If one mole of a gas (mol.wt-40) occupies a volume of 20litres, under the same conditions of temperature and pressure the volume occupied by 2 moles of gas B (mol.wt=80) is

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′x′ moles of N2 gas at S.T.P. conditions occupy a volume of 10 litres, then the volume of ′2x′ moles of CH4 at 273oC and 1.5 atm is:

The correct option is D 26.6 litAs we know that P V=n RTAt STP, for gas 22.4 L/molMolecule of N2=25/26The molecule of CH4 will be = 50/56 PV=nRT V=nRT/P(50/56)(0.08206)(273+273.15)/1.5 in K=26.666 or 27 L Approximately.

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

The correct option is D 26.6 lit
As we know that
P V=n RT
At STP, for gas 22.4 L/mol
Molecule of N2=25/26
The molecule of CH4 will be = 50/56
PV=nRT
V=nRT/P
(50/56)(0.08206)(273+273.15)/1.5 in K
=26.666 or 27 L Approximately.