Question

A 2.0 L-closed container at 300 K holds 100 cm³ of H₂O(l). The equilibrium vapor pressure is found to be 0.0328 atm. What pressure of Ar gas must be placed in the container to double the equilibrium water vapor pressure to 0.0656 atm?

Answer 1

We have equilibrium pressure created by water at the given conditions and we also know that Ar is non dissolvable in water as most of the noble gases follow the same

we have total volume = 2.0L

volume of water = 100 cm³ = 0.1 L

volume of gas over it = 1.9 L

T = 300K

P = 0.0328 atm

lets calculate moles of water vapours = n = PV/RT = 0.0328 atm.1.9L / 0.08206 atm.L/mol.K.300K

= 0.002531481 mol

new pressure = 0.0656 atm

new moles of gases = n' = 0.0656 atm .1.9L / 0.08206 atm.L/mol.K x 300K = 0.005062962 mol

mole fraction of Ar = 0.005062962 - 0.002531481 = 0.002531481mol

mole fraction of Ar = 0.002531481/0.005062962 = 0.5

Pressure due to argon = 0.0656 atm x 0.5 = 0.0328 atm

This could have been simply solved by adding gas pressure as gas pressure is additive in nature

pressure due to Ar = total final pressure needed - pressure due to water = 0.0656 atm - 0.0328 atm = 0.0328 atm