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A cylinder contains hydrogen gas at pressure of 162 kPa and temperature $127^{O} C$ . Its density is : ( $R = 8.3 {Jmol}^{- 1} K^{- 1}$ )

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$0 .2 kg / m^{3}$

$0 .1 kg / m^{3}$

$0 .02 kg / m^{3}$

$0 .5 kg / m^{3}$

answer is B.

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Detailed Solution

$\begin{array}{l} PV = nRT \\ d = \frac{PM}{RT} = \frac{162 \times 10^{3} \times 2 \times 10^{- 3}}{8 .3 \times 400} Kg / m^{3} \\ d = 0.1 Kg / m^{3} \end{array}$

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