Physical Chemistry 2 2007Homework assignment 1, Solutions

Problem 1:

The coefficient of isothermal compressibility is defined as

κT = −

1

V

(

∂V

∂P

)

T

and the coefficient of thermal expansion is defined as

α =1

V

(

∂V

∂T

)

P

Derive expressions for the coefficients of isothermal compressibility and thermal expansionusing the equation of state

(a) for an ideal gas,

(b) given by the expression P (Vm − b) = RT,

(c) of the van der Waals form.

Solution

(a) For an ideal gas, PV = nRT , the volume is given by V = nRT/P , so taking thepartial derivatives gives

κT = −

1

V

(

∂V

∂P

)

T

= −

1

VnRT

(

−

1

P 2

)

=1

P

α =1

V

(

∂V

∂T

)

P

=1

V

nR

P=

1

T

(b) For a gas with the equation of state P (Vm − b) = RT where Vm = V/n, the molarvolume is given by Vm = RT/P + b. Taking the partial derivative with respect to P gives

κT = −

1

Vm

(

∂Vm

∂P

)

T

= −

1

Vm

(

−

RT

P 2

)

=Vm − b

VmP=

1

P−

b

VmP

where the correction to the ideal gas result is clearly ∆κT = −b/VmP .Similarly, taking the partial derivative with respect to T gives

α =1

Vm

(

∂Vm

∂T

)

P

=1

Vm

(

R

P

)

=1

Vm

(

Vm − b

T

)

=1

T−

b

TVm

1

where the correction to the ideal gas result is clearly ∆α = −b/TVm. So, both thecompressibility and the thermal expansion are reduced by the deviation of this equationof state from the ideal gas equation of state.

(c) For a van der Waals gas, the equation of state is (P + a/V 2m)(Vm − b) = RT . This can

be rewritten to isolate P in the following way

P =RT

(Vm − b)−

a

V 2m

The compressibility coefficient can be written in terms of the partial derivative of P withrespect to Vm

κT = −

1

Vm

1(

∂P∂Vm

)

T

By evaluating this partial derivative, one gets

κT =1

Vm

(

RT(Vm−b)2

−2aV 3

m

)

Similarly, the thermal expansion coefficient can be written in terms of a partial deriva-tive of T with respect to Vm

α =1

Vm

1(

∂T∂Vm

)

P

The temperature, T , van be isolated from the equation of state in the following way

T =PVm

R−

Pb

R+

a

RVm−

ab

RV 2m

and after differentiating this expression with respect to Vm at constant P , one obtains

α =1

Vm

1(

PR− 0 −

aRV 2

m

+ 2baRV 3

m

) =R

(

PVm −a

Vm

+ 2baV 2

m

) =1

(

T + bPR

−2a

VmR+ 3ab

V 2m

R

)

which can be rewritten as a cubic equation in the molar volume

PV 3m − (bP + RT )V 2

m + aVm − ab = 0

By calculating the partial derivative of this equation with respect to P holding T fixed,one gets

1 · V 3m + P · 3V 2

m

(

∂Vm

∂P

)

T

− bV 2m − (bP + RT )2Vm

(

∂Vm

∂P

)

T

+ a

(

∂Vm

∂P

)

T

= 0

2

After dividing by V 3m this equation becomes

0 = 1 + 3P (−κT ) −

b

Vm−

2(bP + RT )

Vm(−κT ) +

a

Vm(−κT )

After bringing the terms which contain κT to the left hand side, the equation becomes

κT

(

3P −

bP + RT

Vm+

a

V 2m

)

= 1 −

b

Vm

Isolating κT gives

κT =1 −

bVm

(

3P −bP+RT

Vm

+ aV 2

m

)

Similarly, to get the thermal expansion coefficient, calculate the partial derivative ofthe equation with respect to T. This gives

α = =R

3PVm − 2(bP + RT ) + a/Vm

Problem 2:

The heat of combustion of caffeine was determined by first burning benzoic acid andthen caffeine. In both cases the calorimeter was filled with 466 g of distilled water. When0.0717 g of benzoic acid, C7H6O2(s), were burned as well as 1.1 cm of the iron wire used toignite the sample, the temperature increased from 22.615 ◦C to 23.487 ◦C. When 0.0624g of caffeine, C8H10O2N4(s), were burned, along with 1.8 cm of the wire, the temperatureincreased from 22.714 ◦C to 23.346 ◦C. What is the heat capacity of the calorimeter withand without water, and what is the energy change, ∆U , per mol for the combustion ofcaffeine? (The energy change for combustion of benzoic acid is ∆U◦ = −26.434 kJ/g, thecombustion of the wire gives 9.62 J/cm, and the specific heat of water is 1 cal/g deg).

Solution:

A. Find heat capacity of the calorimeter: Burn 0.0171 g of benzoic acid and 1.1 cm of wire.This gives temperature rise of ∆T=23.487-22.615 deg = 0.872 deg. The heat released is

qV = 0.0717g · 26434J/g + 1.1cm · 9.62J/g = 1895J + 11J = 1906J

The total heat capacity is

CV = CbombV + Cprod

V + CwaterV = 1906J/0.872deg = 2186J/deg

Subtract heat capacity of the water, CwaterV = 466 g · 1cal/g·deg ·4.184 J/cal = 1950 J/deg,

which means that CbombV + Cprod

V = (2186 − 1950) J/deg = 236 J/deg

B. Heat of combustion for caffeine: Burn 0.0624 g of caffeine as well as 1.8 cm of wire.This gives temperature rise of ∆T=22.714-23.346 deg = 0.632 deg. Heat released due to

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combustion of coffeine is ∆U = ∆Utot - ∆Uwire = 0.632deg ·2186 J/deg - 1.8cm·9.62J/cm= (1382 - 17)J = 1365 J from this sample.

