Consider an ideal gas with the following distribution of speeds:

Speed $(m/s)$	$\%$ of molecules
$200$	$10$
$400$	$20$
$600$	$40$
$800$	$20$
$1000$	$10$

$(a)$ Calculate $v_{rms}$ and hence $T$. (Given mass of one molecule $m = 3.0 \times 10^{-26} \ kg$, Boltzmann constant $k_B = 1.38 \times 10^{-23} \ J/K$)
$(b)$ If all the molecules with speed $1000 \ m/s$ escape from the system, calculate the new $v_{rms}$ and hence the new $T$.

(N/A) The root mean square speed is given by $v_{rms} = \sqrt{\frac{\sum n_i v_i^2}{\sum n_i}}$.
Using the given data:
$v_{rms} = \sqrt{\frac{10(200)^2 + 20(400)^2 + 40(600)^2 + 20(800)^2 + 10(1000)^2}{10+20+40+20+10}}$
$v_{rms} = \sqrt{\frac{10(4 \times 10^4) + 20(16 \times 10^4) + 40(36 \times 10^4) + 20(64 \times 10^4) + 10(100 \times 10^4)}{100}}$
$v_{rms} = \sqrt{\frac{10^4(40 + 320 + 1440 + 1280 + 1000)}{100}} = \sqrt{408000} \approx 638.75 \ m/s$.
Using $v_{rms} = \sqrt{\frac{3k_B T}{m}}$, we get $T = \frac{m v_{rms}^2}{3k_B} = \frac{3.0 \times 10^{-26} \times (638.75)^2}{3 \times 1.38 \times 10^{-23}} \approx 0.098 \ K$.
$(b)$ Removing molecules with $v = 1000 \ m/s$, the new distribution has $90$ molecules total.
$v_{rms}' = \sqrt{\frac{10(200)^2 + 20(400)^2 + 40(600)^2 + 20(800)^2}{90}}$
$v_{rms}' = \sqrt{\frac{10^4(40 + 320 + 1440 + 1280)}{90}} = \sqrt{\frac{3080000}{90}} \approx 585.18 \ m/s$.
$T' = \frac{m (v_{rms}')^2}{3k_B} = \frac{3.0 \times 10^{-26} \times (585.18)^2}{3 \times 1.38 \times 10^{-23}} \approx 0.082 \ K$.