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  • Consider an ideal gas with the following distribution of speeds.

Consider an ideal gas with the following distribution of speeds.

Speed % of molecules
200 10
400 20
600 40
800 20
1000 10

 
(i) Calculate V rms and hence T. (m = 3.0 \times 10^{-26} kg)
(ii) If all the molecules with speed 1000 m/s escape from the system, calculate new Vrms and hence T.

Answers (1)

(i) 

\\v^{2} (rms) = \frac{10 \times (200)^{2} + 20(400)^{2} + 40(600)^{2} + 20(800)^{2} + 10(1000)^{2}}{10+20+40+20+10}\\=4.08\times10^{5}

 

\frac{1}{2}m v^{2} (rms)=\frac{3}{2} Kb T

T = \frac{m v^{2} (rms)}{3 Kb} = \frac{3 \times 10^{-26} \times 10^{5} \times 4.08}{3\times 1.38\times 10^{-23}} = 296K

(ii)

Molecules escape at rate of 1000m/s, we can calculate the v2 (rms) as follows

v^{2} (rms) = \frac{10 \times (200)^{2} + 20(400)^{2} + 40(600)^{2} + 20(800)^{2}}{10+20+40+20}

v^2 (rms) = \frac{105[1\times 4 + 2 \times 16 + 4 \times 36 + 2 \times 64]}{90}

v(rms) = \frac{100}{3}\sqrt{ 308} \sim 585 m/s

T = \frac{1}{3} (\frac{m v^{2} (rms)}{Kb}) = \frac{3 \times 10-26 \times (585)2}{3 \times 1.28 \times 10^{-23}} = 284.04 K

Posted by

infoexpert24

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