JEE Main · 2016 · 03 Apr 2016 · Offline · States of Matter

Two closed bulbs of equal volume (

V

) containing an ideal gas initially at pressure

p_{i}

and temperature

T_{1}

are connected through a narrow tube of negligible volume as shown in the figure below. The temperature of one of the bulbs is then raised to

T_{2}

. The final pressure

p_{f}

Answer: (B)

Solution

The gas begins in both bulbs at

p_{i}

and

T_{1}

, holding a fixed total of

n = \frac{2 p _{i} V}{R T _{1}}

moles. When one bulb is heated to

T_{2}

the gas settles to a common final pressure

p_{f}

, redistributing while the total count stays fixed:

\frac{p _{f} V}{R T _{1}} + \frac{p _{f} V}{R T _{2}} = \frac{2 p _{i} V}{R T _{1}}

. Cancelling

\frac{V}{R}

p_{f} (\frac{1}{T _{1}} + \frac{1}{T _{2}}) = \frac{2 p _{i}}{T _{1}}

, so

p_{f} = \frac{2 p _{i} T _{2}}{T _{1} + T _{2}}

JEE Main · 2016 · 03 Apr 2016 · Offline · Single correct option