A gas in a flexible container initially has a volume of $32.0 L$ .
If we decrease the pressure of the gas from $4.0 a t m$ to $1.0 a t m$ and decrease the temperature from $273$ degrees Celsius to $0$ degrees Celsius, what is the new volume of the gas?

0 L

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32.0 L

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16.0 L

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64.0 L

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8.0 L

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Solution

The correct option is B $64.0 L$
Using the ideal gas equation we have,
$P V = n R T$
$⟹ \frac{P V}{T} = c o n s t a n t$
$⟹ \frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$
$⟹ \frac{4 \times 32}{273 + 273} = \frac{1 \times V_{2}}{0 + 273}$
$⟹ V_{2} = 64 L$

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