IChO 2004 · Theory problem 1

Thermodynamics

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For his \(18^{\mathrm{th}}\) birthday party in February Peter plans to turn a hut in the garden of his parents into a swimming pool with an artificial beach. In order to estimate the costs for heating the water and the house, Peter obtains the data for the natural gas composition and its price.

1.1 Write down the chemical equations for the complete combustion of the main components of natural gas, methane and ethane, given in Table 1. Assume that nitrogen is inert under the chosen conditions.

Calculate the reaction enthalpy, the reaction entropy, and the Gibbs energy under standard conditions (\(1.013\cdot10^5\ \mathrm{Pa}\), \(25.0\,^\circ\mathrm C\)) for the combustion of methane and ethane according to the equations above assuming that all products are gaseous.

The thermodynamic properties and the composition of natural gas can be found in Table 1.

1.2 The density of natural gas is \(0.740\ \mathrm{g\,dm^{-3}}\) (\(1.013\times10^5\ \mathrm{Pa}\), \(25.0\,^\circ\mathrm C\)) specified by PUC, the public utility company.

a) Calculate the amount of methane and ethane (in moles) in \(1.00\ \mathrm{m^3}\) of natural gas (natural gas, methane, and ethane are not ideal gases!).

b) Calculate the combustion energy which is released as thermal energy during the burning of \(1.00\ \mathrm{m^3}\) of natural gas under standard conditions assuming that all products are gaseous. (If you do not have the amount from 1.2a) assume that \(1.00\ \mathrm{m^3}\) natural gas corresponds to \(40.00\ \mathrm{mol}\) natural gas.)

According to the PUC the combustion energy will be \(9.981\ \mathrm{kWh}\) per \(\mathrm{m^3}\) of natural gas if all products are gaseous. How large is the deviation (in percent) from the value you obtained in b)

The swimming pool inside the house is \(3.00\ \mathrm m\) wide, \(5.00\ \mathrm m\) long and \(1.50\ \mathrm m\) deep (below the floor). The tap water temperature is \(8.00\,^\circ\mathrm C\) and the air temperature in the house (dimensions given in the figure below) is \(10.0\,^\circ\mathrm C\). Assume a water density of \(\rho=1.00\ \mathrm{kg\,dm^{-3}}\) and air behaving like an ideal gas.

Figure 1

1.3 Calculate the energy (in MJ) which is required to heat the water in the pool to \(22.0\,^\circ\mathrm C\) and the energy which is required to heat the initial amount of air (\(21.0\ \%\) of \(\ce{O2}\), \(79.0\ \%\) of \(\ce{N2}\)) to \(30.0\,^\circ\mathrm C\) at a pressure of \(1.013\times10^5\ \mathrm{Pa}\).

In February, the outside temperature is about \(5\,^\circ\mathrm C\) in Northern Germany. Since the concrete walls and the roof of the house are relatively thin (\(20.0\ \mathrm{cm}\)) there will be a loss of energy. This energy is released to the surroundings (heat loss released to water and/or ground should be neglected). The heat conductivity of the wall and roof is \(1.00\ \mathrm{W\,K^{-1}\,m^{-1}}\).

1.4 Calculate the energy (in MJ) which is needed to maintain the temperature inside the house at \(30.0\,^\circ\mathrm C\) during the party (12 hours).

\(1.00\ \mathrm{m^3}\) of natural gas as delivered by PUC costs 0.40 € and \(1.00\ \mathrm{kWh}\) of electricity costs 0.137 €. The rent for the equipment for gas heating will cost him about 150.00 € while the corresponding electrical heaters will only cost 100.00 €.

1.5 What is the total energy (in MJ) needed for Peter’s “winter swimming pool” calculated in 1.3 and 1.4? How much natural gas will he need, if the gas heater has an efficiency of \(90.0\ \%\)? What are the different costs for the use of either natural gas or electricity? Use the values given by PUC for your calculations and assume \(100\ \%\) efficiency for the electric heater.

