This gas law conversions practice problem set covers a variety of topics, including converting between different units of pressure, temperature, volume, and mass. The problems also require the use of the ideal gas law and Boyle's law.
Problem 1:
Convert 1.5 atm to kPa.
1 atm = 101.325 kPa
1.5 atm * 101.325 kPa/atm = 152 kPa
Problem 2:
Convert 13 °C to Kelvin.
K = °C + 273
13 °C + 273 = 286 K
Problem 3:
Convert 785 mmHg to atm.
760 mmHg = 1 atm
785 mmHg * 1 atm/760 mmHg = 1.03 atm
Problem 4:
Convert 100 mL to L.
1000 mL = 1 L
100 mL * 1 L/1000 mL = 0.1 L
Problem 5:
An inflated balloon has a volume of 0.55 L at standard pressure of 1 atm and is allowed to rise to a height of 6.5 km, where the pressure is about 0.9 atm. Assuming that the temperature remains constant, what is the final volume of the balloon?
Boyle's law states that P₁V₁ = P₂V₂
P₁ = 1 atm
V₁ = 0.55 L
P₂ = 0.9 atm
V₂ = ?
(1 atm)(0.55 L) = (0.9 atm)(V₂)
V₂ = 0.61 L
Problem 6:
A sample of chlorine gas occupies a volume of 946 mL at a pressure of 760 mmHg. Calculate the pressure of the gas if the volume is reduced to 154 mL. Assume that the temperature remains constant.
Boyle's law states that P₁V₁ = P₂V₂
P₁ = 760 mmHg
V₁ = 946 mL
P₂ = ?
V₂ = 154 mL
(760 mmHg)(946 mL) = (P₂)(154 mL)
P₂ = 4310 mmHg
Problem 7:
The ideal gas law states that PV = nRT, where P is pressure, V is volume, n is the number of moles of gas, R is the ideal gas constant, and T is temperature. The ideal gas constant is equal to 0.0821 L atm/mol K.
Use the ideal gas law to calculate the number of moles of gas in a 5.0 L container at a pressure of 1.0 atm and a temperature of 25 °C.
PV = nRT
n = PV/RT
n = (1.0 atm)(5.0 L)/(0.0821 L atm/mol K)(298 K)
n = 0.20 mol
Problem 8:
Use the ideal gas law to calculate the volume of a 0.50 mol sample of gas at a pressure of 2.0 atm and a temperature of 100 °C.
PV = nRT
V = nRT/P
V = (0.50 mol)(0.0821 L atm/mol K)(373 K)/(2.0 atm)
V = 8.6 L