What is the ratio of kinetic energies of 3 g | vedclass

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(C) The kinetic energy $(KE)$ of an ideal gas is given by the formula: $KE = \frac{3}{2} nRT$. Since the temperature $(T)$ is constant,the kinetic ene

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$12: 1$

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(C) The kinetic energy $(KE)$ of an ideal gas is given by the formula: $KE = \frac{3}{2} nRT$. Since the temperature $(T)$ is constant,the kinetic energy is directly proportional to the number of moles $(n)$,i.e.,$KE \propto n$. Therefore,the ratio of kinetic energies is: $\frac{KE_{H_2}}{KE_{O_2}} = \frac{n_{H_2}}{n_{O_2}}$. Calculate the number of moles for each gas: $n_{H_2} = \frac{	ext{mass}}{	ext{molar mass}} = \frac{3 \ g}{2 \ g/mol} = 1.5 \ mol$. $n_{O_2} = \frac{	ext{mass}}{	ext{molar mass}} = \frac{4 \ g}{32 \ g/mol} = 0.125 \ mol$. Now,find the ratio: $\frac{KE_{H_2}}{KE_{O_2}} = \frac{1.5}{0.125} = \frac{12}{1}$. Thus,the ratio is $12: 1$.

What is the ratio of kinetic energies of $3 \ g$ of hydrogen and $4 \ g$ of oxygen at a certain temperature?

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