Ncert Exercises & Solutions

The molecules of a given mass of gas have a root mean square speeds of 100 $m s^{- 1}$ at 27°C and 1.0 atmospheric pressure. What will be the root mean square speeds of the molecules of the gas at 127°C and 2.0 atmospheric pressure?

Hint: The root-mean-square speed does not depend on the pressure.

Step 1: Write the formula of the root-mean-square speed.

We know that for a given mass of a gas,

v_{r m s} = \sqrt{\frac{3 R T}{M}}

where R is gas constant.
T is the temperature in kelvin.
M is molar massof the gas.
Clearly,

v_{r m s} \propto \sqrt{T}

Step 2: Find the final RMS speed.

\frac{{(v_{rms})}_{1}}{{(v_{rms})}_{2}} = \sqrt{\frac{T_{1}}{T_{2}}}

{(v_{r m s})}_{1} = 100 m / s

T_{1} = 27 ° C = 27 + 273 = 300 K

T_{2} = 127 ° C = 127 + 273 = 400 K

\frac{100}{{(v_{rms})}_{2}} = \sqrt{\frac{300}{400}} = \frac{\sqrt{3}}{2}

{(v_{r m s})}_{2} = \frac{2 \times 100}{\sqrt{3}} = \frac{300}{\sqrt{3}} m / s

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