Question

A piston-cylinder device initially at 0.4-m3 contains nitrogen gas at 600 kPa and 300 K. Now the gas is compressed isothermally to a volume of 0.2 m3. The work done on the gas during this compression process is _____ kJ.

Answer 1

Answer:

W = 166.35 KJ

Explanation:

P₁ = 600 KPa

V₁ = 0.4 m³

V₂ = 0.2 m³

T = 300 K

W = ?

We can find the solution from the standard relation for work in an isothermal process

W = -n*R*T*Ln (Vf / Vi)

We know that

n*R*T = P*V  ⇒    P₁*V₁ = P₂*V₂ = n*R*T = 600 *0.4 = 240

Now, we use the equation

W = -n*R*T*Ln (Vf / Vi) = - P₁*V₁*Ln (Vf / Vi)

⇒   W = -240*Ln (0.2 / 0.4) = 166.35 KJ