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Step-by-Step Solution
Step 1: Identify the Given Data
Mass of one hydrogen molecule, $m = 3.32 \times 10^{-27}\,\text{kg}$.
Number of hydrogen molecules striking the wall per second, $n = 10^{23}\,\text{s}^{-1}$.
Speed of the molecules, $v = 10^3\,\text{m/s}$.
Angle of incidence to the normal, $45^\circ$.
Area of the wall, $A = 2\,\text{cm}^2 = 2 \times 10^{-4}\,\text{m}^2$.
The collisions are elastic (coefficient of restitution $e = 1$).
Step 2: Determine the Momentum Change for One Molecule
Since the molecule strikes at an angle of $45^\circ$ to the normal, its velocity components along the wall (the $y$-axis if the normal is along the $x$-axis) do not contribute to the average force on the wall. We only consider the component of velocity normal to the wall for momentum transfer.
Before collision, let:
Normal (x-direction) component of velocity: $v_x = \frac{v}{\sqrt{2}}$ (towards the wall).
Tangential (y-direction) component of velocity: $v_y = \frac{v}{\sqrt{2}}$.
After collision (elastic and reversing normal component), we have:
Normal (x-direction) component of velocity becomes $-\frac{v}{\sqrt{2}}$ (rebound in opposite direction with same magnitude).
Tangential (y-direction) component remains $-\frac{v}{\sqrt{2}}$ (if we consider the same sign convention as initial).
The change in momentum along the normal direction for one molecule is:
$ \Delta P = p_{\text{final}} - p_{\text{initial}} = m \left(-\frac{v}{\sqrt{2}}\right) - m\left(\frac{v}{\sqrt{2}}\right) = -\frac{2mv}{\sqrt{2}}.$
The magnitude of the change in momentum is
$ \left|\Delta P\right| = \frac{2mv}{\sqrt{2}}.$
Step 3: Calculate the Total Momentum Change per Second
If $n$ molecules strike the wall per second, the total change in momentum per second (which is effectively the net force on the molecules) is:
$ F = \frac{\Delta P}{\Delta t} = \left(\frac{2 m v}{\sqrt{2}}\right) \times n. $
By Newton's third law, the wall experiences an equal and opposite force.
Step 4: Determine the Pressure on the Wall
Pressure $P$ is force per unit area:
$ P = \frac{F}{A} = \frac{\left(\frac{2 m v}{\sqrt{2}}\right) \times n}{A}. $
Substituting the given values:
$ P = \frac{2 \times 3.32 \times 10^{-27}\,\text{kg} \times 10^{3}\,\text{m/s} \times 10^{23}\,\text{s}^{-1}}{\sqrt{2} \times 2\times 10^{-4}\,\text{m}^2} \approx 2.35 \times 10^{3} \,\text{N m}^{-2}.
$
Final Answer
$\displaystyle 2.35 \times 10^{3}\,\text{N m}^{-2}
$
