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Ideal Gas Equation

A gas is encl...

Question

A gas is enclosed in a sphere of volume V at tempreature T and Pressure 'P'. 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 closed. If stopcock is opened the temperature of gas in second sphere becomes $T_{2}$ When the final temperature is $T_{1}$ what is final pressure $^{'} P_{1}^{'}$ within the apparatus

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Solution

Use gaseous equation, $P V = n R T$

In case 1st sphere ,
Volume = V,
Pressure = P,
Temperature = $T_{1}$
So, mole of gas in 1st sphere $= \frac{P V}{n R T_{1}}$
Now stop is opened , then pressure of 1st sphere becomes $P_{!}$
So, mole in 1st sphere $= \frac{P_{1} V}{R T}$

Similarly, in case of 2nd sphere,
Volume $= \frac{V}{2}$
pressure $= P_{1}$
Temperature $= T_{1}$
So, mole in 2nd sphere = $\frac{P_{2} \times \frac{V}{2}}{R T_{2}} = \frac{P_{2} V}{2 R T_{2}}$

Now, total mole of gas = mole of gas in 1st sphere + mole of gas in 2nd sphere.
$\frac{P V}{R T_{1}} = \frac{P_{1} V}{R T_{1}} + \frac{P_{1} V}{2 R T_{2}}$
$⟹ P_{1} = \frac{2 P T_{2}}{2 T_{2} + T_{1}}$

Hence, pressure within the apparatus is $P_{1} = \frac{2 P T_{2}}{2 T_{2} + T_{1}}$

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