Answer
414.6k+ views
Hint: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 $P_1$ and $T_1$ respectively. The two vessels are connected with narrow open tubes. The final temperature of vessel A and B is $T_1$ and $T_2$ respectively with $T_1$>$T_2$. 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,
${{{P}}_2}{{{V}}_{}}{{ = (n - x)R}}{{{T}}_{{1}}}$ for vessel A and ${{{P}}_2}{{{V}}_{}}{{ = (n + x)R}}{{{T}}_2}$
This gives (n - x) R$T_1$ = (n + x) R$T_2$
x = $\dfrac{{{{n(}}{{{T}}_{{1}}}{{ - }}{{{T}}_{{2}}}{{)}}}}{{{{{T}}_{{1}}}{{ + }}{{{T}}_{{2}}}}}$
Considering the initial temperature, volume and pressure we will get that
$P_1$ X 2V = 2nR$T_1$ => V = $\dfrac{{{{nR}}{{{T}}_1}}}{{{{{P}}_1}}}$
Putting the value of V and x in ${{{P}}_2}{{{V}}_{}}{{ = (n - x)R}}{{{T}}_{{1}}}$ we get,
\[P_A \times \dfrac nRT_1P_1 = (n - \dfrac n(T_1- T_2T_1 + T_2)RT_1\]
Solving this we will get $P_2$ = \[\dfrac{{{{2}}{{{P}}_{{1}}}{{{T}}_{{2}}}}}{{{{{T}}_{{1}}}{{ + }}{{{T}}_{{2}}}}}\].
So, the answer to this question is option B.\[\dfrac{{{{2}}{{{P}}_{{1}}}{{{T}}_{{2}}}}}{{{{{T}}_{{1}}}{{ + }}{{{T}}_{{2}}}}}\].
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.
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 $P_1$ and $T_1$ respectively. The two vessels are connected with narrow open tubes. The final temperature of vessel A and B is $T_1$ and $T_2$ respectively with $T_1$>$T_2$. 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,
${{{P}}_2}{{{V}}_{}}{{ = (n - x)R}}{{{T}}_{{1}}}$ for vessel A and ${{{P}}_2}{{{V}}_{}}{{ = (n + x)R}}{{{T}}_2}$
This gives (n - x) R$T_1$ = (n + x) R$T_2$
x = $\dfrac{{{{n(}}{{{T}}_{{1}}}{{ - }}{{{T}}_{{2}}}{{)}}}}{{{{{T}}_{{1}}}{{ + }}{{{T}}_{{2}}}}}$
Considering the initial temperature, volume and pressure we will get that
$P_1$ X 2V = 2nR$T_1$ => V = $\dfrac{{{{nR}}{{{T}}_1}}}{{{{{P}}_1}}}$
Putting the value of V and x in ${{{P}}_2}{{{V}}_{}}{{ = (n - x)R}}{{{T}}_{{1}}}$ we get,
\[P_A \times \dfrac nRT_1P_1 = (n - \dfrac n(T_1- T_2T_1 + T_2)RT_1\]
Solving this we will get $P_2$ = \[\dfrac{{{{2}}{{{P}}_{{1}}}{{{T}}_{{2}}}}}{{{{{T}}_{{1}}}{{ + }}{{{T}}_{{2}}}}}\].
So, the answer to this question is option B.\[\dfrac{{{{2}}{{{P}}_{{1}}}{{{T}}_{{2}}}}}{{{{{T}}_{{1}}}{{ + }}{{{T}}_{{2}}}}}\].
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.
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