: A gas contained in a piston–cylinder assembly undergoes two processes, A and B, between the same end states, 1 and 2, where p1 =1bar, V1 = 1m3 , U1 =400kJ and p2 = 10 bar, V2 = 0.1 m3 , U2 = 450 kJ: Process A: Constant-volume process from state 1 to a pressure of 10 bar, followed by a constantpressure process to state 2. Process B: Process from 1 to 2 during which the pressure-volume relation is pV = constant. Kinetic and potential effects can be ignored. For each of the processes A and B, (a) sketch the process on p–V coordinates, (b) evaluate the work, in kJ, and (c) evaluate the heat transfer, in kJ.

b) W = 9.09 10² kJ ,  W = -2.3 nRT

c)  Q_total = -409 kJ ,   Q = 232 kJ

Explanation:

a) we have an outline of the two processes in the attached

Process A

        Isocoric - Isobaric

B process

        Isothermal

b) This is a thermodynamic process where we can find work with the expression

         W = ∫P dV

A Process  

In the constant volume part ΔV = 0, so the work is zero

            W₁ = 0

In the constant pressure part

        W = P DV

Let's reduce the pressure to SI units

       P = 10 Bar (1.01 105 Pa / 1 Bar) = 10.1 10⁵ Pa

        W₂ = 10.1 10⁵ (1 -0.1)

        W₂ = 9.09 10⁵ J= 9.09 10² kJ

The total work is

      W = W₁ + W₂

      W = 9.09 10² kJ

B process

       PV = d

Where d is a constant, we assume an ideal gas

          PV = nRT

        W = nRT ∫ dV / V

        W = nRT ln V₂ / V₁

        W = nrT ln 0.1 / 1

        W = -2.3 nRT

c) Let's use the first law of thermodynamics

         DU = Q-W

Process A

The first part of the process is at constant volume, so the work is zero

        ΔU₁ = Q₁

        Q₁ = 450 -400

        Q₁ = 50 kJ

The second part

        ΔU₂ = Q₂ + W₂

        Q₂ = ΔU₂ - W₂

        Q₂ = 450 - 909

         Q₂ = -459 kJ

The total heat in the process is

        Q_total = Q₁ + Q₂

         Q_total = 50 - 459

         Q_total = -409 kJ

The negative sign indicates that this heat is released

B process

       ΔU = Q + W

       Q = ΔU - W

       Q = (450-400) - (-2.3 d)

        Q = 50 + 2.3 d

let's use the initial values to calculate the constant

        P V = d

         1.01 10⁵ 1 = d

        Q = 50 + 2.3 1.01 105

         Q = 2.3 10⁵ J

         Q = 2.323 10² kJ

less than one you can't create energy

no sign of dia of fuel cell