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(D) The total amount of gas in the system remains constant. Let $V$ be the volume of each bulb and $P_0$ be the initial pressure at $N.T.P.$ $(T_0 = 2

Vedclass · Accepted Answer

$546$

Vedclass · Answer

(D) The total amount of gas in the system remains constant. Let $V$ be the volume of each bulb and $P_0$ be the initial pressure at $N.T.P.$ $(T_0 = 273 \ K)$.Initially,the total number of moles $\mu = \mu_1 + \mu_2 = \frac{P_0 V}{R T_0} + \frac{P_0 V}{R T_0} = \frac{2 P_0 V}{R T_0}$.Finally,the pressure becomes $P = 1.5 P_0$. One bulb is at $T_1 = 273 \ K$ and the other is at $T_2 = T$.The final number of moles $\mu' = \mu_1' + \mu_2' = \frac{P V}{R T_1} + \frac{P V}{R T_2} = \frac{1.5 P_0 V}{R (273)} + \frac{1.5 P_0 V}{R T}$.Since $\mu = \mu'$,we have:$\frac{2 P_0 V}{R (273)} = \frac{1.5 P_0 V}{R (273)} + \frac{1.5 P_0 V}{R T}$Dividing by $\frac{P_0 V}{R}$,we get:$\frac{2}{273} = \frac{1.5}{273} + \frac{1.5}{T}$$\frac{0.5}{273} = \frac{1.5}{T}$$T = 273 	imes 3 = 819 \ K$.Converting to Celsius: $T(^oC) = 819 - 273 = 546^\circ C$.

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.	$(ii)$ $\frac{3}{2}RT$
	$(iii)$ $\frac{3}{2}k_BT$

Two identical glass bulbs are interconnected by a thin glass tube. $A$ gas is filled in these bulbs at $N.T.P.$ If one bulb is placed in ice and another bulb is placed in a hot bath,then the pressure of the gas becomes $1.5$ times. The temperature of the hot bath will be ....... $^oC$

Similar Questions

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

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