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(D) Let the initial pressure be $P_0$ and the volume of each bulb be $V$. The initial temperature of both bulbs is $T_0 = 273 \ K$.The total number of

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$546$

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(D) Let the initial pressure be $P_0$ and the volume of each bulb be $V$. The initial temperature of both bulbs is $T_0 = 273 \ K$.The total number of moles of gas is constant:$n_{total} = n_1 + n_2 = \frac{P_0 V}{R T_0} + \frac{P_0 V}{R T_0} = \frac{2 P_0 V}{R T_0}$When one bulb is in ice $(T_1 = 273 \ K)$ and the other is in a hot bath $(T_2 = T)$,the new pressure is $P = 1.5 P_0$.The new number of moles is:$n'_{total} = \frac{P V}{R T_1} + \frac{P V}{R T_2} = \frac{1.5 P_0 V}{R (273)} + \frac{1.5 P_0 V}{R T}$Since $n_{total} = n'_{total}$:$\frac{2 P_0 V}{R (273)} = \frac{1.5 P_0 V}{R (273)} + \frac{1.5 P_0 V}{R T}$Dividing both sides by $\frac{P_0 V}{R}$:$\frac{2}{273} = \frac{1.5}{273} + \frac{1.5}{T}$$\frac{0.5}{273} = \frac{1.5}{T}$$T = \frac{1.5 	imes 273}{0.5} = 3 	imes 273 = 819 \ K$Converting to Celsius:$T(^{\circ}C) = 819 - 273 = 546^{\circ}C$.

Similar Questions

One mole of an ideal gas passes through a process where pressure and volume obey the relation $P = P_0 \left[ 1 - \frac{1}{2} \left( \frac{V_0}{V} \right)^2 \right]$. Here $P_0$ and $V_0$ are constants. Calculate the change in the temperature of the gas if its volume changes from $V_0$ to $2V_0$.

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By using the ideal gas equation $PV = \mu RT$,write the value,unit,and dimension of $R$.

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