Question

A sample of 1.95 mol H2 (Cv = 20.5 J K-1 mol-1) at 21°C and 1.50 atm undergoes a reversible adiabatic compression until the final pressure is 4.50 atm. Calculate the final volume of the gas sample and the work associated with this process. Assume that the gas behaves ideally.

Answer 1

ANSWER:

Given,

Number of moles of H₂ = 1.95 mol

Gas constant, R = 0.082057 L atm mol^-1K^-1

Initial temperature, T₁ = 21 ^oC

Initial pressure, P₁ = 1.50 atm

Final pressure, P₂ = 4.50 atm

Since, gas behaves idially. So,

PV =nRT

Now,

(1.50 atm) V₁ = 1.95 mol x 0.082057 L atm mol^-1K^-1 x 294.15 K

initial volume, V₁ = 31.38 L

And, same gas at different condition follows

P₁V₁ = P₂V₂

1.50 atm x 31.38 L = 4.50 atm x V₂

V₂ = 10.46 L

Hence, the final volume of the gas sample and the work associated with this process is 10.46 L.