Question

Hydrogen gas is contained in a closed vessel at 1 atm (100 kPa) and 300 K. (a) Calculate the mean speed of the molecules. (b) Suppose the molecules strike the wall with this speed making an average angle of $45^\circ$ with it. How many molecules strike each (square metre) of the wall per second ?

Answer 1

(a) The mean speed of the molecules is approximately $1782 \, \text{m/s}$.

(b) The number of molecules striking each square metre of the wall per second is approximately $1.08 \times 10^{28} \, \text{molecules m}^{-2} \text{s}^{-1}$.

Answer 2

The problem involves concepts from the Kinetic Theory of Gases. We need to calculate the mean speed of hydrogen molecules and the number of molecules striking a unit area of the wall per second.

Part (a): Calculate the mean speed of the molecules.

The mean speed ($v_{avg}$) of gas molecules is given by the formula: $v_{avg} = \sqrt{\frac{8RT}{\pi M}}$ Where:

• $R$ is the ideal gas constant = $8.314 \, \text{J mol}^{-1} \text{K}^{-1}$
• $T$ is the absolute temperature = $300 \, \text{K}$
• $M$ is the molar mass of hydrogen gas (H$_2$) = $2 \, \text{g/mol} = 2 \times 10^{-3} \, \text{kg/mol}$

Substituting the given values: $v_{avg} = \sqrt{\frac{8 \times 8.314 \, \text{J mol}^{-1} \text{K}^{-1} \times 300 \, \text{K}}{\pi \times 2 \times 10^{-3} \, \text{kg/mol}}}$ $v_{avg} = \sqrt{\frac{19953.6}{3.14159 \times 0.002}}$ $v_{avg} = \sqrt{\frac{19953.6}{0.00628318}}$ $v_{avg} = \sqrt{3175659.5}$ $v_{avg} \approx 1782 \, \text{m/s}$

Part (b): How many molecules strike each (square metre) of the wall per second?

The number of molecules striking a unit area of the wall per second ($Z_w$) is given by the formula: $Z_w = \frac{1}{4} n v_{avg}$ Where:

• $n$ is the number density of molecules (number of molecules per unit volume)
• $v_{avg}$ is the mean speed calculated in Part (a)

First, we need to calculate the number density ($n$). We can use the ideal gas law in terms of molecules: $PV = NkT$ So, the number density $n = \frac{N}{V} = \frac{P}{kT}$ Where:

• $P$ is the pressure = $1 \, \text{atm} = 100 \, \text{kPa} = 100 \times 10^3 \, \text{Pa}$
• $k$ is the Boltzmann constant = $1.38 \times 10^{-23} \, \text{J K}^{-1}$
• $T$ is the absolute temperature = $300 \, \text{K}$

Substituting the values for $n$: $n = \frac{100 \times 10^3 \, \text{Pa}}{1.38 \times 10^{-23} \, \text{J K}^{-1} \times 300 \, \text{K}}$ $n = \frac{10^5}{414 \times 10^{-23}}$ $n = \frac{10^5}{4.14 \times 10^{-21}}$ $n \approx 2.4155 \times 10^{25} \, \text{molecules/m}^3$

Now, substitute $n$ and $v_{avg}$ into the formula for $Z_w$: $Z_w = \frac{1}{4} \times (2.4155 \times 10^{25} \, \text{molecules/m}^3) \times (1782 \, \text{m/s})$ $Z_w = 0.25 \times 2.4155 \times 10^{25} \times 1782$ $Z_w \approx 1.076 \times 10^{28} \, \text{molecules m}^{-2} \text{s}^{-1}$

The statement "Suppose the molecules strike the wall with this speed making an average angle of $45^\circ$ with it" is a descriptive detail that is inherently accounted for in the statistical mechanical derivation of the $Z_w = \frac{1}{4} n v_{avg}$ formula, which averages over all angles of incidence. Therefore, no additional angle-dependent factor is needed for the calculation using this standard formula.