Question

Consider a container with a frictionless piston that contains a given amount of an ideal gas. Assume the initial volume of the gas is 7 L, the initial temperature of the gas is 22.1 °C, and the system is in equilibrium with an external pressure of 1.1 bar. In step 1, the gas is cooled reversibly to a final temperature -29.9 °C. The external pressure remains constant at all times. In step 2 the gas is heated at constant volume to a final temperature 4.1 °C.

A) What is the final pressure of the gas?

B) What is the final volume?

C) Calculate w for the overall process (steps 1 + 2)

Answer 1

A)

In step 2 the gas is heated at constant volume to a final temperature 4.1 °C.

V₂ = V₃ = 5.767 L

So,T₃ = 4.1 °C =277.1K

T₂ =-29.9 °C. = 243.1K

Pressure   P₂= 1.1 bar.

Final pressure of the gas P₃ = ?

P₂V₂/T₂ = P₃V₃/T₃

P₃ = P₂V₂T₃/ V₃T₂

P₂ = (1.1 x 5.767 x 277.1)/(5.767 x 243.1) = 1.25 bar

Final pressure of the gas is 1.25 bar

B)

Initial volume of the gas V₁ = 7 L

The initial temperature of the gasT₁ = 22.1 °C = 295.1K

The final temperature T₂ -29.9 °C =243.1K

Now calculate the final volume V₂ by using equation

P₁V₁/T₁ = P₂V₂/T₂

V₂ = P₁V₁T₂ /P₂T₁

(1.1bar x7L x 243.1K)/(1.1bar x 295.1K) = 5.767 L

Final volume = 5.767L

C) Calculate w for the overall process (steps 1 + 2)

W for the steps 1 =

W₁ = P ( V₂ - V₁)

      = 1.1 (5.767- 7.0 ) = - 1.356 KJ

W₂ for the steps 2 = 0

Because the volume is constant in step 2, soW₂ = 0

w for the overall process (steps 1 + 2) = - 1.36 KJ+ 0 = - 1.36 KJ