The mass of a hydrogen molecule is $$3.32 \times 10^{-27}$$ kg. If $$10^{23}$$ hydrogen molecules strike, per second, a fixed wall of the area 2 cm$$^2$$ at an angle of 45$$^\circ$$ to the normal, and rebound elastically with a speed of $$10^3$$ m s$$^{-1}$$, then the pressure on the wall is nearly:

We begin with the well-known relation

$$\text{Pressure }(P)=\frac{\text{Force }(F)}{\text{Area }(A)}$$

So, to obtain the pressure, we first have to calculate the force exerted on the wall by the stream of molecules. Force is the rate of change of momentum, therefore

$$F=\frac{\Delta p_{\text{total}}}{\Delta t}$$

Since the data are already “per second”, we may take $$\Delta t = 1\ \text{s}$$, and then

$$F = \text{(change of momentum per molecule)}\times(\text{number of molecules per second}).$$

Let us work out the change of momentum for one molecule. The given velocity of each molecule is $$v = 10^{3}\ \text{m s}^{-1}$$, and the molecule strikes the wall making an angle $$\theta = 45^\circ$$ with the normal. Hence the component of velocity normal to the wall is

$$v_n = v\cos\theta = 10^{3}\cos45^\circ\ \text{m s}^{-1}.$$

Because $$\cos45^\circ = \dfrac{1}{\sqrt2}$$, we have

$$v_n = 10^{3}\times\dfrac{1}{\sqrt2}\ \text{m s}^{-1}.$$

The molecule rebounds elastically, so its speed is unchanged but its normal component reverses direction. Thus

$$p_{\text{initial}} = m v_n,\qquad p_{\text{final}} = -m v_n.$$

Therefore the change in normal momentum for one molecule is

$$\Delta p = p_{\text{final}} - p_{\text{initial}} = (-m v_n) - (m v_n) = -2 m v_n.$$

We shall use the magnitude only, so

$$|\Delta p| = 2 m v_n = 2 m v\cos\theta.$$

Substituting the numerical values one by one:

Mass of one hydrogen molecule

$$m = 3.32\times10^{-27}\ \text{kg},$$

velocity

$$v = 10^{3}\ \text{m s}^{-1},$$

and $$\cos\theta = \dfrac{1}{\sqrt2}\approx0.7071.$$

Hence

$$|\Delta p| = 2 \times (3.32\times10^{-27}) \times (10^{3}) \times 0.7071$$ $$= 2 \times 3.32 \times 0.7071 \times 10^{-24}$$ $$= 4.695\times10^{-24}\ \text{kg m s}^{-1}.$$

Now, the number of molecules striking the wall each second is

$$N = 10^{23}\ \text{s}^{-1}.$$

Therefore the total change of momentum per second (which is the force) becomes

$$F = N\,|\Delta p| = (10^{23})\,(4.695\times10^{-24})$$ $$= 4.695\times10^{-1}\ \text{N}$$ $$= 0.4695\ \text{N}.$$

Next we convert the area of the wall to SI units. The given area is

$$A = 2\ \text{cm}^2.$$

Because $$1\ \text{cm} = 10^{-2}\ \text{m},$$ we have

$$A = 2 \times (10^{-2}\ \text{m})^{2} = 2 \times 10^{-4}\ \text{m}^2.$$

Finally, substituting into the pressure formula, we get

$$P = \frac{F}{A} = \frac{0.4695\ \text{N}}{2\times10^{-4}\ \text{m}^2}$$ $$= \frac{0.4695}{0.0002}\ \text{N m}^{-2}$$ $$= 2347.5\ \text{N m}^{-2}.$$

Written in scientific notation,

$$P \approx 2.35 \times 10^{3}\ \text{N m}^{-2}.$$

Hence, the correct answer is Option B.