Question

1.14 mol sample of argon gas at a temperature of 12.0 °C is found to occupy a volume of 30.0 liters. The pressure of this gas sample is ______ mm Hg.

A sample of krypton gas collected at a pressure of 385 mm Hg and a temperature of 284 K has a mass of 94.7 grams. The volume of the sample is _____L.

A 0.238 gram sample of hydrogen gas has a volume of 973 milliliters at a pressure of 2.06 atm. The temperature of theH₂ gas sample is _____°C

Answer 1

We will assume ideal gas conditions to solve this problem.

Ideal gas law: PV = nRT

P = Pressure

V = Volume

n = number of moles

R = Gas constant = 0.0821

T = Temperature

1) 1.14 mol sample of argon gas at a temperature of 12.0 °C is found to occupy a volume of 30.0 liters

V = 30 L

n = 1.14 mol

T = 12 degree C = 285 K

PV = nRT

=> P x 30 = 1.14 x 0.0821 x 285

=> P = 0.889 atm

We know that 1 atm = 760 mm of Hg

=> Pressure (in mm of Hg) = 0.889x 760 = 675.6 mm of Hg

2) A sample of krypton gas collected at a pressure of 385 mm Hg and a temperature of 284 K has a mass of 94.7 grams

P = 385 mm Hg

=> P = (385 / 760) = 0.5066 atm

T = 284 K

m = 94.7 g

We know that molar mass of Kr = 83.8 g / mol

=> moles (n) = 94.7 / 83.8 = 1.13

PV = nRT

=> 0.5066 x V = 1.13 x 0.0821 x 284

=> V = 52.01 L

3) A 0.238 gram sample of hydrogen gas has a volume of 973 milliliters at a pressure of 2.06 atm

P = 2.06 atm

V = 973 mL = 0.973 L

m = 0.238 g

We know that molar mass of H2 = 2 g / mol

=> Moles (n) pf H2 = 0.238 / 2 = 0.119

PV = nRT

=> 2.06 x 0.973 = 0.119 x 0.0821 x T

=> T = 205.16 K = - 67.84 degree C