IChO 2015 · Theory problem 1

New and well-forgotten old refrigerants

The problem of choosing a refrigerant for refrigeration and air conditioning systems attracted the attention of scientists and technologists throughout the last century. It has been suggested that during this time refrigerants progressed through four generations.

Ammonia, which was ascribed to the first generation, had been used in most of the oldest refrigeration units. It was later replaced by chlorofluorocarbons (CFCs) – derivatives of methane and ethane with the hydrogen atoms replaced by fluorine and chlorine.

In Baku, at "Bakkonditsioner" factory, production of the first Soviet serial household air conditioners BK-1500 had been launched. A second-generation refrigerant chlorodifluoromethane \(\ce{CHF2Cl}\) was used in them. In this problem, we compare various refrigerants in terms of thermodynamics.

Figure 1

First air conditioner of Baku factory in souvenir shop in the Old City (“Icheri Sheher”).

Thermodynamic properties of various refrigerants:

Refrigerant “Generation” \(\Delta H_{\mathrm{vap}}/\mathrm{kJ\,mol^{-1}}\) (at 280 K) \(C_{V(\mathrm{gas})}/\mathrm{J\,K^{-1}\,mol^{-1}}\)
\(\ce{NH3}\) 1 21.3 26.7

Table continued:

\(\ce{CHF2Cl}\) 2 20.0 48.8
\(\ce{CF3CH2F}\) 3 22.1 79
\(\ce{CF3CF=CH2}\) 4 19.1 120

Consider a model refrigeration cycle consisting of 4 steps schematically shown below in the pressure (\(p\)) – internal energy (\(U\)) coordinates.

Figure 2

Diagram 1. Dashed line indicates the phase boundaries

During the first step of the cycle (line 0-1 in diagram 1), a liquid refrigerant is boiling at constant pressure \(p_1\) and temperature \(T_1\) (boiling temperature) until it completely evaporates. At this step, the refrigeration unit absorbs heat from surrounding objects. At the second step, the refrigerant undergoes reversible adiabatic compression and heats up to temperature \(T_2\) (line 1-2). After that the compressed refrigerant is cooled in a condenser at constant pressure \(p_2\) (line 2-3) and then returns to the initial state (line 3-0).

Let the cycle involve 1 mole of refrigerant, which is initially (point 0) completely liquid, \(T_1=280\ \mathrm{K}\), \(T_2=380\ \mathrm{K}\), assume that the vapor of any refrigerant behaves like an ideal gas. The thermodynamic characteristics of refrigerants are listed in the table above.

1.1 For each of refrigerants, ammonia and chlorodifluoromethane, calculate the amount of heat \(Q\) absorbed by refrigeration unit during heat exchange (line 0-1) and the work \(W\) required to compress its vapor adiabatically (line 1-2).

1.2. Which quantity(ies) remain(s) constant during the adiabatic compression step? Indicate by the circle(s).

\[ U\qquad H\qquad S\qquad G\qquad V \]

To compare the energy efficiency of refrigeration cycles with different parameters and refrigerants, the coefficient of performance (\(COP\)) is used, which is defined as a ratio of heat removed from a cooled system to the work of compressor: \(COP=Q/W\).

1.3 Calculate the values of \(COP\) in a considered cycle for ammonia and chlorodifluoromethane.

1.4 Why was ammonia replaced by CFCs in household refrigeration units? (Choose only one option)

a) to increase the energy efficiency of refrigeration cycles,

b) because the density of ammonia is less than that of air under the same conditions,

c) for user safety reasons.

A search for replacement of CFCs as refrigerants started when it was shown that their use can cause irreparable damage to the protective ozone layer of the atmosphere. The third, ozone-friendly generation of refrigerants came on the scene. Its typical representatives are fluoroalkanes.

1.5 What is the cause of the damage made by CFCs to the ozone layer? (Choose only one option.)

a) Ozone molecule easily adds to \(\ce{C-F}\) bond,

b) \(\ce{C-F}\) bond is easily broken by radiation, which leads to the formation of free radicals,

c) ozone molecule easily adds to \(\ce{C-Cl}\) bond,

d) \(\ce{C-Cl}\) bond is easily broken by radiation, which leads to the formation of free radicals.

