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Mean Free Path, Collision Frequency

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Question

Calculate the temperature values at which the molecules of the first two members of the homologous series, $C_{n} H_{2 n + 2}$ will have the same rms speed as $C O_{2}$ gas at 770 K. The normal b.p. of n $-$ butane is 273 K. Assuming ideal gas behavior of n $-$ butane up to this temperature, calculate the mean velocity and the most probable velocity of its molecules at this temperature.

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Solution

$v_{r m s} = \sqrt{\frac{3 R T}{M}}$ , given $v_{{r m s}_{C H_{4}}} = v_{{r m s}_{C O_{2}}}$
$v_{{r m s}_{C_{2} H_{6}}} = v_{{r m s}_{C O_{2}}}$
$\Rightarrow \sqrt{\frac{T_{C H_{4}}}{M_{C H_{4}}}} = \sqrt{\frac{T_{C O_{2}}}{M_{C O_{2}}}}; \sqrt{\frac{T_{C_{2} H_{6}}}{M_{C_{2} H_{6}}}} = \sqrt{\frac{T_{C O_{2}}}{M_{C O_{2}}}}$
$T_{C H_{4}} = \frac{770 K \times 16}{44}; T_{C_{2} H_{6}} = \frac{770 K \times 30}{44}$
$T_{C H_{4}} = 280 K; T_{C_{2} H_{6}} = 525 K$
n-butane $= C_{4} H_{10} (58.12 g / m o l)$
Mean velocity $= \sqrt{\frac{8 R T}{Π M}} = 3.157 \times 10^{2} m / s$
Most probable velocity $= \sqrt{\frac{2 R T}{M}} = 2.798 \times 10^{2} m / s$

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