Table 1: Composition of natural gas

Chemical Substance mol fraction \(x\) \(\Delta_{\mathrm f}H^0\,(\mathrm{kJ\,mol^{-1}})^{-1}\) \(S^0\,(\mathrm{J\,mol^{-1}\,K^{-1}})^{-1}\) \(C_p^0\,(\mathrm{J\,mol^{-1}\,K^{-1}})^{-1}\)
\(\ce{CO2(g)}\) 0.0024 −393.5 213.8 37.1
\(\ce{N2(g)}\) 0.0134 0.0 191.6 29.1
\(\ce{CH4(g)}\) 0.9732 −74.6 186.3 35.7
\(\ce{C2H6(g)}\) 0.0110 −84.0 229.2 52.5
\(\ce{H2O(l)}\) – −285.8 70.0 75.3
\(\ce{H2O(g)}\) – −241.8 188.8 33.6
\(\ce{O2(g)}\) – 0.0 205.2 29.4

Equation:

\[ J=E\times(A\times\Delta t)^{-1} =\lambda_{\mathrm{wall}}\times\Delta T\times d^{-1} \]

\(J\) energy flow \(E\) along a temperature gradient (wall direction \(z\)) per area \(A\) and time \(\Delta t\)

\(d\) wall thickness

\(\lambda_{\mathrm{wall}}\) heat conductivity

\(\Delta T\) difference in temperature between the inside and the outside of the house

Official English text of the problem.

Answers

  • 1.1 The balanced combustion equations are \[ \ce{CH4(g) + 2 O2(g) -> CO2(g) + 2 H2O(g)} \] \[ \ce{C2H6(g) + 7/2 O2(g) -> 2 CO2(g) + 3 H2O(g)}. \] The thermodynamic quantities at \(298.15\ \mathrm K\), per mole of fuel, are:
Fuel \(\Delta_{\mathrm r}H^\circ\) / \(\mathrm{kJ\,mol^{-1}}\) \(\Delta_{\mathrm r}S^\circ\) / \(\mathrm{J\,mol^{-1}\,K^{-1}}\) \(\Delta_{\mathrm r}G^\circ\) / \(\mathrm{kJ\,mol^{-1}}\)
Methane C −802.5 −5.3 −800.9
Ethane CC −1428.4 +46.6 −1442.3

  • 1.2(a) In \(1.00\ \mathrm{m^3}\) of natural gas: \[ \boxed{n(\ce{CH4})=43.9\ \mathrm{mol}},\qquad \boxed{n(\ce{C2H6})=0.496\ \mathrm{mol}}. \]

  • 1.2(b) The heat released is \[ \boxed{Q=35.9\ \mathrm{MJ}=9.97\ \mathrm{kWh}}. \] Using unrounded intermediate values, PUC’s value is approximately \(0.086\%\) higher, or about \(0.1\%\).

  • 1.3 \[ \boxed{Q_{\mathrm{water}}=1.32\times10^3\ \mathrm{MJ}},\qquad \boxed{Q_{\mathrm{air}}=12.0\ \mathrm{MJ}}. \]

  • 1.4 \[ \boxed{Q_{\mathrm{loss}}=1.56\times10^3\ \mathrm{MJ}}. \]

  • 1.5 The total heat requirement and natural-gas volume are \[ \boxed{Q_{\mathrm{total}}=2.88\times10^3\ \mathrm{MJ}},\qquad \boxed{V_{\mathrm{gas}}=89.2\ \mathrm{m^3}}. \] Including equipment rental, the estimated costs are €185.68 for gas and €209.77 for electricity. Gas heating saves approximately €24.09.

Solution

The molecular identities used throughout are methane C (\(\ce{CH4}\)), ethane CC (\(\ce{C2H6}\)), oxygen O=O (\(\ce{O2}\)), carbon dioxide O=C=O (\(\ce{CO2}\)), water O (\(\ce{H2O}\)), and nitrogen N#N (\(\ce{N2}\)). SMILES do not specify physical phase; gaseous and liquid water are distinguished by the phase labels.

1.1 — Combustion equations and thermodynamic quantities

Complete combustion converts the carbon atoms into carbon dioxide and the hydrogen atoms into water. Balancing carbon, hydrogen, and then oxygen gives

\[ \ce{CH4(g) + 2 O2(g) -> CO2(g) + 2 H2O(g)} \]

and

\[ \ce{C2H6(g) + 7/2 O2(g) -> 2 CO2(g) + 3 H2O(g)}. \]

Nitrogen remains unchanged under the stated assumption. Each equation is written for one mole of fuel.

