Assuming ideal behaviour,the ratio of kinetic | vedclass

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Question

(D) The kinetic energy $(KE)$ of an ideal gas is given by the formula $KE = \frac{3}{2} n R T$ (for translational kinetic energy of any ideal gas).Num

Vedclass · Accepted Answer

$12: 1$

Vedclass · Answer

(D) The kinetic energy $(KE)$ of an ideal gas is given by the formula $KE = \frac{3}{2} n R T$ (for translational kinetic energy of any ideal gas).Number of moles of $H_2$ $(n_{H_2})$ = $\frac{3 \ g}{2 \ g/mol} = 1.5 \ mol$.Number of moles of $O_2$ $(n_{O_2})$ = $\frac{4 \ g}{32 \ g/mol} = 0.125 \ mol = \frac{1}{8} \ mol$.Since the temperature $T$ is the same for both,the ratio of kinetic energies is:$\frac{KE_{H_2}}{KE_{O_2}} = \frac{\frac{3}{2} n_{H_2} R T}{\frac{3}{2} n_{O_2} R T} = \frac{n_{H_2}}{n_{O_2}} = \frac{1.5}{0.125} = \frac{1.5 	imes 8}{1} = \frac{12}{1}$.Thus,the ratio is $12: 1$.

Assuming ideal behaviour,the ratio of kinetic energies of $3 \ g$ of $H_2$ and $4 \ g$ of $O_2$ at any temperature is

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