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(D) Given that the gases in containers $A$ and $B$ are in thermal equilibrium,their temperatures must be equal,so $T_A = T_B = T$.The ideal gas equati

Vedclass · Accepted Answer

$P_A V_A = P_B V_B$

Vedclass · Answer

(D) Given that the gases in containers $A$ and $B$ are in thermal equilibrium,their temperatures must be equal,so $T_A = T_B = T$.The ideal gas equation is given by $PV = nRT$,where $n$ is the number of moles.The number of moles is defined as $n = m/M$,where $m$ is the mass and $M$ is the molar mass of the gas.For container $A$: $P_A V_A = (m/M_A) RT$.For container $B$: $P_B V_B = (m/M_B) RT$.Since the masses $m$ are equal,we have $P_A V_A / M_A = P_B V_B / M_B$.If the gases are different,$M_A 
eq M_B$,then $P_A V_A$ does not necessarily equal $P_B V_B$. However,if the gases are the same,then $M_A = M_B$,which implies $P_A V_A = P_B V_B$. Since the question asks what 'can' be true,option $D$ is a valid possibility.

Column-$I$	Column-$II$
$(a)$ Kinetic energy per unit mole of gas.	$(i)$ $\frac{1}{2}RT$
$(b)$ Kinetic energy per one molecule of gas.	$(ii)$ $\frac{3}{2}RT$
	$(iii)$ $\frac{3}{2}k_BT$

$A$ gas in container $A$ is in thermal equilibrium with another gas in container $B$. Both contain equal masses of the two gases in the respective containers. Which of the following can be true?

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Column-$I$ represents physical quantity and Column-$II$ represents formula. Match them correctly:
Column-$I$ Column-$II$
$(a)$ Kinetic energy per unit mole of gas. $(i)$ $\frac{1}{2}RT$
$(b)$ Kinetic energy per one molecule of gas. $(ii)$ $\frac{3}{2}RT$
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