Question

Solve the following.

A 20 L container holds 0.650 mol of He gas at 37°C at a pressure of 628.3 bar. What will be new pressure inside the container if the volume is reduced to 12 L. The temperature is increased to 177°C and 1.25 mol of additional He gas was added to it?

Answer 1

Given:

V₁ = Initial volume = 20 L,
n₁ = Initial number of moles = 0.650 mol
P₁ = Initial pressure = 628.3 bar
T₁ = Initial temperature = 37°C = 37 + 273.15 K = 310.15 K
n₂ = Final number of moles = 0.650 + 1.25 = 1.90 mol,
V₂ = Final volume = 12 L
T₂ = Final temperature = 177°C = 177 + 273.15 K = 450.15 K,
R = 0.0821 L atm K^–1 mol^–1

To find: P₂ = Final pressure

Formula: PV = nRT

Calculation:

According to the ideal gas equation,

P₂V₂ = n₂RT₂

∴`P_2 = ("n"_2"RT"_2)/"V"_2`

∴`P_2 =(1.90xx0.0821xx450.15)/12`

∴P₂ = 5.852 atm

The final pressure of the gas is 5.852 atm.