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Q.

120 g of an ideal gas of molecular weight 40 is confined to a volume of 20 litre at 400 K, then the pressure of gas is

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a

490 atm

b

4.92 atm

c

2236 atm

d

22.4 atm

answer is B.

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Detailed Solution

$P = \frac{nRT}{V}$

$P = \frac{3 \times 0 .082 \times 400}{20} \Rightarrow P = 4 .92 atm$

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