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Question: Two closed vessels A and B of equal volume containing air at pressure \(P_1\)​ and temperature T1​ a...

Two closed vessels A and B of equal volume containing air at pressure P1P_1​ and temperature T1​ are connected to each other through a narrow open tube. If the temperature of one is now maintained at T1T_1​ and another at T2T_2​ (where T1T_1​>T2T_2​) then what will be the final pressure?
T12P1T2\dfrac{{{T_1}}}{{2{P_1}{T_2}}},
A.T12P1T2\dfrac{{{{{T}}_{{1}}}}}{{2{{{P}}_{{1}}}{{{T}}_{{2}}}}}
B.2P1T1T1+T2\dfrac{{{{2}}{{{P}}_{{1}}}{{{T}}_{{1}}}}}{{{{{T}}_{{1}}}{{ + }}{{{T}}_{{2}}}}}
C.2P1T12T1T2\dfrac{{{{2}}{{{P}}_{{1}}}{{{T}}_{{{12}}}}}}{{{{{T}}_{{1}}}{{ - }}{{{T}}_{{2}}}}}
D.2P1T1+T2\dfrac{{{{2}}{{{P}}_{{1}}}}}{{{{{T}}_{{1}}}{{ + }}{{{T}}_{{2}}}}}

Explanation

Solution

First find the equation for how many moles are transferred and then find the equation for volume. Then put these two in the initial equations.

Complete step by step answer: Let’s start with understanding the question and for understanding the questions we need to analyse all the given parameters. We are given that
Initial pressure and temperature of two closed vessels is equal and is P1P_1 and T1T_1 respectively. The two vessels are connected with narrow open tubes. The final temperature of vessel A and B is T1T_1 and T2T_2 respectively with T1T_1>T2T_2. Since both the vessels are connected then the final pressure in both the vessels will be the same. Let the final pressure be P.
Let’s assume that x amount of mole is transferred from A to B, now applying the Ideal gas equation we get,
P2V=(nx)RT1{{{P}}_2}{{{V}}_{}}{{ = (n - x)R}}{{{T}}_{{1}}} for vessel A and P2V=(n+x)RT2{{{P}}_2}{{{V}}_{}}{{ = (n + x)R}}{{{T}}_2}
This gives (n - x) RT1T_1 = (n + x) RT2T_2
x = n(T1T2)T1+T2\dfrac{{{{n(}}{{{T}}_{{1}}}{{ - }}{{{T}}_{{2}}}{{)}}}}{{{{{T}}_{{1}}}{{ + }}{{{T}}_{{2}}}}}
Considering the initial temperature, volume and pressure we will get that
P1P_1 X 2V = 2nRT1T_1 => V = nRT1P1\dfrac{{{{nR}}{{{T}}_1}}}{{{{{P}}_1}}}
Putting the value of V and x in P2V=(nx)RT1{{{P}}_2}{{{V}}_{}}{{ = (n - x)R}}{{{T}}_{{1}}} we get,
PA×nRT1P1=(nn(T1T2T1+T2)RT1P_A \times \dfrac nRT_1P_1 = (n - \dfrac n(T_1- T_2T_1 + T_2)RT_1
Solving this we will get P2P_2 = 2P1T2T1+T2\dfrac{{{{2}}{{{P}}_{{1}}}{{{T}}_{{2}}}}}{{{{{T}}_{{1}}}{{ + }}{{{T}}_{{2}}}}}.
So, the answer to this question is option B. 2P1T2T1+T2\dfrac{{{{2}}{{{P}}_{{1}}}{{{T}}_{{2}}}}}{{{{{T}}_{{1}}}{{ + }}{{{T}}_{{2}}}}}.

Note: Diffusion is the concept which is largely studied in many fields to get an idea about how different glasses and liquids diffuse into each other. One of the most famous experiments which has been performed by almost everyone is that when you spray the perfume in one corner of the room the smell of the perfume spreads at every corner of the room. This is caused by diffusion of perfume molecules in the air.