Answer to Question #197884 in Molecular Physics | Thermodynamics for asim

Question

1. (a) Explain what is meant by a reversible process. Why are real processes not reversible? (5 marks)

(b) A cylinder contains a unit mass of a fluid. The initial pressure is 10 bar and the initial volume is 0.03 m^3 . The fluid expands reversibly behind a piston according to the equation pV^2 = constant until the volume is doubled. The fluid is then cooled reversibly at constant pressure until the piston goes back to its initial position. Heat is then supplied with the piston firmly locked in position until the pressure increases and reaches the initial value of 10 bar.

(i) Sketch the cycle on a labelled pressure-volume diagram. (5 marks)

(ii) Calculate the net work done by the fluid. (15 marks)

Answer 1

1 (a): A reversible process is the sum of all possible changes of parameters (volume, pressure, temperature, etc.) where the new equilibrium state can be converted or returned to the original state and parameters conditions by infinitesimal changes. The real processes are not reversible because the new equilibrium state differs from the original one and this is also limited by the second law of thermodynamics because of the fact that the environment will also be affected when we undergo that process, making it more difficult to reach the start conditions if we try to reverse it.

1 (b): The net work done by the cycle is shown as:

After that we find that the net work done happens only at the first two steps (because the last process occurs at constant volume thus no work is performed), we start to analyze each part of the path

for path A-B: From PV²=K we rearrange to have an expression for pressure to substitute in the equation:

• $W_{AB} = -\int P dV = - \intop (\frac{K}{V^2})dV = - K \intop \frac{dV}{V^2} = \frac{K}{V}$

We substitute using K = P₁V₁² and we evaluate the integral from V₁ to V₂=2V₁

• $W_{AB} =K( \frac{1}{V_2}-\frac{1}{V_1}) = P_1V_1^2( \frac{1}{2V_1}-\frac{1}{V_1}) = -\frac{P_1V_1}{2}$
• $W_{AB} = -\frac{P_1V_1}{2} = -0.150\, bar\cdot m^3$

Then, for path B-C the pressure P₂ is constant so we evaluate the work as

• $W_{BC} = - P_2\intop dV = - P_2 (V_f - V_i)$
• $W_{BC} = - P_2 (V_1-2V_1) = P_2V_1 = 0.075\, bar\cdot m^3$

Then we find the total work done by the fluid by addition:

$W_T = W_{AB} \,+ \, W_{BC} \, = (-0.150+0.075) \, bar\cdot m^3$

$W_T = - \, 0.075 \, \bcancel{bar\cdot m^3} \, * \frac{10^5\,\bcancel{Pa}}{1\,\bcancel{bar}} * \frac{1\,kJ}{10^3\,\bcancel{Pa*m^3}} = - 7.5\,kJ$

1 (b) The net work done is - 7.5 kJ or -7500 J

References:

• Castellan, G. W. (1967). Physical Chemistry, Gilbert W. Castellan (No. 541.3 P49 1964.), p 389.