Question

Two closed vessel $A$ and $B$ of equal volume containing air at pressure $P_1$ and temperature $T_1$ are connected to each other through a narrow open tube. If the temperature of one is now maintained at $T_1$ and other at $T_2$ (where $T_1>T_2$) then that what will be the final pressure?

• A. $\frac{T_1}{2P_1{T}_2}$
• B. $\frac{2P_1{T}_2}{T_1+T_2}$
• C. $\frac{2P_1{T}_1}{T_1-T_2}$
• D. $\frac{2P_1}{T_1+T_2}$

Answer 1

Answer: (B)

$\frac{2P_1{T}_2}{T_1+T_2}$

Let $T_1>T_2$; final pressure will be same. Let $x$ mole transfer from $A$ to $B$ vessel.
$\therefore P_{A}V=(n-x)R{T}_1$
and $P_{A}V=(n+x)R{T}_2$
$\therefore x=\frac{n(T_1-T_2)}{T_1+T_2}$;
finally, $P_1\times 2V=2nR{T}_1;V=\frac{nR{T}_1}{P_1}$
$\therefore P_A\times \frac{nR{T}_1}{P_1}=(n-\frac{n(T_1-T_2)}{T_1+T_2})R{T}_1$
$P_A=\frac{2P_1{T}_2}{T_1+T_2}$

Hence, the correct option is $\text{B}$