Case -1: An ideal gas of molecular weight M a | vedclass

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(B) The average kinetic energy $(KE_{avg})$ of an ideal gas is given by $\frac{3}{2}kT$. Since the temperature $T$ is halved in Case $-2$, the average

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Both the averages are halved

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(B) The average kinetic energy $(KE_{avg})$ of an ideal gas is given by $\frac{3}{2}kT$. Since the temperature $T$ is halved in Case $-2$, the average kinetic energy is also halved.The average speed $(C_{avg})$ is given by $\sqrt{\frac{8RT}{\pi M}}$.For Case $-1$: $C_1 \propto \sqrt{\frac{T}{M}}$.For Case $-2$: $C_2 \propto \sqrt{\frac{T/2}{2M}} = \sqrt{\frac{T}{4M}} = \frac{1}{2} \sqrt{\frac{T}{M}}$.Thus, $C_2 = \frac{1}{2} C_1$.Therefore, both the average kinetic energy and the average speed are halved in the second case.

Case $-1:$ An ideal gas of molecular weight $M$ at temperature $T$.
Case $-2:$ Another ideal gas of molecular weight $2M$ at temperature $T/2$.
Identify the correct statement in the context of the above two cases.

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Case $-1:$ An ideal gas of molecular weight $M$ at temperature $T$.Case $-2:$ Another ideal gas of molecular weight $2M$ at temperature $T/2$.Identify the correct statement in the context of the above two cases.