Two closed bulbs of equal volume $(V)$ containing an ideal gas initially at pressure $μ$ 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 $p_{f}$ is :

2 p_{i} (\frac{T_{1}}{T_{1} + T_{2}})

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2 p_{i} (\frac{T_{2}}{T_{1} + T_{2}})

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2 p_{i} (\frac{T_{1} T_{2}}{T_{1} + T_{2}})

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p_{i} (\frac{T_{1} T_{2}}{T_{1} + T_{2}})

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Solution

The correct option is B $2 p_{i} (\frac{T_{2}}{T_{1} + T_{2}})$
Initially, the number of moles of gas in each bulb is

$n_{1} = \frac{P_{i} V}{R T_{1}}$ and $n_{2} = \frac{P_{i} V}{R T_{1}}$

After the temperature of the second bulb is raised to $T_{2}$ then the number of moles of gas in both the bulbs are

$n_{1}^{^{'}} = \frac{P_{f} V}{R T_{1}}$ and $n_{2}^{^{'}} = \frac{P_{f} V}{R T_{2}}$

Now, the total number of moles of gas in both the bulbs remains the same in both cases.

$n_{1} + n_{2} = n_{1}^{^{'}} + n_{2}^{^{'}}$

$2 \frac{P_{i} V}{R T_{1}} = \frac{P_{f} V}{R T_{1}} + \frac{P_{f} V}{R T_{2}}$

$\Rightarrow 2 \frac{P_{i} V}{R T_{1}} = \frac{P_{f} V}{R} (\frac{T_{2} + T_{1}}{T_{1} T_{2}})$

$P_{f} = \frac{2 P_{i} T_{2}}{T_{1} + T_{2}}$

Option B is correct.

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Similar questions

Q. 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}

is :

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 $p_{f}$ is: