Question

0.881 Mol sample of helium gas at a temperature of 9.00°C is found to occupy a volume of 28.2 liters. The pressure of this gas sample is ___mm Hg.

A 0.949 gram sample of hydrogen gas has a volume of 856 milliliters at a pressure of 3.10 atm. The temperature of the H2 gas sample is ____°C

Answer 1

Apply ideal gas Law:

PV = nRT

substitute data

P = nRT/V

T = 9C = 9+273 = 282 K

V = 28.2 L

P = (0.881)(0.082)(282)/(28.2) = 0.72242 atm

note that we need mm Hg

so

1 atm = 760 mm HG

so

P = 0.72242 atm * 760 mm Hg / atm = 549.03 mm Hg

Q2.

m = 0.949 g of H2 g

V = 856 mL = 0.856 L

P = 3.10 atm

T = ? in C

MW of H2 = 2 g/mol

so

mol of H2 = mass/MW = 0.949/2 = 0.4745 mol of H2

apply ideal gas law

PV = nRT

T = PV/(RT)

T = (3.1)(0.856)/(0.082*0.4745)

T = 68.20015 K

T = 68.20015-273 = -204.79985 °C