Question

An ideal gas expands from 17.0 L to 64.0 L at a constant pressure of 1.00 atm. Then, the gas is cooled at a constant volume of 64.0 L back to its original temperature. It then contracts back to its original volume without changing temperature. Find the total heat flow, in joules, for the entire process.

Answer 1

The total heat flow is Q = -24.46 J.

Step-by-step explanation:

Transformation 1:

V₁ = 17.0 L

V₂ = 64.0 L

P = 1.00 atm

T₁ → T₂

T₁ < T₂

ΔU₁ = Q₁ + W₁

Transformation 2:

V = V₂ = 64.0 L

T₂ → T₁

T₁ < T₂

ΔU₂ = Q₂ + W₂, W₂ = 0 ∴ ΔU₂ = Q₂

Transformation 3:

T = T₁

V₂ → V₁

ΔU₃ = Q₃ + W₃, ΔU₃ = 0 ∴ Q₃ = - W₃

ΔU is a state function, therefore the total ΔU = 0.

ΔU = ΔU₁ + ΔU₂ + ΔU₃ = 0

Q₁ + W₁ + Q₂ + 0 = 0

Q₁ + Q₂ = W₁

Q₁ + Q₂ = - PΔV

Q₁ + Q₂ = -1.00 atm x (64.0 - 17.0)L

Q₁ + Q₂ = - 47.0 J

Q₃ = -W₃

Q₃ = - nRTln(V₁/V₂)

PV = nRT

nRT = 1.00 atm x 17.0L

Q₃ = - 17 x ln (17/64)

Q₃ = 22.54 J

Q = Q₁ + Q₂ + Q₃

Q = -47.0 + 22.54

Q = -24.46 J