Question

A gas sample containing 0.3525 moles of a compound is trapped in a 2.641 liter vessel at 28.4 °C. What is the pressure inside the vessel? Assume ideality. A sample of a gas was isolated in a gas containment chamber. The volume of the chamber was 1.245 cubic meters. The temperature was 22.80 °C, and the pressure was 88.26 kPa. What volume would this gas sample occupy at STP? Assume ideality. A student obtained the following data for an unknown gas: 6.155 grams of the gas occupied 1484 mL at 27.3 °C and 747.2 torr. Calculate the molar mass of the gas, assuming ideal gas behavior. What is the mole fraction of hydrogen in a gaseous mixture that consists of 8.00 g of hydrogen and 12.00 g of neon in a 3.50 liter container maintained at 35.20 °C?

Answer 1

1.

Number of moles of gas sample, n = 0.3525 mol

Volume of the Vessel i.e. volume of the gas = V = 2.641 L

Temperature of the gas = T =

28.4°C = 28.4 + 273.15 K = 301.55 K

Assuming ideality of the gas, we can solve for the pressure inside the vessel using the ideal gas equation as follows:

PV = nRT

Where R is the gas constant with value of .

R=0.08206L atm K-1 mol-1

Hence, the pressure can be calculated as

PV = nRT nRT 0.3525 mol x 0.08206L atm K-1 mol-1 x 301.55 K 2.641 L 3.303 atm

Hence, the pressure inside the vessel is approximately 3.303 atm.

2.

Volume of the gas chamber = V =

1.245 m

Temperature =

T = 22.80'C = 22.80 + 273.15 K = 295.95 K

Pressure, P = 88.26 kPa = 88260 Pa

Hence, using ideality of the gas, we can calculate the number of moles of gas present as follows:

PV = nRT 88260 Pa x 1.245 m3 " RT 8.314 m3 Pa K-1 mol-1 x 295.95 K →n-^2 44.658 mol

Now, at STP

T=273.15 K

P = 1 atm

Hence, the volume occupied by 44.658 mol of the gas at STP can be calculated as follows

nRT 44.658 mol x 0.08206 L atm K-1 mol-1 x 273.15 K 1 atm 1000. L = 1.000 m3 V

Hence, the Volume of the gas at STP is about 1.000 m^3.

3.

Volume of the gas , V= 1484 mL = 1.484 L

Temperature =

T = 27.3°C = 27.3 + 273.15 K = 300.45 K

Pressure, P = 747.2 torr

Hence, using ideality, the number of moles of gas can be calculated as

747.2 torr x 1.484 L 62,363 L torr K-1 mol-1 x 300.45 K 00.45 0.05918 mol RT

Hence, 0.05918 mol of the gas weighs 6.155 g.

Hence, the mass of 1 mol of gas can be calculated as

6.155 g 0.05918 mol G = 104.0 g/mol

Hence, the molar mass of the gas is approximately 104.0 g/mol.

4.

Amount of Hydrogen gas in the gaseous mixture = 8.00 g

Molar mass of hydrogen gas, H2 = 2.00 g/mol

Hence, number of moles of H2 gas in the mixture is

8 mass "Hemolar mass 8.00 g = 4.00 mol 2.00g/mol

Amount of Neon gas = 12.00 g

Molar mass of Neon = 20.18 g/mol

Hence, number of moles of Neon gas in the mixture is

mass "Ne molar mass 12.00 g 20.18g/mol z = 0.594 mol

Now, the mole fraction of H2 in the mixture can be calculated as

| TH2 nH+mNe 4.00 mol 4.00 mol +0.594 mol

Hence, the mole fraction of H2 gas in the mixture is 0.87.