Download The heat of combustion of caffeine was determined by first burning be

Physical Chemistry 2
2006
Homework assignment 2
Problem 1:
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, C7 H6 O2 (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, C8 H10 O2 N4 (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
CV = CVbomb + CVprod + CVwater = 1906J/0.872deg = 2186J/deg
Subtract heat capacity of the water, CVwater = 466 g · 1cal/g·deg ·4.184 J/cal = 1950 J/deg,
which means that CVbomb + CVprod = (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
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 2:
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
C̄P = a + bT
1
where a = 22.34 JK −1 mol−1 and b = 48.1 × 10−3 JK −2 mol−1 . Calculate ∆H̄ and ∆Ū
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.
∆H =
∂H
∂T
Z
T2
dT =
Z
T2
Cp (T )dT =
T1
P
T1
T
b 2 2
(a + bT )dT = aT + T
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 3:
(a) Show that the first law expression for heat flow (assuming only PV-work), δq =
dU + P dV , can be rewritten as δq = CV dT + RT d ln V for an ideal gas.
(b)
Using the expression above, show that δq is not an exact differential.
(c) Show that by multiplying δq by 1/T , an exact differential is formed, i.e. show that
δq
= CV d ln T + R d ln V
T
is an exact differential. This shows that
function is defined as entropy.
δq
T
is the differential of a state function. This state
Solution:
(a) The first law dU = δq
+ δw becomes dU = δq − pdV when the work is only PV-work.
∂U
For ideal gas dU = ∂T V dT = CV dT and P dV = nRT dV /V = nRT d ln V . Together,
this gives
δq = CV dT + nRT d ln V
(b) For δq
to be an exact differential, the derivative of the coefficient of dT with respect to
V
V , ∂C
would need to be equal to the derivative of the coefficcient of dV with respect
∂V
T
/V )
V
to T , ∂(nRT
. But, ∂C
∂T
∂V T = 0 for ideal gas since volume does not affect the energy
V
of an ideal
gas (or
any derivative of the energy) due to the lack of molecular interactions,
∂(nRT /V )
while
= nR/V . So, the the condition for an exact differential is not satisfied.
∂T
V
2
(c) For δq/T to be an exact differential,
∂(CV /T )
∂V
would need to be equal to
T
∂(nR/V )
∂T
.
V
Both are equal to zero, so δq/T is an exact differential. This illustrates for the particular case of an ideal gas that the entropy, which is defined by dS = δq/T for a reversible
process, is a state function, while δq is not.
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
so ∆H = ∆U + ∆(P V ) = ∆U + ∆(nRT ) = ∆U since neither n nor T changes. So,
RV
∆H = 0. The work done in a reversible isothermal process is w = − V12 Pext dV =
−nRT ln V2 /V1 = nRT ln P2 /P1 = 1 mol · 8.314 J/Kmol · 298 K · ln 1/5 = −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
T 2 = T1
P2
P1
(γ−1)/γ
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
is 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:
(a) One mol of a perfect, monatomic gas expands reversibly and isothermally at 300 K
from a pressure of 10 atm to a pressure of 2 atm. Determine the value of q, w ∆H and
∆U for this process.
(b) Now assume the gas expands by the same amount again isothermally but now irreversibly against 1 atm pressure (instead of reversible expansion) and calculate again q, w,
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∆U and ∆H. Compare with the values obtained in part (a) and explain qualitatively the
difference.
(c) Compare now the isothermal expansion in (a) to an adiabatic, reversible expansion.
Sketch how the pressure changes as a function of volume in the two cases (label the curves
clearly) and calculate the final temperature of the gas assuming the initial temperature is
300 K.
Problem 6:
The enthalpy change is ∆Hvap = 40 kJ/mol when water boils at 100 ◦ C and 1 atm.
This accounts for both the increase in internal energy as the molecules go from the liquid
to vapour phase, as well as the expansion of the volume when water vapor forms. What
fraction of the enthalpy change is due to the volume expansion?
Solution:
The enthalpy change is ∆H = ∆U + ∆(P V ). The second term is the contribution due to
the volume expansion. Since the pressure is constant, ∆(P V ) = P ∆V ≈ P Vvap . It is a
good approximation here to simply neglect the volume of the liquid. Assuming the ideal
gas equation of state is a good enough approximation for the vapor, this can be evaluated as
V = nRT /P and this gives ∆(P V ) = P ·nRT /P = nRT = 1 mol ·8.314 J/Kmol ·398 K =
3.3 kJ. The relative contribution of the volume expansion term to the enthalpy change is
3.3/40 = 0.08 or 8%.
Problem 7:
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
4
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
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.
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