Question

A dry gas occupies 136.5 cm^3 at STP. If the same mass of gas is collected over water at 27 degree Celsius at a total pressure of 725 torr what volume does it occupy

Answer 1

Answer:

For the dry gas at STP :

The volume of the dry gas, V₁ = 136.5 cm³ = 0.1365 L

Standard pressure, P₁ = 1 atm

Temperature, T₁ = 273 K

For the gas collected over the water:

Temperature, T₂ = 27℃ = 27 + 273 = 300 K

The total pressure, Ptotal = 725 torr

We know, at 27℃ the vapour pressure of water, Ph2o = 26.7 torr

By Dalton’s law of partial pressure, we have

Ptotal = P(dry gas) + Ph2o

⇒ P(dry gas) = 725 torr – 26.7 torr = 698.3 torr = 698.7/760 atm = 0.9188 atm [∵ 1 atm = 760 torr]

Also, we are given that the mass of gas is same, therefore, the no. of moles will also be same i.e., n₁ = n₂ = n

Let the volume of the gas collected over the water to be “V₂” and the P(dry gas) be denoted as “P₂”.

Now, by using the ideal gas law, we have

PV = nRT

⇒ nR = PV / T  

∴ [P₁* V₁] / T₁ = [P₂ * V₂] / T₂

⇒ [1 * 0.1365] / 273 = [0.9188 * V₂] / 300

⇒ 0.0005 = 0.00306 * V₂

⇒ V₂ = 0.1633 L ≈ 163.3 cm³

Thus, the volume that the gas occupies is 0.1633 L or 163.3 cm³.