For either reaction,

\[ \Delta_{\mathrm r}H^\circ =\sum_{\mathrm{products}}\nu_i\Delta_{\mathrm f}H_i^\circ -\sum_{\mathrm{reactants}}\nu_i\Delta_{\mathrm f}H_i^\circ, \]

\[ \Delta_{\mathrm r}S^\circ =\sum_{\mathrm{products}}\nu_iS_i^\circ -\sum_{\mathrm{reactants}}\nu_iS_i^\circ, \]

and, at \(T=25.0+273.15=298.15\ \mathrm K\),

\[ \Delta_{\mathrm r}G^\circ =\Delta_{\mathrm r}H^\circ-T\Delta_{\mathrm r}S^\circ. \]

For methane, using the data for gaseous water,

\[ \begin{aligned} \Delta_{\mathrm r}H^\circ &=[-393.5+2(-241.8)]-[-74.6+2(0)]\\ &=\boxed{-802.5\ \mathrm{kJ\,mol^{-1}}}, \end{aligned} \]

\[ \begin{aligned} \Delta_{\mathrm r}S^\circ &=[213.8+2(188.8)]-[186.3+2(205.2)]\\ &=\boxed{-5.3\ \mathrm{J\,mol^{-1}\,K^{-1}}}, \end{aligned} \]

\[ \begin{aligned} \Delta_{\mathrm r}G^\circ &=-802.5-(298.15)(-5.3\times10^{-3})\\ &=\boxed{-800.9\ \mathrm{kJ\,mol^{-1}}}. \end{aligned} \]

For ethane,

\[ \begin{aligned} \Delta_{\mathrm r}H^\circ &=[2(-393.5)+3(-241.8)]-[-84.0+\tfrac72(0)]\\ &=\boxed{-1428.4\ \mathrm{kJ\,mol^{-1}}}, \end{aligned} \]

\[ \begin{aligned} \Delta_{\mathrm r}S^\circ &=[2(213.8)+3(188.8)]-[229.2+\tfrac72(205.2)]\\ &=\boxed{+46.6\ \mathrm{J\,mol^{-1}\,K^{-1}}}, \end{aligned} \]

\[ \begin{aligned} \Delta_{\mathrm r}G^\circ &=-1428.4-(298.15)(46.6\times10^{-3})\\ &=\boxed{-1442.3\ \mathrm{kJ\,mol^{-1}}}. \end{aligned} \]

1.2(a) — Amounts of methane and ethane

The supplied density gives the mass directly:

\[ m=\rho V =(0.740\ \mathrm{g\,dm^{-3}})(1000\ \mathrm{dm^3}) =740\ \mathrm g. \]

Use the conventional molar masses

\[ \begin{aligned} M(\ce{CO2})&=44.01\ \mathrm{g\,mol^{-1}},& M(\ce{N2})&=28.02\ \mathrm{g\,mol^{-1}},\\ M(\ce{CH4})&=16.04\ \mathrm{g\,mol^{-1}},& M(\ce{C2H6})&=30.07\ \mathrm{g\,mol^{-1}}. \end{aligned} \]

The mean molar mass is the mole-fraction-weighted average, including the noncombustible components:

\[ \begin{aligned} \overline M &=\sum_i x_iM_i\\ &=0.0024(44.01)+0.0134(28.02) +0.9732(16.04)+0.0110(30.07)\\ &=16.42199\ \mathrm{g\,mol^{-1}}. \end{aligned} \]

Hence

\[ n_{\mathrm{total}} =\frac{740}{16.42199} =45.0615\ \mathrm{mol}. \]

The individual amounts are

\[ n(\ce{CH4}) =x_{\ce{CH4}}n_{\mathrm{total}} =0.9732(45.0615) =43.8539\ \mathrm{mol} \approx\boxed{43.9\ \mathrm{mol}}, \]

\[ n(\ce{C2H6}) =x_{\ce{C2H6}}n_{\mathrm{total}} =0.0110(45.0615) =0.495677\ \mathrm{mol} \approx\boxed{0.496\ \mathrm{mol}}. \]

These mass and mole-fraction relationships do not require ideal-gas behaviour.

