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Two closed vessels of equal volume containing air at pressure $P_1$ and temperature $T_1$ are connected to each other through a narrow tube. If the temperature in one of the vessels is now maintained at $T_1$ and that in the other at $T_2$, what will be the pressure in the vessels
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The correct answer is:
$\frac{2 P_1 T_2}{T_1+T_2}$
$\frac{P_1}{T_1}+\frac{P_1}{T_1}=\frac{P}{T_1}+\frac{P}{T_2}$
$\frac{2 P_1}{T_1}=P\left(\frac{T_1+T_2}{T_1 T_2}\right)_{;} \quad \therefore P=\frac{2 P_1\left(T_1 T_2\right)}{T_1\left(T_1+T_2\right)}=\frac{2 P_1 T_2}{T_1+T_2}$
$\frac{2 P_1}{T_1}=P\left(\frac{T_1+T_2}{T_1 T_2}\right)_{;} \quad \therefore P=\frac{2 P_1\left(T_1 T_2\right)}{T_1\left(T_1+T_2\right)}=\frac{2 P_1 T_2}{T_1+T_2}$
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