0.740 moles of a gas occupies 20.0 L at 700.0 mm Hg. What will the volume be when four moles of gas are added to the container and the new pressure is 800.0 mm Hg. Assume no change in temperature

Md Ariful Islam

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Md Ariful Islam

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n1 = 0.740 moles

n2 = 0.740 moles + 4.000 moles = 4.740 moles

P1 = 700.0 mm Hg

V1 = 20.0 L

P2 = 800.0 mm Hg

V2 = unknown

T1 = T2 = T = const

Solution:

The Ideal Gas equation can be used.

The Ideal Gas equation can be expressed as: PV = nRT

(1): P1V1 = n1RT

(2): P2V2 = n2RT

Combine equations (1) and (2):

RT = P1V1 / n1 = P2V2 / n2

P1V1n2 = P2V2n1

To find the final volume, solve the equation for V2:

V2 = P1V1n2 / P2n1

V2 = (700.0 mm Hg × 20.0 L × 4.740 mol) / (800.0 mm Hg × 0.740 mol) = 112.095 L = 112 L

V2 = 112 L

Answer: the volume will be 112 liters.

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