First slide

Rms speed

Question

At what temperature will the rms speed of oxygen molecules become just sufficient for escaping from
the earth's atmosphere?
(Take, mass of oxygen molecule,
${m = 2.76 × 10}^{- 26} kg$ and Boltzmann's constant,
$k_{B} = 1.38 \times 10^{- 23} J K^{- 1}$ )

Difficult

A

5.016 $\times$ 10⁴ K
B

8.326 $\times$ 10⁴ K
C

2.508 $\times$ 10⁴ K
D

1.254 $\times$ 10⁴ K

Solution

The minimum velocity with which the body must be projected vertically upwards, so that it could escape from the
earth's atmosphere is its escape velocity (v_e ).
As, $v_{e} = \sqrt{2 g R}$
Substituting the value of $g (9.8 m s^{- 2})$ and radius of earth $(R = 6.4 \times 10^{6} m)$ , we get
   $ν_{e} = \sqrt{2 \times 9.8 \times 6.4 \times 10^{6}}$
   $≅ 11.2 km s^{- 1} = 11.2 \times 10^{3} m s^{- 1}$
Let the temperature of molecule be T when it attains v_e.
According to the question, v_rms =v_e.
where, v_rms is the rms speed of the oxygen molecule
$\Rightarrow \sqrt{\frac{3 k_{B} T}{m_{O_{2}}}} = 11.2 \times 10^{3}$
or    $T = \frac{{(112 \times 10^{3})}^{2} (m_{O_{2}})}{(3 k_{B})}$
Substituting the given values, i.e.
   $k_{B} = 1.38 \times 10^{- 23} J K^{- 1}$
and $m_{O_{2}} = m = 2.76 \times 10^{- 26} k g,$
we get $T = \frac{{(11.2 \times 10^{3})}^{2} (276 \times 10^{- 26})}{(3 \times 1.38 \times 10^{- 23})} ≃ 8.326 \times 10^{4} K$

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