Question

Asked in: jee main-2016

Two closed bulbs of equal volume (V) containing an ideal gas initially at pressure pi and temperature T1 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 T2.  The final pressure pf is :

 Your Option is correct !!

A.

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

B.

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

C.

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

D.

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

Answers (1)
P Pankaj Sanodiya

As we know,

Initially:

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

Finally :

n_{1}^{\prime}=\frac{p_{f} V}{R T_{1}} \quad n_{2}^{\prime}=\frac{p_{f} V}{R T_{2}}

Now,

\begin{array}{l}{n_{1}+n_{2}=n_{1}^{1}+n_{2}^{1}} \\ {\frac{p_{i} V}{R T_{1}}+\frac{p_{i} V}{R T_{1}}=\frac{p_{f} V}{R T_{1}}+\frac{p_{f} V}{R T_{2}}} \\ {\frac{2 p_{i} V}{R T_{1}}=\frac{p_{f} V}{R}\left(\frac{T_{1}+T_{2}}{T_{1} T_{2}}\right)} \\ {P_{f}=\frac{2 p_{i} T_{2}}{T_{1}+T_{2}}}\end{array}

Hence option (c) is correct.

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