} \right)\left( {300\,{\text{K}}} \right)}}{{32 \times {{10}^{ - 3}}\,{\text{kg}}}}}$

${v_{rms}} = 483.4\,{\text{m/s}}$

Hence, the root mean square velocity of the oxygen gas is $483.4\,{\text{m/s}}$.

Each gas molecule in the vessel moves along both positive and negative directions of the X, Y and X axes inside the vessel.

Hence, the contribution of each gas molecule in collision with the area of the wall is (1/6)th.

Therefore, the number of collisions $N$ of the gas molecules on the given area of the wall is
$N = \dfrac{1}{6}n{v_{rms}}$

Substitute $2.44 \times {10^{25}}\,{{\text{m}}^{ - 3}}$ for $n$ and $483.4\,{\text{m/s}}$ for ${v_{rms}}$ in the above equation.
$N = \dfrac{1}{6}\left( {2.44 \times {{10}^{25}}\,{{\text{m}}^{ - 3}}} \right)\left( {483.4\,{\text{m/s}}} \right)$
$\Rightarrow N = 1.97 \times {10^{27}}$

Therefore, the number of collisions by the gas molecules on the wall is $1.97 \times {10^{27}}$.

Hence, the value of $x$ is $1.97$.

Note:
The value of the product of the Avogadro’s number and the Boltzmann constant is equal to the gas constant. Hence, the value of Boltzmann constant in the formula for root mean square velocity is replaced by the product of Avogadro’s number and the Boltzmann constant is equal to the gas constant.