1.2(b) — Heat of combustion and comparison with PUC

At constant pressure, the heat released is the negative of the combustion enthalpy. Only methane and ethane contribute to the combustion heat:

\[ \begin{aligned} Q_{\mathrm{released}} &=-\left[n(\ce{CH4})\Delta_{\mathrm r}H^\circ_{\ce{CH4}} +n(\ce{C2H6})\Delta_{\mathrm r}H^\circ_{\ce{C2H6}}\right]\\ &=(43.8539)(802.5)+(0.495677)(1428.4)\\ &=35900.8\ \mathrm{kJ}\\ &=35.9008\ \mathrm{MJ}. \end{aligned} \]

Since \(1\ \mathrm{kWh}=3.600\ \mathrm{MJ}\),

\[ Q_{\mathrm{released}} =\frac{35.9008}{3.600} =9.97243\ \mathrm{kWh}. \]

Thus, to three significant figures,

\[ \boxed{Q_{\mathrm{released}}=35.9\ \mathrm{MJ}=9.97\ \mathrm{kWh}}. \]

Taking the calculated value as the reference, as requested,

\[ \begin{aligned} \text{deviation} &=\frac{Q_{\mathrm{PUC}}-Q_{\mathrm{calculated}}} {Q_{\mathrm{calculated}}}\times100\%\\ &=\frac{9.981-9.97243}{9.97243}\times100\%\\ &=\boxed{0.0859\%}. \end{aligned} \]

PUC’s value is therefore approximately \(0.09\%\) higher; this very small difference is about \(0.1\%\).

1.3 — Heating the water and the initial air

The supplied heat capacities are treated as constant over the temperature intervals.

For the pool water,

\[ V_{\mathrm{water}} =(3.00)(5.00)(1.50) =22.5\ \mathrm{m^3} =2.25\times10^4\ \mathrm{dm^3}. \]

Its mass is

\[ m_{\mathrm{water}} =(1.00\ \mathrm{kg\,dm^{-3}}) (2.25\times10^4\ \mathrm{dm^3}) =2.25\times10^4\ \mathrm{kg}. \]

Using \(M(\ce{H2O})=18.02\ \mathrm{g\,mol^{-1}}\),

\[ n_{\mathrm{water}} =\frac{2.25\times10^7\ \mathrm g} {18.02\ \mathrm{g\,mol^{-1}}} =1.24861\times10^6\ \mathrm{mol}. \]

The temperature increase is

\[ \Delta T_{\mathrm{water}}=22.0-8.00=14.0\ \mathrm K. \]

Therefore,

\[ \begin{aligned} Q_{\mathrm{water}} &=n_{\mathrm{water}}C_{p,\mathrm{water}}\Delta T_{\mathrm{water}}\\ &=(1.24861\times10^6)(75.3)(14.0)\ \mathrm J\\ &=1316.29\ \mathrm{MJ}\\ &\approx\boxed{1.32\times10^3\ \mathrm{MJ}}. \end{aligned} \]

For the air, the building consists of a rectangular prism topped by a triangular prism. Its volume is

\[ \begin{aligned} V_{\mathrm{air}} &=(8.00)(15.00)(3.00) +\frac12(8.00)(2.00)(15.00)\\ &=360+120\\ &=480\ \mathrm{m^3}. \end{aligned} \]

The pool is below floor level, so its volume is not subtracted.

At the initial temperature \(T_i=283.15\ \mathrm K\), the ideal-gas law, with \(R=8.314\ \mathrm{J\,mol^{-1}\,K^{-1}}\), gives

\[ n_{\mathrm{air}} =\frac{pV}{RT_i} =\frac{(1.013\times10^5)(480)} {(8.314)(283.15)} =2.06549\times10^4\ \mathrm{mol}. \]

The molar heat capacity of the mixture is

\[ \begin{aligned} C_{p,\mathrm{air}} &=0.210\,C_p(\ce{O2})+0.790\,C_p(\ce{N2})\\ &=0.210(29.4)+0.790(29.1)\\ &=29.163\ \mathrm{J\,mol^{-1}\,K^{-1}}. \end{aligned} \]

The question specifies heating the initial amount of air at constant pressure, so \(C_p\) applies:

\[ \begin{aligned} Q_{\mathrm{air}} &=n_{\mathrm{air}}C_{p,\mathrm{air}}(30.0-10.0)\\ &=(2.06549\times10^4)(29.163)(20.0)\ \mathrm J\\ &=12.0472\ \mathrm{MJ}\\ &\approx\boxed{12.0\ \mathrm{MJ}}. \end{aligned} \]

1.4 — Heat loss during the party

Heat passes through the four walls, including the two triangular gables, and the two sloping roof faces.

