At what temperature, the volume of 'V' of a certain mass of a gas at $37^{o} C$ will be doubled, keeping the pressure constant?

$327^{o} C$

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$347^{o} C$

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$527^{o} C$

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$54^{o} C$

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Solution

The correct option is B
$347^{o} C$

$\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}$
$V_{1} = V, T_{1} = 37 + 273 = 310 k$
$V_{2} = 2 V, T_{2} = ?$
$\frac{V}{310} = \frac{2 V}{T_{2}}$
$T_{2} = 620 K = 347^{o} C$

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