Problem Set 7 Solutions for Chemistry 3321, Exercises of Physical Chemistry

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Problem Set 7

Solutions

Due 10-30-

14 pts.

1. ∆G for Change in P (2 pts)

Calculate the change in Gibbs energy of 35 g of ethanol (mass density 0.789 g cm

− 3 ) when

the pressure is increased isothermally from 1 atm to 3000. atm. Report your answer to 2

significant figures.

GIVEN: ASSUME incompressible fluid

m = 35 g ⇒ 2 Sig Figs

ρ = 0. 789 g cm

− 3

V =

m ρ

35 g

1. 789 g cm−^3

= 44. 35995 cm

3 = 4. 435995 × 10

− 5 m

3

isothermal, ∆T = 0, dT = 0

Pi = 1 atm Pf = 3000 atm

∆P = 2999 atm = 3. 038737 × 10

8 P a

FIND: ∆G(P )

Start with the Fundamental equation for G

dG = −S dT + V dP

isothermal, so dT = 0

dG = +V dP

integrate

∆G = +

Pf ∫

Pi

V dP

V is constant since we are assuming incompressible fluid

Therefore, V comes out of the integral

∆G = +V ∆P

= 4. 435995 × 10

− 5 m

3 × 3. 038737 × 10

8 P a

= +1. 347982 × 10

4 P a m

3 = +1. 347982 × 10

4 J

∆G = +1. 3 × 10

4 J = +1. 3 kJ

3. Phase Diagram (2 pts)

Use the phase diagram below to state what would be observed when a sample of carbon

dioxide, initially at 1.0 atm and 298 K, is subjected to the following cycle: (a) isobaric

heating to 320. K, (b) isothermal compression to 100. atm, (c) isobaric cooling to 210. K, (d)

isothermal decompression to 1.0 atm, (e) isobaric heating to 298 K.

(a) Gas is heated above Tc, but the pressure is still well below Pc, therefore, the sample

remains a gas. There is NO phase transition.

(b) The pressure is increased above Pc and temperature remains above Tc. Therefore the gas

slowly becomes a supercritical fluid. There is NO phase transition.

(c) The temperature is lowered below Tc. Initially, the supercritical fluid will become a liquid

without a phase transition occurring. Then, as the sample is cooled further, the liquid

eventually becomes a solid in a phase transition.

(d) The pressure is lowered below the solid–gas coexistence curve. The solid sample sublimes

to gas around 3.5 atm. This IS a phase transition.

(e) The gas is heated to return to its original state. There is NO phase transition.

4. ∆f usH and ∆f usS (2 pts)

The molar volume of a certain solid is 161. 0 cm

3 mol

− 1 at 1. 00 atm and 350.75 K, it’s melting

temperature. The molar volume of the liquid at this temperature and pressure is 163. 3 cm

3 mol

− 1 .

At 100. atm the melting temperature changes to 351.26 K. Calculate the enthalpy and entropy

of fusion of the solid.

GIVEN:

P = 1. 00 atm = 101325 P a ⇒ 3 Sig Figs

T = 350. 75 K

V (^) sol = 161. 0 cm

3 mol

− 1

V (^) liq = 163. 3 cm

3 mol

− 1

∆f usV = V (^) liq − V (^) sol = (163. 3 − 161 .0) cm

3 mol

− 1 = 2. 3 cm

3 mol

− 1

= 2. 3 × 10

− 6 m

3 mol

− 1

Pf = 100 atm = 10132500 P a

Tf = 351. 26 K

FIND: ∆f usS and ∆f usH at Pf and Tf

Start by rearranging the Claperyon equation

dP

dT

∆f usS

∆f usV

dP =

∆f usS

∆f usV

dT

∆f usV can be assumed T independent

however, ∆f usS cannot be independent of T

instead, substitute ∆f usS =

∆f usH

T

because ∆f usH can be assumed to be T independent

dP =

∆f usH

T ∆f usV

dT

integrate

∫^ Pf

P

dP =

∫^ Tf

T

∆f usH

T ∆f usV

dT

∆f usV and ∆f usH come out of the integral, assumed T independent

Pf ∫

P

dP =

∆f usH

∆f usV

Tf ∫

T

T

dT

5. Ttrs(P ) (2 pts)

Calculate the melting point of ice under a pressure of 50. bar. Assume that the density of ice

under these conditions is approximately 0.92 g cm

− 3 and that of liquid water is 1.00 g cm

− 3 .

