'x' moles of $N_{2}$ gas at S.T.P conditions occupy a volume of 10 litres, then the volume of '2x' moles of $C H_{4}$ at $273^{0} C$ and 1.5 atm is

20 lit

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26.6 lit

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5 lit

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16.6 lit

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Solution

The correct option is B
26.6 lit

$\frac{P_{1} V_{1}}{n_{1} T_{1}} = \frac{P_{2} V_{2}}{n_{2} T_{2}}$

We know that $273^{0} C$ = $546 K$ and STP = $0^{0} C$ i.e 273K.

Therefore, $\frac{1 \times 10}{x \times 273} = \frac{1.5 \times V_{2}}{2 x \times 546}$

$V_{2} = 26.6$ litres

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