Question

5 moles of ideal gas is initially at 300 K and 5 bar. It is compressed to 10 bar at 300 K. This change is carried out by two different reversible processes:

A: heating at constant volume followed by cooling at constant pressure

B: cooling at constant pressure followed by heating at constant volume

Depict these processes on a PT graph. (Hand drawn on engineering sheet would suffice). Calculate ΔU, ΔH, Q, and W requirements for each path. Cv=20.78 J/mol.K and Cp=29.10 J/mol.K.

Answer 1

5 moles at 300 K and 5 bar, the volume of gas is

V= (5 mol x 300 K x 0.082 L.atm/K.mol) / 4.9346 atm

V= 24.92 L = 0.02492 m³

5 moles at 300 K and 10 bar, the volume of gas is

V= (5 mol x 300 K x 0.082 L.atm/K.mol) / 9.8692 atm

V= 12.46 L = 0.01246 m³

a) In the first step the gas is heated at a constant volume equal to its initial valve until the final pressure of 10 bar is reached.

The gas temperature at the end of this step is:

T´= T₁ (P₂/P₁) = 300 x (10/5) = 600 K

For this step the volume is constant

Q = ΔU = Cv ΔT = 20.78 x (600-300) = 6234 J

During the second step the gas is cooled at the constant pressure of 10 bar to its final state

Q = ΔH = Cp ΔT = 29.10 x (300-600) = -8730 J

Also, ΔU = ΔH – Δ(PV) = ΔH – P ΔV = -8730 – (2x10⁶)x(0.01246-0.02492) = -6234 J

For the two steps combined

Q = 6234 -8730 = -2496 J

ΔU = 6234 – 6234 = 0 J

W = ΔU – Q = 0 – (-2496) = 2496 J

ΔH = ΔU = 0 J

b) During first step the gas is cooled at the constant pressure of 5 bar until the final volume of 0.01246 m³ is reached.

The temperature of the gas at the end of this cooling step is:

T´= T₁ (V₂/V₁) = 300 x (0.01246/0.02492) = 150 K

Q = ΔH = Cp ΔT = 29.10 x (150-300) = -4365 J

Also, Q = ΔH – Δ(PV) = ΔH – P ΔV = -4365 – (5x10⁵)x(0.01246-0.02492) = -3117 J

During the second step the volume is held constant at while the gas is heated to its final state

ΔU = Q = Cv ΔT = 20.78 x (300-150) = 3117 J

The complete process represent the sum of its steps

Hence: Q = - 4365 + 3117 = -1248 J

          ΔU = - 3117 + 3117 = 0 J

            Q + W = 0 à W = - Q = 1248 J

          ΔH = ΔU = 0 J