Per gram of caffeine, the heat released is ∆U = 1365 J/0.0624 g = 21.870 kJ/g.Per mol of caffeine, the heat released is 21.870 kJ/g · 194 g/mol = 4243 kJ/mol

where 194 is the molecular weight (8 · 12 + 10 + 2 · 16 + 4 · 14 = 194).

Problem 3:

Methane gas is heated from 25 ◦ until the volume has doubled. The pressure is constantat 1 bar. The variation in molar heat capacity with temperature has been measured andthe results of those experiments can be summarized by

CP = a + bT

where a = 22.34 JK−1mol−1 and b = 48.1 × 10−3 JK−2mol−1. Calculate ∆H and ∆Uper mol, (you can estimate the final volume temperature assuming ideal gas behaviour).

Solution:

The final volume is twice the initial volume, so V2 = 2V1. The pressure is constant,so P1 = P2. The initial temperature is T1 = 298K. The final temperature can be foundassuming ideal gas behaviour, T2/T1 = PV2/PV1 = 2. So, T2 = 2T1 = 596 K and∆T = 298 K.

∆H =

(

∂H

∂T

)

P

dT =

∫ T2

T1

Cp(T )dT =

∫ T2

T1

(a + bT )dT =

[

aT +b

2T 2

]T2

T1

= 6.657 kJ/mol + 6.407 kJ/mol = 13.06 kJ/mol

Find change in internal energy,

∆U = ∆H − P∆V = ∆H − R∆T = (13.06 − 2.48)kJ/mol = 10.59 kJ/mol

Problem 4:

(a) One mole of a perfect, monatomic gas expands reversibly and isothermally at 25◦Cfrom an initial pressure of 5 atm to 1 atm. Determine the value of q, w, ∆U and ∆H forthis process.

(b) One mole of a perfect, monatomic gas expands reversibly and adiabatically from 5 atmto 1 atm. The initial temperature is 25◦C. Determine the final temperature of the gas, aswell as q, w and ∆U .

Solution:

(a) Since the temperature does not change (isothermal) and the gas is ideal, the internalenergy does not change, i.e. ∆U = 0. By definition of the enthalpy H = U + PV

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so ∆H = ∆U + ∆(PV ) = ∆U + ∆(nRT ) = ∆U since neither n nor T changes. So,

∆H = 0. The work done in a reversible isothermal process is w = −

∫ V2

V1

PextdV =

−nRT lnV2/V1 = nRT lnP2/P1 = 1 mol · 8.314 J/Kmol · 298 K · ln 1/5 = −3988 J . Bythe first law, q + w = ∆U = 0 so q = 3988 J .

(b) For an adiabatic process, q = 0. As the gas expands against the external pressure itdoes work and looses energy and, since no heat flows in, the temperature drops. The finaltemperature can be found from the equation

T2 = T1

(

P2

P1

)(γ−1)/γ

where γ = CP /CV . For an ideal gas CP = CV + nR and for one mol of monatomicideal gas CV = 3R/2. Therefore, γ = 5/3 and (γ − 1)/γ = 2/5. The final temperatureis T2 = 298 K · (1/5)2/5 = 156 K. The change in internal energy gan then be obtained,∆U = (3/2)R∆T = 1.5 · 8.314 J/Kmol · (298 − 156) K = 1770 J . By the first law,w = ∆U − q = 1770 − 0 = 1770 J . Finally, the enthalpy change is ∆H = ∆U + nR∆T =1770 J + 8.314 J/Kmol · (298 − 156) K = 2950 J .

Problem 5:

The constant volume heat capacity of a gas consisting of P4(g) molecules has beenmeasured to be 67.15 J/Kmol. The P4 molecules are non-linear.

(a) Assuming ideal gas behaviour, what is the expected heat capacity ratio, γ = CP /CV ,based on the experimental data?

(b) What value of the heat capacity ratio, γ, would one expect from the high temper-ature approximations to translational, rotational and vibrational partition functions (the”equipartition theorem” results)?

(c) Often, the vibrational degrees of freedom are not active at room temperature and aretherefore skipped when the heat capacity is estimated. What is the heat capacity ratio,γ, one would expect from the high temperature approximations if vibrational degrees offreedom are skipped? Which is more accurate in this case: (i) including vibration or (ii)skipping it? Explain briefly the results you obtain for P4(g).

Solution:

(a) For an ideal gas CP = CV + R so γ = CP /CV = 1 + R/CV = 1 + 8.314/67.15 = 1.124

(b) The molecule is non-linear, so the total number of vibrational degrees of freedomis 3N − 6 = 6 and the high temperature approximation (equipartition theorem) predictsCV = CV,tr + CV,rot + CV,vib = (3/2)R + (3/2)R + 6R = 9R and a heat capacity ratio ofγequip = 1 + 1/9 = 1.111

(c) Skipping the vibrational degrees of freedom in the high temperature approximationgives CV = CV,tr+CV,rot = (3/2)R+(3/2)R = 3R and a heat capacity ratio of γequip−vib =1 + 1/3 = 1.333. In this case it is more accurate to include the vibrational contribution

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to the heat capacity even though the high temperature approximation is used. This isbecause the vibrational frequencies of P4 are very low, the molecule is ’floppy’. At roomtemperature the vibrational degrees of freedom are to large extent thermally excited andcontribute to the heat capacity.

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