If $10^{22}$ gas molecules each of mass $10^{-26}$ kg collide with a surface (perpendicular to it) elastically per second over an area $1\, m^2$ with a speed $10^4$ m/s, the pressure exerted by the gas molecules will be of the order of :

Question

In a mixture of gases, the average number of degrees of freedom per molecule is 6. If the rms speed of the molecule is $c$, what is the velocity of sound in the gas?

Answer 1

To determine the pressure exerted by the gas molecules, we begin by understanding the principles of momentum and force. When gas molecules collide elastically with a surface, there is a change in momentum due to the reversal of velocity. This change in momentum results in a force, which in turn relates to pressure.

The change in momentum for one molecule during an elastic collision is given by:

$\Delta p = mv - (-mv) = 2mv$

Where:

m = 10^{-26} \, \text{kg} (mass of one molecule)
v = 10^4 \, \text{m/s} (velocity of the molecule)

Now, we calculate the force exerted by all molecules using:

F = \frac{\Delta p}{\Delta t} \times \text{Number of molecules}

Substitute the known values:

F = 2mv \times 10^{22} \, \text{molecules/second}

= 2 \times 10^{-26} \, \text{kg} \times 10^4 \, \text{m/s} \times 10^{22}

= 2 \times 10^0 \, \text{kg m/s}^2

= 2 \times 1 \, \text{N}

The force exerted is 2 \, \text{N}. The area given is 1 \, \text{m}^2. Thus, the pressure is:

P = \frac{F}{A} = \frac{2 \, \text{N}}{1 \, \text{m}^2} = 2 \, \text{N/m}^2

Therefore, the pressure exerted by the gas molecules is of the order of 10^3 \, \text{N/m}^2.

The correct option is:

10^3 \; \text{N/m}^2

Answer 2