The rectangular wall area is

\[ A_{\mathrm{rectangular}} =2(15.00)(3.00)+2(8.00)(3.00) =138.0\ \mathrm{m^2}. \]

The combined area of the triangular gables is

\[ A_{\mathrm{gables}} =2\left[\frac12(8.00)(2.00)\right] =16.0\ \mathrm{m^2}. \]

The sloping distance from an eave to the ridge is

\[ s=\sqrt{(8.00/2)^2+(2.00)^2} =\sqrt{20.00} =4.47214\ \mathrm m. \]

Thus the roof area is

\[ A_{\mathrm{roof}} =2(15.00)(4.47214) =134.164\ \mathrm{m^2}. \]

The total area conducting heat to the outside is

\[ A=138.0+16.0+134.164 =288.164\ \mathrm{m^2}. \]

The temperature difference, thickness, and duration are

\[ \Delta T=30.0-5=25\ \mathrm K,\qquad d=0.200\ \mathrm m,\qquad \Delta t=12(3600)=43200\ \mathrm s. \]

Rearranging the supplied heat-flow equation,

\[ Q_{\mathrm{loss}} =\frac{\lambda A\Delta T\,\Delta t}{d}. \]

Therefore,

\[ \begin{aligned} Q_{\mathrm{loss}} &=\frac{(1.00)(288.164)(25)(43200)}{0.200}\ \mathrm J\\ &=1.556086\times10^9\ \mathrm J\\ &=1556.09\ \mathrm{MJ}\\ &\approx\boxed{1.56\times10^3\ \mathrm{MJ}}. \end{aligned} \]

This is the heat that must be supplied during the party to maintain the indoor temperature.

1.5 — Total energy, gas consumption, and costs

Adding the results without prematurely rounding,

\[ \begin{aligned} Q_{\mathrm{total}} &=Q_{\mathrm{water}}+Q_{\mathrm{air}}+Q_{\mathrm{loss}}\\ &=1316.287+12.047+1556.086\\ &=2884.42\ \mathrm{MJ}\\ &\approx\boxed{2.88\times10^3\ \mathrm{MJ}}. \end{aligned} \]

For gas heating, use PUC’s specified energy content:

\[ q_{\mathrm{gas}} =9.981\ \mathrm{kWh\,m^{-3}} =35.9316\ \mathrm{MJ\,m^{-3}}. \]

At \(90.0\%\) efficiency, each cubic metre supplies

\[ q_{\mathrm{useful}} =0.900(35.9316) =32.33844\ \mathrm{MJ\,m^{-3}}. \]

Hence

\[ V_{\mathrm{gas}} =\frac{2884.42}{32.33844} =89.1948\ \mathrm{m^3} \approx\boxed{89.2\ \mathrm{m^3}}. \]

The fuel charge is

\[ C_{\mathrm{gas,fuel}} =(89.1948\ \mathrm{m^3})(0.40\ \text{€}\,\mathrm{m^{-3}}) =35.68\ \text{€}. \]

Including equipment rental,

\[ \boxed{C_{\mathrm{gas,total}}\approx35.68+150.00=185.68\ \text{€}}. \]

For electric heating, the \(100\%\) efficiency means the required electrical energy is

\[ E_{\mathrm{electricity}} =\frac{2884.42}{3.600} =801.228\ \mathrm{kWh}. \]

The electricity charge is

\[ C_{\mathrm{electricity,energy}} =(801.228)(0.137) =109.77\ \text{€}. \]

Including equipment rental,

\[ \boxed{C_{\mathrm{electricity,total}}\approx109.77+100.00=209.77\ \text{€}}. \]

The estimated saving from choosing gas is

\[ \boxed{209.77-185.68=24.09\ \text{€}}. \]

The solution was checked against the official solution.