Question

A fluid contained in the cylinder receives 150 KJ of work by means of paddle wheel, together with 50 KJ in the form of heat. At the same time, a piston in the cylinder moves in such a way that pressure remains constant at 200 KPa during the fluid expansion from 2 m3 to 5 m3. Find the change in internal energy and change in enthalpy of the fluid.

Answer 1

     Enthalpy H of a system is the total energy associated with it. It is the sum of internal heat energy U and the product of its pressure P and its volume V.

  H = U + P V

Given total energy input = ΔQ = 150 kJ + 50 kJ = 200 kJ

As per the law of thermodynamics :

      ΔQ = ΔU + W

Work done = W = P ΔV = 200 kPa * (5 - 2) m³

    W = 600 kJ

So change in internal energy = ΔU = ΔQ - W = 200 - 600 = - 400 kJ

Change in enthalpy = ΔH = ΔU + Δ(P V) = ΔU + P ΔV ,   as ΔP = 0.

  =>  ΔH = -400 + 200 * (5-2) = 200 kJ

  Same as the total energy given to the system in all forms.

Internal energy depends on temperature.

Answer 2

     Enthalpy H of a system is the total energy associated with it. It is the sum of internal heat energy U and the product of its pressure P and its volume V.

  H = U + P V

Given total energy input = ΔQ = 150 kJ + 50 kJ = 200 kJ

As per the law of thermodynamics :

      ΔQ = ΔU + W

Work done = W = P ΔV = 200 kPa * (5 - 2) m³

    W = 600 kJ

So change in internal energy = ΔU = ΔQ - W = 200 - 600 = - 400 kJ

Change in enthalpy = ΔH = ΔU + Δ(P V) = ΔU + P ΔV ,   as ΔP = 0.

  =>  ΔH = -400 + 200 * (5-2) = 200 kJ

  Same as the total energy given to the system in all forms.

Internal energy depends on temperature.