22.1

Entropy Changes in Reversible ProcessesFor reversible processes the entropy change can be calculated by starting with the expression:

S = dS = dqrev / T

The particular problem will dictate how the right hand integral will be developed.

Phase changes are viewed as occuring at a constant temperature and pressure. If phases are in equilibrium at the particular temperature and pressure then the phase change will be a reversible process.

The normal boiling point of water is 100.0 oC. What is the entropy of vaporization of 1.00 mole of water at 100.0 oC and 1.00 atm?

100.0 o

C H2O (l) -----------> H2O (g) Hvap, 373.2 K = 9,720 cal/mol

1.00 atm

Why is this phase change reversible?

Since the phase change occurs at a constant temperature of 373.2 K and a constant pressure of 1.00 atm, we can calculate the entropy change as follows:

Svap= dqrev / T = qrev, p / T = Hvap, 373.2 K / T

= + 9,720 cal / 373.2 K = + 26.05 cal / K = + 26.05 e.u.

An e.u. or entropy unit is an older, but occassionaly encountered unit, and 1 e.u. 1 cal / K.

22.2

Could you calculate the entropy of the melting of ice at - 5.0 oC and 1.00 atm:

- 5.0 o

C H2O (s) -----------> H2O (l) Hmelt, 268.2 K

1.00 atm

in a manner similar to the calculation we just did, assuming you had a value for Hmelt, 268.2 K , and why or why not?

The normal freezing point of cesium is 28.7 oC. What is the entropy of freezing for 1.00 mole of Cs?

28.7 o

C Cs (l) -----------> Cs (s) Hfreezing, 301.9 K = - 500 cal/mol 1.00 atm

Does the sign on your calculated entropy change make sense?

22.3

3.00 moles of an ideal monatomic gas initially at 100 oC and 10.0 atm expands adiabatically and reversibly to 2.00 atm. What is S for the gas?

S = dS = dqrev / T = 0 / T = 0

If the gas had expanded irreversibly, would the result be the same?

In an adiabatic expansion of a gas, the expansion contributes to an increase in entropy for the gas, while the cooling that accompanies the adiabatic expansion contributes to an entropy decrease for the gas. What can you say about these two contributions to the entropy of the gas in an adiabatic reversible expansion?

22.4

6.00 moles of the ideal gas O2 (g) are isothermally and reversibly

compressed at 55 oC from 1.00 atm to 10.0 atm. What is S for the O2 (g)?

S = dS = dqrev / T = (1/T) dqrev (Justify this step?)

= (1/T) - dwrev (Why were we able to set dqrev = - dwrev?)

= (1/T) P dV (Why is dwrev = - P dV?)

= (1/T) (n R T / V) dV

= n R ln (V2 / V1)

= n R ln (P1 / P2)

= (6.00 moles) (8.314 J / mole K) ln (1.00 atm / 10.0 atm)

= - 115 J / K

Why is the entropy change for the gas negative and is this what you would expect?

22.5

4.00 moles of He are heated at constant volume from 25.0 oC to 100.0 oC. What is S for the He?

All heatings and coolings can be considered to occur reversibly.

S = dS = dqrev, v / T

= dE / T (Can you justify this step?)

= n Cv dT / T

= n 3/2 R dT / T (How did we know that Cv for He

was 3/2 R?)

= n 3/2 R ln (T2 / T1)

= (4.00 moles) 3/2 (8.314 J / mole K) ln (373.2 K / 289.2 K)

= + 11.2 J / K

2.000 moles of solid molybdenum (IV) oxide, MoO2 (s), are cooled

from 500.0 oC to 25.0 oC at a constant pressure of 1.00 atm. The heat capacity of MoO2 (s) is:

Cp [MoO2 (s)] (cal/mole K ) = 16.2 + 3.00x10-3 T - 3.00x10+5 T-2

What is S for the MoO2 (s)?