1.6 However, under the 1997 Kyoto Protocol, fluoroalkanes also had to be replaced because they accumulate in the atmosphere and rapidly absorb infrared radiation, causing a rise in temperature of the atmosphere (the greenhouse effect). The refrigerants of the fourth generation such as 2,3,3,3-tetrafluoropropene \(\ce{CF3CF=CH2}\) have been suggested and are coming into use. Why does this compound enhance the greenhouse effect less than fluoroalkanes? (Choose only one option)

a) It is more reactive and easier to decompose,

b) it easily reacts with ozone,

c) it is better soluble in water.

1.7 Calculate the values of the \(COP\) in the refrigeration cycle considered above for two refrigerants of the third and fourth generations – \(\ce{CF3CH2F}\) and \(\ce{CF3CF=CH2}\). Did the energy efficiency improve in comparison with \(\ce{CHF2Cl}\)? Choose “Yes” or “No”.

Unlike household appliances, industrial refrigeration systems are often still using ammonia. It does not contribute to the greenhouse effect nor does it destroy the ozone layer. Industrial units can have a huge size and a large cost. Prior to their construction, they should be carefully modeled taking into account many different factors. In real systems, some part of the refrigerant at the start of the heat exchange with the environment is in the vapor phase (point 0 in the diagram below), and at the end (point 1) it is always overheated above the boiling point.

Figure 3

Diagram 2. Dashed line indicates the phase boundaries

Consider a cycle with 1 mole of ammonia. Its thermodynamic properties are the following: enthalpy of vaporization \(\Delta H_{\mathrm{vap}}=23.35\ \mathrm{kJ\,mol^{-1}}\) at \(T_{\mathrm{vap}}=239.8\ \mathrm{K}\) (boiling temperature at 1 bar pressure). Heat capacity of the liquid phase \(C_{V(\mathrm{liq})}=77\ \mathrm{J\,K^{-1}\,mol^{-1}}\), of the gas phase \(C_{V(\mathrm{gas})}=26.7\ \mathrm{J\,K^{-1}\,mol^{-1}}\). Assume that the heat capacities are temperature-independent and the vapor behaves like an ideal gas. The temperature dependence of the saturated vapor pressure of ammonia can be described by the empirical equation

\[ \log(p/\mathrm{bar})=4.87-1114/(T/\mathrm{K}-10.4). \]

During the first step of the cycle (line 0-1 in diagram 2), the equilibrium mixture of liquid refrigerant and its vapor receives heat from the environment at constant pressure \(p_1=3.0\ \mathrm{bar}\). The refrigerant completely evaporates and overheats up to the temperature \(T_1=275\ \mathrm{K}\). In the beginning of the process (point 0), the molar fraction of gaseous ammonia is \(x=0.13\).

1.8 Calculate the initial temperature of refrigerant \(T_0\), its volume change \(\Delta V\) and the amount of heat \(Q\) absorbed by refrigeration unit during this step. Take into account that the dependence of \(\Delta H_{\mathrm{vap}}\) from the temperature cannot be neglected. Then the refrigerant is reversibly and adiabatically compressed. It heats up to the temperature \(T_2=393\ \mathrm{K}\) (line 1-2).

1.9 Find the work \(W\) required for compression and the \(COP\) of the system. If you were not able to find \(Q\) in 4.1, use \(Q=20.15\ \mathrm{kJ}\).

At the next step corresponding to the line 2-3 in diagram, the compressed refrigerant is cooled in a condenser at constant pressure. Then it returns to the initial state through adiabatic expansion with zero work (line 3-0).

1.10 Determine the temperature \(T_3\) at point 3 to which the refrigerant is cooled in a condenser.

In the production of refrigeration units it is necessary to consider climatic factors. If a condenser is cooled by atmospheric air, the temperature \(T_3\) increases as the air temperature increases.

1.11 How will the \(COP\) change if \(T_3\) increases while \(T_0,T_1,T_2\) remain the same?

a) increase,

b) remain the same,

c) decrease.

Official English text of the problem.

