Thermodynamics
Question Sheet 1
1. [td1-a]
(a) 1 mol of an ideal gas (initially at a pressure of 1 bar) is reversibly
compressed to a quarter of its initial volume at a constant temperature of 298 K. What is its new pressure? State the change in internal
energy of the gas, ∆U , and calculate the work done on the gas, w,
and the heat exchanged with the surroundings, q.
(b) The same sample of gas is now heated whilst confined to its new
volume until its temperature is 398 K. Calculate the new pressure
and the quantities ∆U , w and q. [Reminder: the molar heat capacity
of an ideal gas, CV,m = 23 R].
(c) Next, the gas is allowed to expand reversibly and isothermally at
398 K until its volume has returned to its original value. Calculate
the new pressure and ∆U , w and q.
(d) Finally, the gas is allowed to cool at constant volume back to 298 K.
Its pressure therefore returns to 1 bar. Calculate ∆U , w, and q.
(e) Draw a p − V diagram for this whole cycle, and calculate ∆U , w and
q.
2. [td1-b]
(a) Given the following data, calculate the enthalpy change due to the
reaction between magnesium and carbon dioxide at 298 K:
2Mg(s) + CO2 (g) → 2MgO(s) + C(s)
∆f H (MgO) = −601.70 kJ mol−1 ;
∆f H (CO2 (g)) = −393.51 kJ mol−1
Why is it a bad idea to attempt to put out a magnesium fire with a
CO2 fire extinguisher?
(b) A sample of 10 g of magnesium is placed on a large block of solid CO2 .
When the sample is heated, it begins to react as above. Neglecting
the heat initially applied to start the reaction, determine change in
mass of the CO2 block after all the magnesium has reacted. Assume
that none of the heat released sublimes the unreacted CO2 . The
molar masses of Mg and CO2 are 24.31 g mol−1 and 44.01 g mol−1
respectively.
3. [td1-c]
The standard enthalpy of formation of naphthalene, C10 H8 at 298 K is
78.53 kJ mol−1 . Assuming that the following molar heat capacities can be
represented by expressions of the form
Cp,m = A + BT +
C
T2
(1)
with the coefficients A, B and C given below, calculate ∆f H at 350 K.
A /J K−1 mol−1
3
10 B /J K−2 mol−1
105 C /J K mol−1
C(s, gr)
16.86
4.77
-8.54
H2 (g)
27.28
3.26
0.50
C10 H8 (s)
-115.9
3920.0
0.0
[Hint: calculate
∆r Cp = ∆r A + ∆r BT +
∆r C
T2
for the reaction forming C10 H8 . Then integrate, according to Kirchoff’s
law (being careful when evaluating the upper and lower limits).]
4. [td1-d]
A 0.5 g sample of benzoic acid, C6 H5 CO2 H, was combusted in a bomb
calorimeter in the presence of excess oxygen. The temperature of the
calorimeter rose by 0.880 K from 298 K. In two separate experiments in
the same apparatus, the combustion of 0.992 g of oleic acid (1) and 1.045 g
of eliadic acid (2) gave temperature increases of 2.609 K and 2.718 K
respectively.
(a) From the benzoic acid calibration, determine the heat capacity of the
calorimeter. The molar mass of benzoic acid is 122.12 g mol−1 and
its molar internal energy change on combustion is −3227 kJ mol−1 .
(b) Determine the molar internal energy of combustion of the fatty acids.
The molar mass of each fatty acid is 282.46 g mol−1 .
(c) Estimate the molar enthalpy of combustion of the fatty acids.
(d) The standard enthalpies of formation of H2 O(l) and CO2 (g) are
−285.8 kJ mol−1 and −393.5 kJ mol−1 respectively. Determine the
enthalpies of formation of the two fatty acids.
(e) Justify any difference you find in ∆f H for oleic and eliadic acid.
O
OH
1
O
OH
2
5. [td1-e]
The mineral gypsum, CaSO4 · 2H2 O, exists in equilibrium with andhydrite, CaSO4 , in the presence of liquid water. Given the data below, determine the equilibrium constant and show that gypsum is stable at 298 K.
Assuming that ∆trs H and ∆trs S are independent of temperature, determine the temperature at which anhydrite becomes more stable. How
may the temperature-dependence of these enthalpy and entropy changes
be taken into account?
H2 O(l)
andhydrite
gypsum
∆f Hm /kJ mol−1
-285.8
-1434.1
-2022.6
Sm
/J K−1 mol−1
69.9
106.69
194.14
6. [td1-m]
The following table provides data for two allotropes of carbon at 298 K.
Vs = 1/ρ is the specific volume and
1 ∂V
κT = −
V
∂p T
is the isothermal compressibility.
∆f G /kJ mol−1
Vs /cm3 g−1
κT /kPa−1
Graphite
0
0.444
3.04 × 10−8
Diamond
+2.8678
0.284
0.187 × 10−8
(a) Which is the more stable allotrope at 298 K and 1 bar?
(b) Stating your assumptions, calculate the pressure required to convert
graphite into diamond at 298 K.
[Hints: First attempt the calculation on the assumption that both
diamond and graphite are incompressible (κT = 0). Then calculate
the pressure-dependence of Vm for each phase, and use the approximation e−x ≈ 1 − x for small x].