Question

At what temperature does the rms speed of oxygen molecules equal the escape velocity of - (i) Earth, (ii) the Moon? Assume gravitational acceleration at the moon is 0.6g.

• A. $5.6\times {10}^{5}K;7.0\times {10}^{5}K$
• B. $2.4\times {10}^{6}K;7.0\times {10}^{3}K$
• C. $3.5\times {10}^{4}K;700K$
• D. $1.6\times {10}^{5}K;7.0\times {10}^{3}K$

Answer 1

The correct option is D

$1.6\times {10}^{5}K;7.0\times {10}^{3}K$

Putting the values of R and gravitational acceleration in the expression of $v_e$,

$v_e^{Earth}=11200m/s$

$v_e^{Moon}=2380m/s$

The molar mass of $O_2$ is 32gms, or 0.032kgs. Now from kinetic theory,

$v_{rms}=\sqrt{\frac{3RT}{M}}$

$\Rightarrow T=\left[\frac{M{v}_{rms}^2}{3R}\right].$

(i) For $v_{rms}$ of $O_2$ to be $v_e^{Earth-}$

$T=\left[\frac{0.032\times (11200{)}^2}{3\times 8.314}\right]K$

$=1.61\times {10}^{5}K$.

(ii) For $v_{rms}$ of $O_2$ to be $v_e^{Moon-}$

$T=\left[\frac{0.032\times (2380{)}^2}{3\times 8.314}\right]K$

$=7.27\times {10}^{3}K$.