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Gas laws from KTG

A volume V of...

Question

A volume V of a gas at a temperature $T_{1}$ , and a pressure p is enclosed in a sphere. It is connected to another sphere of volume $\frac{V}{2}$ by a tube and stopcock. The second sphere is initially evacuated and the stopcock is closed. If the stopcock is opened the temperature of the gas in the second sphere becomes $T_{2}$ . The first sphere is maintained at a temperature $T_{1}$ . What is the final pressure $p_{1}$ within the apparatus ?

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Solution

Using ideal gas equation,

$P V = n R T$

$\Rightarrow n = \frac{P V}{R T}$

Case I:- When the stopcock is closed.

Pressure enclosed in sphere $= p$

Temperature of sphere $= T_{1}$

Volume $= V$

$∴$ total moles of gas $(n) = \frac{p V}{R T_{1}}$

Case II:- When the stopcock is opened.

Pressure within the apparatus, i.e., final pressure $= p_{1}$

Temperature of first sphere $= T_{1}$

Temperature of second sphere $= T_{2}$

Volume of first sphere $= V$

Volume of second sphere $= \frac{V}{2}$

$∴$ Moles of gas in first sphere $(n_{1}) = \frac{p_{1} V}{R T_{1}}$

$∴$ Moles of gas in second sphere $(n_{2}) = \frac{p_{1} \frac{V}{2}}{R T_{2}} = \frac{p_{1} V}{2 R T_{2}}$

Now,

Total moles of gas = moles of gas in first sphere + moles of gas in second sphere

$n = n_{1} + n_{2}$

$\frac{p V}{R T_{1}} = \frac{p_{1} V}{R T_{1}} + \frac{p_{1} V}{2 R T_{2}}$

$\Rightarrow p_{1} = \frac{2 p T_{2}}{2 T_{2} + T_{1}}$

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