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Gay Lussac's Law

The density o...

Question

The density of methane at $0.20$ atm pressure and $27^{o} C$ is:

0.13 g

L^{- 1}

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0.26 g

L^{- 1}

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45.67 g

L^{- 1}

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26.0 g

L^{- 1}

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Solution

The correct option is B $45.67 g$ $L^{- 1}$
Given $P = 0.2 a t m$ $T = 27 + 273 = 300 k$ $R = 0.0821$

$M = C H_{4} = 12 + 1 \times 4 = 16 g l m o l$ $D = ?$

$D = \frac{P M}{R T} = \frac{0.2 \times 300}{0.821 \times 16} = \frac{60}{1.3136}$

$D = 45.676 g L^{- 1}$

Hence, the correct option is C.

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The density of a gas is 2 $\frac{g}{L}$ at 1 atm and $27^{0} C$ . The density of the same gas at 2 atm and $127^{0} C$ is:

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