Question

# Two vessels each of volume V have air at pressure ${ P }_{ 1 }$ and temperature ${ T }_{ 1 }$. They are connected through a narrow tube and in one of the vessels the temperature changes from ${ T }_{ 1 }$ to ${ T }_{ 2 }$, keeping the other at${ T }_{ 1 }$. Find the pressure of the gas. Assume the volume of gas is negligible.(A) $\dfrac { { 2P }_{ 1 }{ T }_{ 1 } }{ { T }_{ 1 }+{ T }_{ 2 } }$ (B) $\dfrac { { 2P }_{ 1 } }{ { T }_{ 1 }+{ T }_{ 2 } }$(C) $\dfrac { { 2P }_{ 1 }{ T }_{ 2 } }{ { T }_{ 1 }+{ T }_{ 2 } }$(D) None of these

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Hint: Ideal gas equation is given by PV=nRT, in which both n and R are constants , then we will get $\dfrac { PV }{ T }$ as a constant. When both the vessels are connected by a tube , then the sum of initial conditions of both the vessels will be equal to the sum of final conditions which is against a constant. Then, by equating both the equations we will get the value of ${ P }_{ 2 }$.

Complete Step by Step Solution:
Given in the question , that there are two vessels, each of volume V at pressure ${ P }_{ 1 }$ and Temperature${ T }_{ 1 }$, and they have been connected together by a narrow tube.
So, we can name each vessel as A and B respectively.
After connecting, vessel B is changing it’s temperature to ${ T }_{ 2 }$and therefore we need to find the final pressure of both the vessels.
For that, we need to use Ideal Gas Equation,
Which is given by PV=nRT
Where P = pressure
T= temperature
R= universal gas constant
V=volume of the gas
Initially,
For vessel A , ${ P }_{ 1 }{ V }_{ 1 }={ nRT }_{ 1 }$ and for vessel B, ${ P }_{ 1 }{ V }_{ 1 }={ nRT }_{ 1 }$ (since the volume are same in both the vessels , we are taking it as V)
Finally, when both the vessels are connected together, then the temperature of vessel B changes to${ T }_{ 2 }$.
For vessel A , ${ P }_{ 2 }{ V }_{ 1 }={ nRT }_{ 1 }$and for vessel B, ${ P }_{ 2 }{ V }_{ 1 }={ nRT }_{ 2 }$(since the volume are same in both the vessels , we are taking it as V)
We need to find the expression of pressure${ P }_{ 2 }$,
After rearranging the ideal gas equation, we will get $\dfrac { PV }{ T }$= nR, which is a constant.
n is the number of moles which remain constant in both vessels and R is the universal gas constant.
We know that when the vessels are connected together by a narrow tube, it attains an equilibrium state, and so, the sum of initial conditions of both the vessels should be equal to the sum of final conditions of both the vessels.
Hence, $\dfrac { { P }_{ 1 }{ V } }{ { T }_{ 1 } } +\dfrac { { P }_{ 1 }{ V } }{ { T }_{ 1 } } =\dfrac { { P }_{ 2 }{ V } }{ { T }_{ 1 } } +\dfrac { { P }_{ 2 }{ V } }{ { T }_{ 2 } } \\ \Rightarrow \dfrac { { P }_{ 1 }\left( 2V \right) }{ { T }_{ 1 } } =\quad { P }_{ 2 }{ V }\quad \left( \dfrac { 1 }{ { T }_{ 1 } } +\dfrac { 1 }{ { T }_{ 2 } } \right) \\ \Rightarrow \dfrac { { 2P }_{ 1 } }{ { T }_{ 1 } } =\quad { P }_{ 2 }\left( \dfrac { 1 }{ { T }_{ 1 } } +\dfrac { 1 }{ { T }_{ 2 } } \right) \\ \Rightarrow \quad \dfrac { { 2 }P_{ 1 } }{ { T }_{ 1 } } ={ { P }_{ 2 } }\left( \dfrac { { T }_{ 2 }+{ T }_{ 1 } }{ { T }_{ 1 }{ T }_{ 2 } } \right) \\ \Rightarrow { \quad 2P }_{ 1 }={ { P }_{ 2 } }\left( \dfrac { { T }_{ 2 }+{ T }_{ 1 } }{ { T }_{ 2 } } \right) \\ \Rightarrow \quad { P }_{ 2 }=\dfrac { { 2P }_{ 1 }{ T }_{ 2 } }{ \left( { T }_{ 1 }+{ T }_{ 2 } \right) }$

Therefore, the answer is Option (C) $\dfrac { { 2P }_{ 1 }{ T }_{ 2 } }{ { T }_{ 1 }+{ T }_{ 2 } }$

Note: Here, the constant pressure at the end, in both the vessels is obtained by the fact that, while opening the tap, no gas is leaking out of the narrow tube. And this is leading to the constant pressure in both the vessels at the end. Ideal gas equation is the key equation that we are using at the both final and initial conditions.