Question

1) What is the pressure in a 18.2-L cylinder filled with 27.0g of oxygen gas at a temperature of 329K ?

2) A gas mixture with a total pressure of 765mmHg contains each of the following gases at the indicated partial pressures: CO2, 129mmHg ; Ar, 208mmHg ; and O2, 191mmHg . The mixture also contains helium gas.

What mass of helium gas is present in a 12.4-L sample of this mixture at 276K ?

Answer 1

PROBLEM #1

Values:

P= ?

V= 18.2 L

m= 27.0 g O₂ Fw= 32 g.moL^-1 n= 27g/32 g.moL^-1= 0.8437 moL

T= 329 K

R = 0.0821 atm.L.moL^-1.K^-1

P.V = n.R.T

P = (n.R.T)/V = (0.8437 moL. 0.0821 atm.L.moL^-1.K^-1.329K)/18.2 L

P = 1.25 atm

PROBLEM #2

P_t= 765 mmHg

P_^CO2= 129 mmHg

P_^Ar= 208 mmHg

P_^O2= 191 mmHg

PHe= 237 mmHg

Mass (Helium)= ?

T= 276 K

V= 12.4 L

a) How much moles of mixture exist?

P.V = n.R.T

(P.V) = n.(R.T)

n =(P.V)/(R.T)

n = (765 mmHg.12.4 L)/(0.0821 atm.L.moL^-1.K^-1.276 K) _x 1atm/760 mmHg

n_t = 0.5508 moL of mixture

b) How is molar fraction of Helium gas?

X_He= n_He/n_t

X_He= (Mass_He/Fw_He)/n_t

X_He= (Mass_He/(n_txFw_He)

X_He= (Mass_He/(0.5508 moL_x4 g.moL^-1)

X_He= (Mass_He/(2.20 gram))

c) How much is Mass of Helium gas?

P_He = P_t^x X_He

P_He = P_t^x (Mass_He/(2.20 gram))

237 mmHg = 765 mmHg ^_x (Mass_He/(2.20 gram of mixture))

(237 mmHg/765 mmHg) x 2.20 grams = Mass_He

Mass_He =(237 mmHg/765 mmHg) x 2.20 grams

Mass_He =0.68 grams of Helium gas