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Byju's Answer

Standard XII

Chemistry

Ideal Gas Equation

A volume V ...

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 $V / 2$ by a tube and stopcock is opened the temperature of the gas in the second sphere becomes $T_{2}$ . The first sphere is maintained at a temperature $Y_{1}$ . What is the final pressure $p_{1}$ within the apparatus?

\frac{2 p T_{2}}{2 T_{2} + T_{1}}

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\frac{2 p T_{2}}{T_{2} + 2 T_{1}}

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\frac{p T_{2}}{2 T_{2} + T_{1}}

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\frac{2 p T_{2}}{T_{1} + T_{2}}

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Solution

The correct option is A $\frac{2 p T_{2}}{2 T_{2} + T_{1}}$
We have the ideal gas equation as follow:-
$P V = n R T$
The number of moles of gases can be written as:-
$n = \frac{P V}{R T}$
Number of moles present in 1st sphere before stopcock open is
$n = \frac{p V}{R T_{1}}$
Number of moles present in 1st sphere after stopcock open is
$n_{1} = \frac{p_{1} V}{{R T}_{1}}$
Number of moles present in 2st sphere before stopcock open is
$n_{1} = \frac{p_{2} \frac{V}{2}}{R T_{2}}$
Now we have,
$n = n_{1} + n_{2}$

$\frac{p V}{{R T}_{1}} = \frac{p_{1} V}{{R T}_{1}} + \frac{p_{1} V}{{2 R T}_{2}}$
After solving above expression we get final pressure as
$p_{1} = \frac{{2 p T}_{2}}{{2 T}_{2} + T_{1}}$
Correct option is A

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