A 20 L container holds 0.650 mol of He gas at 37°C at a pressure of 6

1.	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» 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 : P = Final pressure Formula : PV = nRT Calculation : According to ideal gas equation, P₂V₂ = n₂RT₂. ∴ P₂ = \(\frac{n_2RT_2}{V_2}\) = \(\frac{1.90\times 0.0821 \times 450.15}{12}\) = 5.852 atm. ∴ The final pressure of the gas is 5.852 atm. [Note : In the above numerical, converting the pressure value to different units, we get : 5.852 atm = 4447.52 torr = 5.928 bar]

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»

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 :

P = Final pressure

Formula :

PV = nRT

Calculation :

According to ideal gas equation,

P₂V₂ = n₂RT₂.

∴ P₂ = \(\frac{n_2RT_2}{V_2}\)

= \(\frac{1.90\times 0.0821 \times 450.15}{12}\)

= 5.852 atm.

∴ The final pressure of the gas is 5.852 atm.

[Note : In the above numerical, converting the pressure value to different units, we get : 5.852 atm = 4447.52 torr = 5.928 bar]