GIVEN:

P = 1 bar = 1 × 10

5 P a

T = 0

◦ C = 273. 15 K

Pf = 50 bar = 5. 0 × 10

6 P a ⇒ 2 Sig Figs

∆P = 5. 0 × 10

6 P a − 1 × 10

5 P a = 4. 9 × 10

6 P a

ρice = 0. 92 g cm

− 3 = 9. 2 × 10

5 g m

− 3 ρwater = 1. 00 g cm

− 3 = 1. 00 × 10

6 g m

− 3

∆f usV = M ∆

1 ρ

= 18. 02 g mol

− 1

1

1. 00 × 106 g m−^3

1

9. 2 × 105 g m−^3

∆f usV = − 1. 566957 × 10

− 6 m

3 mol

− 1

∆f usH = 6. 008 × 10

3 J mol

− 1

FIND: Tf at Pf

Substitute ∆f usS =

∆f usH T

into the Clapeyron equation

ASSUME ∆f usH and ∆f usV are independent of T and integrate

dP =

∆f usH

∆f usV

dT

T

Pf ∫

P

dP =

∆f usH

∆f usV

Tf ∫

T

dT

T

∆P =

∆f usH

∆f usV

ln (

Tf

T

) solve for Tf

ln

Tf

T

∆P ∆f usV

∆f usH

exponentiate both sides

Tf

T

= exp

∆P ∆f usV

∆f usH

Tf = T exp

∆P ∆f usV

∆f usH

= 273. 15 K exp

(4. 9 × 10

6 P a) × (− 1. 566957 × 10

− 6 m

3 mol

− 1 )

6. 008 × 103 J mol−^1

= 2. 728011 × 10

2 K = 272. 8 K

Tf = 2. 7 × 10

2 K

6. Slope of Chemical Potential (3 pts)

Calculate the difference in slope of the chemical potential against pressure on either side of (a)

the normal freezing point of water and (b) the normal boiling point of water. The densities of

ice and water at 0

◦ C are 0.917 g cm

− 3 and 1.000 g cm

− 3 , and those of water and water vapor

at 100.

◦ C are 0.958 g cm

− 3 and 0.598 g dm

− 3 , respectively. By how much does the chemical

potential of water vapor exceed that of liquid water at 1.2 atm and 100.

◦ C?

GIVEN:

MH 2 O = 18. 02 g mol

− 1 = 18. 02 × 10

− 3 kg mol

− 1

◦ C = 273. 15 K

ρsol = 0. 917 g cm

− 3 = 9. 17 × 10

2 kg m

− 3

ρliq = 1. 000 g cm

− 3 = 1. 000 × 10

3 kg m

− 3

◦ C = 373. 15 K

ρliq = 0. 958 g cm

− 3 = 9. 58 × 10

2 kg m

− 3

ρvap = 0. 598 g dm

− 3 = 0. 598 kg m

− 3

FIND: Difference in slope of chemical potential, against pressure, on either side of the

(a) the solid–liquid phase and (b) the liquid–vapor phase, AND compute the difference

in chemical potential between liquid and vapor at 1.2 atm and 100

◦ C.

The difference in the slope against pressure for two phases is just...

∂ μβ

∂ P

T

∂ μα

∂ P

T

= V (^) β − V (^) α = ∆trsV

The other equation that you will need is V =

M ρ

and ∆trsV = M

1 ρβ

1 ρα

(a)

∂ μl

∂ P

T

∂ μs

∂ P

T

= V (^) l − V (^) s = ∆trsV = ∆f usV

= M

ρl

ρs

= 18. 02 × 10

− 3 kg mol

− 1

1. 00 × 103 kg m−^3

9. 17 × 102 kg m−^3

= − 1. 631036 × 10

− 6 m

3 mol

− 1

∂ μl

∂ P

T

∂ μs

∂ P

T

= − 1. 63 × 10

− 6 m

3 mol

− 1