Answers

  • 1.1 Ammonia, \(\ce{NH3}\), SMILES N: \(Q=21.3\ \mathrm{kJ}\), \(W=2.67\ \mathrm{kJ}\). Chlorodifluoromethane, \(\ce{CHF2Cl}\), SMILES FC(F)Cl: \(Q=20.0\ \mathrm{kJ}\), \(W=4.88\ \mathrm{kJ}\).
  • 1.2 Circle \(S\) only.
  • 1.3 \(\mathrm{COP}_{\ce{NH3}}=7.98\); \(\mathrm{COP}_{\ce{CHF2Cl}}=4.10\).
  • 1.4 c) For user safety reasons.
  • 1.5 d) Radiation breaks \(\ce{C-Cl}\) bonds, forming free radicals.
  • 1.6 a) It is more reactive and easier to decompose.
  • 1.7 1,1,1,2-Tetrafluoroethane, \(\ce{CF3CH2F}\), SMILES FCC(F)(F)F: \(\mathrm{COP}\approx2.80\), No. 2,3,3,3-Tetrafluoroprop-1-ene, \(\ce{CF3CF=CH2}\), SMILES C=C(F)C(F)(F)F: \(\mathrm{COP}\approx1.59\), No.
  • 1.8 \(T_0\approx264\ \mathrm{K}\), \(\Delta V\approx+6.67\ \mathrm{L}\), \(Q\approx19.82\ \mathrm{kJ}\).
  • 1.9 \(W\approx3.15\ \mathrm{kJ}\), \(\mathrm{COP}\approx6.29\).
  • 1.10 \(T_3\approx298\ \mathrm{K}\).
  • 1.11 c) Decrease.

Solution

Take \(W>0\) to mean work supplied to the refrigerant, and use

\[ R=8.314\ \mathrm{J\,mol^{-1}\,K^{-1}}. \]

All numerical heat and work values refer to the stated \(n=1\ \mathrm{mol}\). For the liquid, use the implicit approximation of negligible volume and incompressibility, so that its internal energy is pressure-independent and \(C_{p,\mathrm{liq}}\simeq C_{V,\mathrm{liq}}\).

1.1

During constant-pressure vaporization, the heat absorbed equals the enthalpy change:

\[ Q=n\Delta H_{\mathrm{vap}}. \]

During adiabatic compression, \(q=0\). The first law therefore gives

\[ W=\Delta U=nC_{V,\mathrm{gas}}(T_2-T_1). \]

Here,

\[ T_2-T_1=380-280=100\ \mathrm{K}. \]

Thus:

Refrigerant Heat absorbed \(Q\) Compression work \(W\)
Ammonia, \(\ce{NH3}\) \(21.3\ \mathrm{kJ}\) \((1\ \mathrm{mol})(26.7\ \mathrm{J\,mol^{-1}\,K^{-1}})(100\ \mathrm{K})=2.67\ \mathrm{kJ}\)
Chlorodifluoromethane, \(\ce{CHF2Cl}\) \(20.0\ \mathrm{kJ}\) \((1\ \mathrm{mol})(48.8\ \mathrm{J\,mol^{-1}\,K^{-1}})(100\ \mathrm{K})=4.88\ \mathrm{kJ}\)

1.2

The compression is both reversible and adiabatic. Consequently,

\[ dS=\frac{\delta q_{\mathrm{rev}}}{T}=0. \]

Therefore, circle \(S\) only.

The temperature increase causes \(U\) and \(H\) to increase, compression decreases \(V\), and \(G\) is not conserved.

1.3

By definition,

\[ \mathrm{COP}=\frac{Q}{W}. \]

For ammonia,

\[ \boxed{\mathrm{COP}_{\ce{NH3}}=\frac{21.3}{2.67}=7.98.} \]

For chlorodifluoromethane,

\[ \boxed{\mathrm{COP}_{\ce{CHF2Cl}}=\frac{20.0}{4.88}=4.10.} \]

The coefficients of performance are dimensionless.

1.4

The correct choice is c), for user safety reasons.

Ammonia is toxic and strongly irritating, so leakage into occupied rooms is hazardous. The replacement refrigerants offered better safety for household use. The calculated coefficients of performance do not support increased energy efficiency as the reason for replacement.

1.5

The correct choice is d).

Ultraviolet radiation can cleave \(\ce{C-Cl}\) bonds homolytically: one bonding electron goes to each fragment, producing a chlorine radical. Chlorine radicals participate in catalytic ozone-destruction cycles, for example

\[ \ce{Cl + O3 -> ClO + O2}, \]

\[ \ce{ClO + O -> Cl + O2}. \]

Adding these equations gives

\[ \ce{O3 + O -> 2O2}. \]

Chlorine is regenerated, allowing it to destroy additional ozone molecules.

