When the pressure of $5$ L of $N_{2}$ is doubled and its temperature is raised from $300$ K to $600$ K, the volume of the gas would be:

10

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5

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15

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20

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Solution

The correct option is B $5$ L
Let the initial pressure (p1) = $X a t m$
The final pressure (p2) = $2 X a t m$
Given: $V_{1}$ = $5 L$
$V_{2}$ = ?
$T_{1}$ = $300 K$
$T_{2}$ = $600 K$
Now,
$\frac{P_{1} V_{1}}{T_{1}}$ = $\frac{P_{2} V_{2}}{T_{2}}$

$\frac{X \times 5}{300}$ = $\frac{2 X \times V_{2}}{600}$

$5 X = \frac{2 X \times V_{2}}{2}$
$10 X = 2 X$ x $V_{2}$
$V_{2}$ = $5 L$

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