Question

Calculate the temperature values at which the molecules of the first two members of the homologous series, $C_n{H}_{2n+2}$ 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.

Answer 1

$v_{rms}=\sqrt{\frac{3RT}{M}}$, given $v_{{rms}_{C{H}_4}}=v_{{rms}_{C{O}_2}}$
$v_{{rms}_{C_2{H}_6}}=v_{{rms}_{C{O}_2}}$
$\Rightarrow \sqrt{\frac{T_{C{H}_4}}{M_{C{H}_4}}}=\sqrt{\frac{T_{C{O}_2}}{M_{C{O}_2}}};\sqrt{\frac{T_{C_2{H}_6}}{M_{C_2{H}_6}}}=\sqrt{\frac{T_{C{O}_2}}{M_{C{O}_2}}}$
$T_{C{H}_4}=\frac{770K\times 16}{44};T_{C_2{H}_6}=\frac{770K\times 30}{44}$
$T_{C{H}_4}=280K;T_{C_2{H}_6}=525K$
n-butane$=C_4{H}_{10}\left(58.12g/mol\right)$
Mean velocity$=\sqrt{\frac{8RT}{\Pi M}}=3.157\times {10}^{2}m/s$
Most probable velocity$=\sqrt{\frac{2RT}{M}}=2.798\times {10}^{2}m/s$