Question

A 1.65 mol of an ideal gas (Cv=3R/2) at T=14.5 oC and P=0.2 bar undergoes the following two step process: first an isothermal expansion against a constant pressure of 0.1 bar until the volume is doubled; followed by a cooling to -35.6 oC at constant volume. Calculate the following thermodynamic quantities for the total process:

1) Work (w) for step 1.

2) Heat (Q) for step 1.

3) Change in internal energy (U) for step 1.

4) Change in enthalpy (H) for step 1.

5) Change in entropy (S) for step 1.

6) Work (w) for step 2.

7) Heat (Q) for step 2.

8) Change in internal energy (U) for step 2.

9) Change in enthalpy (H) for step 2.

10)Change in entropy (S) for step 2.

11)Change in internal energy (U) for the total process.

12)Change in enthalpy (H) for the total process.

13)Change in entropy (S) for the total process.

Answer 1

For first step

At constant temperature

1) work w = -P(V2-V1)

                 = -0.1 x 2 = -0.2 x 101.33 = -20.266 J

2) q = -w = 20.266 J

3) dU = 0

4) dH = 0

5) dS = 0

For second step

constant volume

6) work w = 0

7) q = Cv(T2-T1) = 1.5 x 8.314(-35.6 - 14.5) = -624.80 J

8) dU = q = -624.80 J

9) dH = CpdT

Cp = Cv + R = 1.5R + R = 2.5R

dH = 2.5 x 8.314 x (-35.6 - 14.5) = -1041.33 J

10) dS = q/T = -624.80/(-35.6 - 14.5 + 273) = 2.80 J/K

11) dH = -1041.33 J

12) dS = 2.80 J/K