The Coefficient of Isothermal Compressibility Is Defined As Κt

Home , Heat capacity ratio, Isothermal process, Thermal expansion

Physical Chemistry 2 2007 Homework assignment 1, Solutions

Problem 1: The coeﬃcient of isothermal compressibility is deﬁned as

1 ∂V κT = − V ∂P T and the coeﬃcient of thermal expansion is deﬁned as

1 ∂V α = V ∂T P

Derive expressions for the coeﬃcients of isothermal compressibility and thermal expansion using 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, P V = nRT , the volume is given by V = nRT/P , so taking the partial derivatives gives

− 1 ∂V − 1 − 1 1 κT = = nRT 2 = V ∂P T V P P

1 ∂V 1 nR 1 α = = = V ∂T P V P T

(b) For a gas with the equation of state P (Vm − b) = RT where Vm = V/n, the molar volume is given by Vm = RT/P + b. Taking the partial derivative with respect to P gives − − 1 ∂Vm − 1 −RT Vm b 1 − b κT = = 2 = = Vm ∂P T Vm P VmP P VmP

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

1 ∂V 1 R 1 V − b 1 b α = m = = m = − Vm ∂T P Vm P Vm T T T Vm

1 where the correction to the ideal gas result is clearly ∆α = −b/TVm. So, both the compressibility and the thermal expansion are reduced by the deviation of this equation of state from the ideal gas equation of state. 2 − (c) For a van der Waals gas, the equation of state is (P + a/Vm)(Vm b) = RT . This can be rewritten to isolate P in the following way

RT a P = − − 2 (Vm b) Vm

The compressibility coeﬃcient can be written in terms of the partial derivative of P with respect to Vm 1 1 κT = − Vm ∂P ∂Vm T By evaluating this partial derivative, one gets

1 κT = RT 2a 2 − 3 Vm (Vm−b) V m Similarly, the thermal expansion coeﬃcient can be written in terms of a partial derivative of T with respect to Vm 1 1 α = Vm ∂T ∂Vm P The temperature, T , van be isolated from the equation of state in the following way

P Vm − Pb a − ab T = + 2 R R RVm RVm and after diﬀerentiating this expression with respect to Vm at constant P , one obtains

1 1 R 1 α = = = V P a 2ba a 2ba bP 2a 3ab m − − 2 3 − 2 − 2 R 0 RV + RV P Vm Vm + V T + R VmR + V R m m m m which can be rewritten as a cubic equation in the molar volume

3 − 2 − P Vm (bP + RT )Vm + aVm ab = 0

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

· 3 · 2 ∂Vm − 2 − ∂Vm ∂Vm 1 Vm + P 3Vm bVm (bP + RT )2Vm + a = 0 ∂P T ∂P T ∂P T

2 3 After dividing by Vm this equation becomes b 2(bP + RT ) a 0=1+3P (−κT ) − − (−κT ) + (−κT ) Vm Vm Vm

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

− bP + RT a − b κT 3P + 2 = 1 Vm Vm Vm

Isolating κT gives − b 1 Vm κT = bP +RT a − 2 3P Vm + V m Similarly, to get the thermal expansion coeﬃcient, calculate the partial derivative of the equation with respect to T. This gives

R α = = 3P Vm − 2(bP + RT ) + a/Vm Problem 2: The heat of combustion of caffeine was determined by first burning benzoic acid and then caffeine. In both cases the calorimeter was filled with 466 g of distilled water. When 0.0717 g of benzoic acid, C7H6O2(s), were burned as well as 1.1 cm of the iron wire used to ignite the sample, the temperature increased from 22.615 ◦C to 23.487 ◦C. When 0.0624 g of caffeine, C8H10O2N4(s), were burned, along with 1.8 cm of the wire, the temperature increased from 22.714 ◦C to 23.346 ◦C. What is the heat capacity of the calorimeter with and without water, and what is the energy change, ∆U, per mol for the combustion of caffeine? (The energy change for combustion of benzoic acid is ∆U ◦ = −26.434 kJ/g, the combustion 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

bomb prod water CV = CV + CV + CV = 1906J/0.872deg = 2186J/deg

water · · · Subtract heat capacity of the water, CV = 466 g 1cal/g deg 4.184 J/cal = 1950 J/deg, bomb prod − which means that CV + CV = (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

3 combustion of coffeine is ∆U = ∆Utot - ∆Uwire =0.632deg·2186J/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 constant at 1 bar. The variation in molar heat capacity with temperature has been measured and the results of those experiments can be summarized by

C¯P = a + bT where a = 22.34 JK−1mol−1 and b = 48.1 × 10−3 JK−2mol−1. Calculate ∆H¯ and ∆U¯ per 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 found assuming ideal gas behaviour, T2/T1 = P V2/P V1 = 2. So, T2 = 2T1 = 596 K and ∆T = 298 K.

T2 T2 T2 ∂H b 2 ∆H = dT = Cp(T )dT = (a + bT )dT = aT + T ∂T P ZT1 ZT1 2 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◦C from an initial pressure of 5 atm to 1 atm. Determine the value of q, w, ∆U and ∆H for this process. (b) One mole of a perfect, monatomic gas expands reversibly and adiabatically from 5 atm to 1 atm. The initial temperature is 25◦C. Determine the ﬁnal temperature of the gas, as well as q, w and ∆U.

Solution: (a) Since the temperature does not change (isothermal) and the gas is ideal, the internal energy does not change, i.e. ∆U = 0. By deﬁnition of the enthalpy H = U + P V

4 so ∆H = ∆U + ∆(P V ) = ∆U + ∆(nRT ) = ∆U since neither n nor T changes. So, − V2 ∆H = 0. The work done in a reversible isothermal process is w = V1 PextdV = −nRT ln V2/V1 = nRT ln P2/P1 = 1 mol · 8.314 J/Kmol · 298 K · ln1/5 =R −3988 J. By the 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 it does work and looses energy and, since no heat flows in, the temperature drops. The final temperature can be found from the equation

(γ−1)/γ P2 T2 = T1 P1 where γ = CP /CV . For an ideal gas CP = CV + nR and for one mol of monatomic ideal gas CV = 3R/2. Therefore, γ = 5/3 and (γ − 1)/γ = 2/5. The ﬁnal temperature 2/5 is T2 = 298 K · (1/5) = 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 been measured 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 temperature 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 are therefore 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 of freedom 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 freedom is 3N − 6 = 6 and the high temperature approximation (equipartition theorem) predicts CV = CV,tr + CV,rot + CV,vib = (3/2)R + (3/2)R +6R =9R and a heat capacity ratio of γequip =1+1/9=1.111 (c) Skipping the vibrational degrees of freedom in the high temperature approximation gives 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

5 to the heat capacity even though the high temperature approximation is used. This is because the vibrational frequencies of P4 are very low, the molecule is ’ﬂoppy’. At room temperature the vibrational degrees of freedom are to large extent thermally excited and contribute to the heat capacity.