Compute the temperature at which the r.m.s. speed of nitrogen molecules is $832 m / s$ . [Universal gas constant $R = 8320 J/k mole K$ , molecular weight of nitrogen $= 28$ ].

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Solution

Given:
$V_{rms} = 832 m s^{- 1}$
$R = 8320 J / kmol K$
Molar mass, $M = 28$
R.M.S velocity, $V_{r m s} = \sqrt{\frac{3 R T}{M}}$
Calculation: From formula,
$832 = \sqrt{\frac{3 \times 8320 \times T}{28}}$
$T = \frac{832 \times 832 \times 28}{3 \times 8320} = 776.5 K$

The R.M.S. speed of nitrogen molecules is $832 m s^{- 1}$ at temperature $776.5 K$

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