The total kinetic energy of 0.40 mol of Ar a | vedclass

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(C) The total kinetic energy $(KE)$ of an ideal gas is given by the formula: $KE = \frac{3}{2} nRT$. Given that the total kinetic energy of $Ar$ is eq

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$533$

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(C) The total kinetic energy $(KE)$ of an ideal gas is given by the formula: $KE = \frac{3}{2} nRT$. Given that the total kinetic energy of $Ar$ is equal to the total kinetic energy of $He$: $\frac{3}{2} n_{Ar} R T_{Ar} = \frac{3}{2} n_{He} R T_{He}$. Canceling the common terms $\frac{3}{2} R$ from both sides,we get: $n_{Ar} 	imes T_{Ar} = n_{He} 	imes T_{He}$. Substituting the given values: $0.40 \ mol 	imes 400 \ K = 0.30 \ mol 	imes T_{He}$. $160 = 0.30 	imes T_{He}$. $T_{He} = \frac{160}{0.30} = \frac{1600}{3} \approx 533.33 \ K$. Thus,the temperature is approximately $533 \ K$.

The total kinetic energy of $0.40 \ mol$ of $Ar$ at $400 \ K$ is equal to the total kinetic energy of $0.30 \ mol$ of $He$ at a temperature of .......... $K$.

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