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Question: 120 g of an ideal gas of molecular weight 40 \(mole^{- 1}\) are confined to a volume of 20 L at 400 ...

120 g of an ideal gas of molecular weight 40 mole−1mole^{- 1} are confined to a volume of 20 L at 400 K. Using R=0.0821LatmK−1mole−1R = 0.0821LatmK^{- 1}mole^{- 1}, the pressure of the gas is

A

4.90 atm

B

4.92 atm

C

5.02 atm

D

4.96 atm

Answer

4.92 atm

Explanation

Solution

120g=12040=3moles120g = \frac{120}{40} = 3moles

P=nRTV=3×0.0821×40020=4.92atm.P = \frac{nRT}{V} = \frac{3 \times 0.0821 \times 400}{20} = 4.92atm.