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(D) The total kinetic energy $(KE)$ of an ideal gas is given by the formula: $KE = \frac{3}{2} n R T$.For the kinetic energies to be equal,we set the

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$533$

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(D) The total kinetic energy $(KE)$ of an ideal gas is given by the formula: $KE = \frac{3}{2} n R T$.For the kinetic energies to be equal,we set the expressions for $He$ and $Ar$ equal to each other:$\frac{3}{2} n_{He} R T_{He} = \frac{3}{2} n_{Ar} R T_{Ar}$.Canceling the common terms $\frac{3}{2}$ and $R$,we get: $n_{He} 	imes T_{He} = n_{Ar} 	imes T_{Ar}$.Given: $n_{He} = 0.30 \ mol$,$n_{Ar} = 0.40 \ mol$,and $T_{Ar} = 400 \ K$.Substituting the values: $0.30 	imes T_{He} = 0.40 	imes 400$.$T_{He} = \frac{0.40 	imes 400}{0.30} = \frac{160}{0.30} = 533.33 \ K \approx 533 \ K$.

At what temperature will the total kinetic energy of $0.30 \ mol$ of $He$ be the same as the total kinetic energy of $0.40 \ mol$ of $Ar$ at $400 \ K$ (in $K$)?

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