Density of an ideal gas at 2 atm and 600 K is 2 g/L.
Calculate relative density of this gas with respect to Ne(g) at same conditions :
(Given : R =1/72 atm L/mol.K)

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2.5

answer is A.

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Detailed Solution

$\begin{array}{l} PM = dRT \Rightarrow \\ M_{Gas} = \frac{2 \times 0 .0821 \times 600}{2} = 49 .26 \end{array}$
At const, P and T $d \propto M$
$\frac{d_{Gas}}{d_{Ne}} = \frac{M_{Gas}}{M_{Ne}} = \frac{49 .26}{20} = 2 .463 ≅ 2 .5$

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