At what temperature the rms velocity is equal to the escape velocity of the hydrogen gas from the surface of earth?

I believe your mistake is with units, and it is the following:

$$T=\dfrac{M[c_{rms}]^2}{3R} = \dfrac{\left( 1 \text{amu} \right) [11.2 \frac{km}{s}]^2}{3 \left( 8.3144621 \frac{\mathrm{J}}{\mathrm{\text{mol} K}} \right)} $$

This doesn't even cancel out because you're left with a $\text{mol}$ unit. Add avogadro's number.

$$T = \dfrac{\left( 1 \text{amu} \right) [11.2 \frac{km}{s}]^2}{3 \left( 8.3144621 \frac{\mathrm{J}}{\mathrm{\text{mol} K}} \right)} \left( 6.022 \times 10^{23} \frac{1}{\text{mol}} \right) = 5,028 K $$

This is the case for Earth. For the Moon:

$$T = \dfrac{\left( 1 \text{amu} \right) [2.4 \frac{km}{s}]^2}{3 \left( 8.3144621 \frac{\mathrm{J}}{\mathrm{\text{mol} K}} \right)} \left( 6.022 \times 10^{23} \frac{1}{\text{mol}} \right) = 230 K $$

This is negative in Celsius units. This is not a problem. It is merely saying that even freezing temperatures are enough for a lone Hydrogen atom to escape the gravity of the moon with. All we required was that this number be less than the temperature of the sun, which it is. Any surface that faces away from the sun will again see those photons within a month, unless it's in a crater on one of the poles, which we know can have ice near the surface. So that makes sense.

Answer

Verified

Hint: Escape velocity is the minimum velocity with which a body should be protected from the surface of the earth so that it can escape from the gravitational field of the earth. The escape velocity on the surface of the earth is 11200m/s. Escape velocity on the surface of the moon is 2370m/s.

Formula used:

To solve this type of question we use the following formula.The root mean square (RMS) velocity is given below.${v_{rms}} = \sqrt {\dfrac{{3RT}}{M}} $, where $R$ is the gas constant, $M$ is mass and $T$ is the temperature.The escape velocity of earth is $11.2km/s$

Complete step by step answer:

According to the question, the RMS velocity of hydrogen is equal to the escape velocity of the earth.Let us use the formula ${v_{rms}} = \sqrt {\dfrac{{3RT}}{M}} $and write the following.$\sqrt {\dfrac{{3RT}}{M}} = 11.2 \times {10^3}$Now we have the values, the Gas constant is $R = 8.314$ and Molecular mass of Hydrogen $M = 2 \times {10^{ - 3}}$ Let us substitute these values,$\Rightarrow \sqrt {\dfrac{{3 \times 8.314T}}{{2 \times {{10}^{ - 3}}}}} = 11.2 \times {10^3}$Let us square both the sides.$\Rightarrow \dfrac{{3 \times 8.314T}}{{2 \times {{10}^{ - 3}}}} = {\left( {11.2 \times {{10}^3}} \right)^2}$Let us simplify it.$\Rightarrow \dfrac{{24.942T}}{{2 \times {{10}^{ - 3}}}} = 125.44 \times {10^6}$Let us further simplify.$\Rightarrow 24.942T = 250.88 \times {10^3} \Rightarrow T = \dfrac{{250.88 \times {{10}^3}}}{{24.942}} = 10.06 \times {10^3}$We can also write it as below.$\Rightarrow T = {10^4}K$

$\therefore $ The temperature of hydrogen gas would be ${10^4}K$. Hence, option (D) correct.

Note:

- The relation between escape velocity and orbital velocity is given below.Escape velocity= $\sqrt 2 \times $ orbital velocity.- The root means square velocity is the square root of the average of the square of the velocity and it has units of velocity. - The RMS velocity is used instead of the average because for a typical gas sample the net velocity is zero since the particles are moving in all directions. This is an important formula as the velocity of the particles determines both the diffusion and effusion rates.

Tardigrade - CET NEET JEE Exam App