Ncert Exercises & Solutions

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 $v_{r m s}$ $v_{r m s}$ and hence T. $(m = 3.0 \times 10^{- 26} k g)$ $(m = 3.0 \times 10^{- 26} k g)$

(b) If all the molecules with a speed of 100 m/s escape from the system, calculate the new $v_{r m s}$ $v_{r m s}$ and hence T.

Hint: The total energy of the molecules depends on the temperature of the gas.

Step 1: Find the RMS speed and temperature of the gas in case 1.

(a) We know that,

v_{r m s}^{2} = \frac{\sum n_{i} v_{i}^{2}}{\sum n_{i}}

$v_{r m s}^{2} = \frac{\sum n_{i} v_{i}^{2}}{\sum n_{i}}$

This is the RMS speed for all molecules collectively.

Now,

v_{r m s} = {(\frac{\sum n_{i} v_{i}^{2}}{\sum n_{i}})}^{\frac{1}{2}}

$v_{r m s} = {(\frac{\sum n_{i} v_{i}^{2}}{\sum n_{i}})}^{\frac{1}{2}}$

= \sqrt{\frac{n_{1} v_{1}^{2} + n_{2} v_{2}^{2} + n_{3} v_{3}^{2} + . . . . . . . . . . + n_{n} v_{n}^{2}}{n_{1} + n_{2} + n_{3} + . . . . . . . + n_{n}}}

$= \sqrt{\frac{n_{1} v_{1}^{2} + n_{2} v_{2}^{2} + n_{3} v_{3}^{2} + . . . . . . . . . . + n_{n} v_{n}^{2}}{n_{1} + n_{2} + n_{3} + . . . . . . . + n_{n}}}$

= \sqrt{\frac{n_{1} v_{1}^{2} + n_{2} v_{2}^{2} + n_{3} v_{3}^{2} + n_{4} v_{4}^{2} + n_{5} v_{5}^{2}}{n_{1} + n_{2} + n_{3} + n_{4} + n_{5}}}

$= \sqrt{\frac{n_{1} v_{1}^{2} + n_{2} v_{2}^{2} + n_{3} v_{3}^{2} + n_{4} v_{4}^{2} + n_{5} v_{5}^{2}}{n_{1} + n_{2} + n_{3} + n_{4} + n_{5}}}$

= \sqrt{\frac{10 \times {(200)}^{2} + 20 \times {(400)}^{2} + 40 \times {(600)}^{2} + 20 \times {(800)}^{2} + 10 \times {(1000)}^{2}}{100}}

$= \sqrt{\frac{10 \times {(200)}^{2} + 20 \times {(400)}^{2} + 40 \times {(600)}^{2} + 20 \times {(800)}^{2} + 10 \times {(1000)}^{2}}{100}}$

= \sqrt{1000 \times (4 + 32 + 144 + 128 + 100)}

$= \sqrt{1000 \times (4 + 32 + 144 + 128 + 100)}$

= \sqrt{408 \times 1000} = 639 m / s

$= \sqrt{408 \times 1000} = 639 m / s$

Now, according to the kinetic theory of gasses,

\frac{1}{2} m v_{r m s}^{2} = \frac{3}{2} k_{B} T

$\frac{1}{2} m v_{r m s}^{2} = \frac{3}{2} k_{B} T$

[

K_{B}

$K_{B}$ = Boltzmann constant, m = mass of gaseous molecules]

$∴ T = \frac{1}{3} \frac{m v_{r m s}^{2}}{k_{B}} = \frac{1}{3} \times \frac{30 \times 10^{- 26} \times 4.08 \times 10^{5}}{1.38 \times 10^{- 23}}$

$= 2.96 \times 10^{2} K = 296 K$

Step 2: Find the RMS speed and temperature of the gas in case 2.

(b) If all the molecules with speed 1000 m/s escape, then,

$v_{r m s}^{2} = \frac{10 \times {(200)}^{2} + 20 \times {(400)}^{2} + 40 \times {(600)}^{2} + 20 \times {(800)}^{2}}{90}$

$= \frac{10 \times 100^{2} \times (1 \times 4 + 2 \times 16 + 4 \times 36 + 2 \times 64)}{90}$

$= 10000 \times \frac{308}{9} = 342 \times 1000 m^{2} / s^{2}$

$v_{r m s} = 584 m / s$

Again,

$T = \frac{1}{3} \frac{m v_{r m s}^{2}}{k_{B}}$

$= \frac{1}{3} \times \frac{3 \times 10^{- 26} \times 10^{5}}{1.38 \times 10^{- 23}}$

$= 2.478 \times 10^{2}$

$= 247.8 = 248 K$

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