Question

You have equal masses of neon gas and argon gas in a 100 L vessel at 3.731 atm and 25*C.

a. What is the total mass of gas in the vessel?

Consider that the gas is heated to 118.5*C in two possible ways.

a. The gas is heated to 118.5*C in a rigid container such that the volume is constant. Determine the partial pressure of the neon gas in such a container.

b. The gas is heated to 118.5*C in a container fitted with a piston such that the pressure is constant. Determine the density of the gas in such a container at 118.5*C.

Answer 1

a)

mass, assuming ideal gas law:

Apply Ideal Gas Law,

PV = nRT

where

P = absolute pressure

V = total volume of gas

n = moles of gas

T = absolute Temperature

R = ideal gas constant

n = PV/(RT)

n = (3.731)(100)/(0.082*298) = 15.2684 moles

1/2*15.2684 = 7.6342 mol of Ne, Ar

mass of Ne = mol*MW = 7.6342*20.1797 =154.05 g

mass of Ar = mol*MW = 7.6342*39.948 = 304.97 g

total mass = 154.05 +304.97 = 459.02 g of gases

B)

if T = 118.5°C fid:

a) if rigid container

P1/T1 = P2/T2

3.731/(298) = P2 / (118.5+273.15)

P2 = 3.731/(298) * (118.5+273.15) = 4.903 atm

P-Ar = P-N2 = 0.5*4.903 = 2.4515 atm

b)

if piston is present, then P = constant

then

Density = mass/V

Find V:

V1/T1 = V2/T2

V2 = T2/T1*V1 = (118.5+273.15) / 298 * 100 = 131.42 L

V half = 1/2*131.42 = 65.71 L

Now..

D-Neon = mass/V = 154.05 /65.71 = 2.344 g/L

D-Argon = mass/V = 304.97 /65.71 = 4.64115 g/L