An ideal gas initially at 600 K and 10 bar undergoes a four-step mechanically reversible cycle in a closed system. In step 12, pressure decreases isothermally to 3 bar; in step 23, pressure decreases at constant volume to 2 bar; in step 34, volume decreases at constant pressure; and in step 41, the gas returns adiabatically to its initial state.
(a) Sketch the cycle on a PV diagram.
(b) Determine (where unknown) both T and P for states 1, 2, 3, and 4.
(c) Calculate Q, W, ?U, and ?H for each step of the cycle.
(a) A rough sketch is given below. We could refine it after doing the calculations.
(b) For state 1, we are given that 
At these conditions, the molar volume is
Since step 12 is isothermal, we also have that 
and it is specified that 
is 
At
these conditions, the molar volume is 
In step 23, 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 
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.
The molar volume is then 
(c) Step 12 is isothermal so ?U = ?H = 0, and Q = -W = RT ln(V2/V1) = 
For step 23, we can, as always, compute ?U and ?H from
At constant volume, W = 0, and Q = ?U =-4157 J/mol.
For step 34, we again compute ?U and ?H from
?U =Cv ?T = 2.5 R*(378.9 K - 400 K) =2.5*8.314 J mol-1K-1 *(-21.1 K) = -438.56 J/mol, and
?H =Cp ?T = 3.5 R*(378.9 K-400 K) =3.5*8.314 J mol-1K-1 *(-21.1 K) = -613.98 J/mol.
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.
For step 41, we again compute ?U and ?H from
?U =Cv ?T = 2.5 R*(600 K-378.9 K) =2.5*8.314 J mol-1K-1 *(221.1 K) = 4595.56 J/mol, and
?H =Cp ?T = 3.5 R*(600 K-378.9 K) =3.5*8.314 J mol-1K-1 *(221.1 K) = 6433.78 J/mol.
This step is adiabatic, so Q = 0, and W = ?U = 4595.56 J/mol.
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