An ideal gas initially at 600 K and 10 bar undergoes a four-step mechanically reversible cycle in a closed system. Pressure decreases isothermally to 3 bar; Pressure decreases at constant volume to 2 bar; Volume decreases at constant pressure;The gas returns adiabatically to its initial state

Question

Take = (7/2)R and = (5/2)R.

(a) Sketch the cycle on a PV diagram.

(b) Determine (where unknown) both T and P states 1, 2, 3, and 4.

(c) Calculate Q, W, ?U, and ?H for each step of the cycle.

Answer 1

a. A rough sketch is given below. We could refine it after doing the calculations.



b. For state 1, We are given that = 600 K and = 10 bar = Pa. At these conditions, the molar volume is



we also have that = 600 K, and it is specified that is 3 bar At

these conditions, the molar volume is



the pressure is reduced from 3 bar to 2 bar by cooling at constant volume. Decreasing The pressure by 2/3 requires decreasing the temperature by 2/3 at constant volume, so

= 400 K and = 2 bar = Pa, while



Finally, we can find the conditions for state 4 from the fact that it can go back to state 1 via adiabatic compression. For an adiabatic process on an ideal gas, we have 0.6314.

So, = 0.6314*600 K = 378.9 K. The molar volume is then





c. So

we can, as always, compute ?U and ?H from

and

At constant volume, W = 0, and Q = ?U =-4157 J/mol.

we again compute ?U and ?H from

and

At constant pressure, Q = ?H = -613.98 J/mol.

Since ?U = Q + W (in general), we have

W = ?U – Q = -438.56 J/mol – (-613.98 J/mol_=175.42 J/mol.

We again compute ?U and ?H from

and

This step is adiabatic, so Q = 0, and W = ?U = 4595.56 J/mol.