Question

Chem Question - Gases

Molar Mass N = 14.01 g/mol
Molar Mass H₂O = 18.016 g/mol
Vapor Pressure of Water at 25 C is 23.76 torr
Vapor Pressure of Water at 65 C is 187.54 torr

(Show all work and calculations, include units in answer. If calculations must be used in several parts, rounding should be made to 6 decimal points to ensure accuracy. Final Answers can be rounded to 3 decimal points.)

A syringe filled with air can be used to test the relationship between the temperature and the volume of a sample of gas. The syringe has a movable plunger that allows an enclosed sample of gas to be held at a constant external pressure,so the gas inside the syringe can be held at a constant total pressure of 1.00 atm while allowing the volume of the gas to increase or decrease.

a) You fill a syringe with 5.0 mL of air at 25°C and 1.00 atm. How many moles of air are contained in the syringe?

b) Assuming that the mole fraction of nitrogen in air is 0.79 and the mole fraction of oxygen in air is 0.21, determine the partial pressure of each gas, and the mass of each gas contained in the syringe.

c) On repeating the experiment, you accidentally trap a small amount of water inside the syringe. (Assume that the water vapor reaches equilibrium inside the syringe, and neglect the volume of the liquid water which will be present.) Given a temperature of 25°C, a volume of 5.0 mL, and a total pressure of 1.00 atm, determine the partial pressure of water vapor and the partial pressure of air (without water vapor) inside the syringe.

d) You increase the temperature of the syringe to 65°C; the total pressure remains 1.00 atm, but the volume increases. Determine the volume of the syringe at the new temperature. (Assume that the water vapor reaches equilibrium at 65°C.)

Answer 1

a)

Volume of air = 5ml, 1000ml =1 L, volume of air, V = 5/1000 L=5*10^-3 L, T =25 deg.c= 25+273= 298K, P= 1atm, R=0.0821 L.atm/mole.K

From gas law, n= no of moles =PV/RT = 1*5*10^-3/ (0.0821*298)=0.000204

1. Air contains 0.21 mole fraction of O2 and 0.79 mole fraction N2. Partial pressure= mole fraction* total pressure, hence partial pressure : O2= 0.21*1= 0.21 atm, N2= 0.79*1=0.79 atm

Molar mass of mixture= 0.21* molar mass of O2+ 0.79* molar mass of N2= 0.21*32+0.79*28=28.84 g/mole

Mass of air = moles of air* molar mass= 0.000204*28.84 gm =0.005883

1 mole of air contains 0.21*32 gm of =6.72 gm of O2 and 0.79*32 = 25.28 gm of N2

Mass of O2 = 0.005803*6.72/28.84= 0.001371 gm and N2=0.005883*25.28/28.84 =0.005157 gm

Partial pressure of air= total pressure- saturation pressure of water

At 25 deg.c, vapor pressure of water= 23.76 mm Hg =23.76/760 atm=0.03126= partial pressure of water vapor since the air is saturated.

Partial pressure of air= 1-0.003126= 0.9687 atm

d) when the temperature is 65 deg,c , partial pressure of air= 760-187.54 mm Hg = 572.46 mm Hg= 572.46/760 atm =0.753 atm

given V1= 5ml, P1= 0.9687 atm, T1= 298 K, P2= 0.753 atm, T2= 65+273= 338K

from gas law, P1V1/T1= P2V2/T2

V2= P1V1T2/(P2T1)= 1* 5*338/(0.753*298)=7.53 ml