Question

If the volume of wet gas collected over water is 78.0 mL at 20 ∘C and 753 mmHg , what is the volume of dry gas at STP conditions? (The vapor pressure of water at 20 ∘C is 17.5 mmHg.)

Answer 1

PV= nRT

P = Pressure in atm                       V= Volume in Liter

n = no of moles               R = 0.0821 L atm K-1 Mol-1

T = Temperature in Kelvin

Using the above formula we can calculate moles of gas

Pgas = 753 mm Hg 17.5 mm Hg = 735.5 mm Hg = 0.9677 atm

V = 78 ml = 0.078 Liter T = 273 + 20 = 293 K

n = PV /RT = 0.9677 atm x 0.078 L / 0.08206 L atm K-1 Mol-1 x 293 K = 0.00313 Moles

Using the moles we can calculate the new volume at STP

P = 1 atm V = ? n =  0.00313 Moles T = 273 K

V = nRT / P = 0.00313 Moles x 0.08206 L atm K-1 Mol-1 x 273 K / 1 atm = 0.07033 Liter or 70.3 ml

Hence volume of dry gas at STP conditions is 70.3 ml