Two containers of equal volume contain the same gas at different pressures $P_{1}$ and $P_{2}$ and absolute temperatures $T_{1}$ and $T_{2}$ , respectively. On joining the vessels, the gas reaches a common pressure $P$ and common temperature $T$ . The ratio $\frac{P}{T}$ is equal to

\frac{P_{1}}{T_{1}} + \frac{P_{2}}{T_{2}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{P_{1} T_{1} + P_{2} T_{2}}{(T_{1} + T_{2})^{2}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{P_{1} T_{2} + P_{2} T_{1}}{(T_{1} + T_{2})^{2}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{P_{1}}{2 T_{1}} + \frac{P_{2}}{2 T_{2}}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $\frac{P_{1}}{2 T_{1}} + \frac{P_{2}}{2 T_{2}}$
From the ideal gas equation,
$P V = n R T$
We can write,
Number of moles in the first vessel
$n_{1} = \frac{P_{1} V}{R T_{1}}$
Number of moles in the second vessel
$n_{2} = \frac{P_{2} V}{R T_{2}}$

On joining the vessels, volume becomes double i.e $2 V$ .
Total number of moles of the gas $n = n_{1} + n_{2}$
$\Rightarrow \frac{P (2 V)}{R T} = \frac{P_{1} V}{R T_{1}} + \frac{P_{2} V}{R T_{2}}$
$\Rightarrow \frac{P}{T} = \frac{P_{1}}{2 T_{1}} + \frac{P_{2}}{2 T_{2}}$
Thus, option (d) is the correct answer.

Suggest Corrections

Similar questions

Two containers of equal volume contain the same gas at pressures $P_{1}$ and $P_{2}$ and absolute temperatures $T_{1}$ and $T_{2}$ respectively. On joining the vessels, the gas reaches a common pressure P and common temperature T. The ratio $\frac{P}{T}$ is equal to