The species in this cycle are chlorine atom/radical, SMILES [Cl]; ozone, SMILES [O-][O+]=O (one resonance representation); chlorine monoxide radical, SMILES [O]Cl; dioxygen, SMILES O=O; and atomic oxygen, SMILES [O]. The equations express the catalytic sequence; the homolytic electron movement is described above.

1.6

The correct choice is a).

2,3,3,3-Tetrafluoroprop-1-ene, C=C(F)C(F)(F)F, contains a carbon–carbon double bond that provides a reactive site for atmospheric degradation. Its shorter atmospheric lifetime limits its accumulation and therefore reduces its contribution to the greenhouse effect.

No \(E/Z\) designation is possible for this alkene because one carbon of the double bond bears two hydrogen atoms.

1.7

For each refrigerant in the first model cycle,

\[ \mathrm{COP} =\frac{n\Delta H_{\mathrm{vap}}} {nC_{V,\mathrm{gas}}(T_2-T_1)}. \]

For 1,1,1,2-tetrafluoroethane,

\[ W=(1\ \mathrm{mol})(79\ \mathrm{J\,mol^{-1}\,K^{-1}}) (100\ \mathrm{K})=7.9\ \mathrm{kJ}, \]

\[ \boxed{\mathrm{COP}_{\ce{CF3CH2F}}=\frac{22.1}{7.9}\approx2.80.} \]

For 2,3,3,3-tetrafluoroprop-1-ene,

\[ W=(1\ \mathrm{mol})(120\ \mathrm{J\,mol^{-1}\,K^{-1}}) (100\ \mathrm{K})=12.0\ \mathrm{kJ}, \]

\[ \boxed{\mathrm{COP}_{\ce{CF3CF=CH2}}=\frac{19.1}{12.0}\approx1.59.} \]

Both values are below the value \(4.10\) for chlorodifluoromethane. Thus, select “No” for both refrigerants.

1.8

Initially, the liquid and vapor are in equilibrium, so \(T_0\) is the saturation temperature at \(3.0\ \mathrm{bar}\). Taking the empirical logarithm as base 10,

\[ \log_{10}(3.0)=4.87-\frac{1114}{T_0/\mathrm{K}-10.4}. \]

Hence,

\[ T_0= \left(10.4+\frac{1114}{4.87-\log_{10}(3.0)}\right)\mathrm{K} =263.992\ \mathrm{K}. \]

Therefore,

\[ \boxed{T_0\approx264\ \mathrm{K}.} \]

Initially, \(xn=0.13\ \mathrm{mol}\) is gaseous. Neglecting the liquid volume,

\[ V_0=\frac{xnRT_0}{p_1}. \]

At point 1, all the refrigerant is gaseous:

\[ V_1=\frac{nRT_1}{p_1}. \]

Using \(p_1=3.0\times10^5\ \mathrm{Pa}\),

\[ V_0=0.951\ \mathrm{L}, \qquad V_1=7.621\ \mathrm{L}. \]

Thus,

\[ \Delta V=\frac{nR(T_1-xT_0)}{p_1} =6.670\times10^{-3}\ \mathrm{m^3}, \]

\[ \boxed{\Delta V\approx+6.67\ \mathrm{L}.} \]

To calculate the heat, first correct the enthalpy of vaporization to \(T_0\). For the ideal gas,

\[ C_{p,\mathrm{gas}}=C_{V,\mathrm{gas}}+R =35.014\ \mathrm{J\,mol^{-1}\,K^{-1}}. \]

Using \(C_{p,\mathrm{liq}}\simeq77\ \mathrm{J\,mol^{-1}\,K^{-1}}\), Kirchhoff’s relation gives

\[ \Delta H_{\mathrm{vap}}(T_0) = \Delta H_{\mathrm{vap}}(T_{\mathrm{vap}}) + (C_{p,\mathrm{gas}}-C_{p,\mathrm{liq}}) (T_0-T_{\mathrm{vap}}). \]

Numerically,

\[ \begin{aligned} \Delta H_{\mathrm{vap}}(T_0) &=23.35\ \mathrm{kJ\,mol^{-1}}\\ &\quad+ (-0.041986\ \mathrm{kJ\,mol^{-1}\,K^{-1}}) (263.992-239.8)\ \mathrm{K}\\ &=22.334\ \mathrm{kJ\,mol^{-1}}. \end{aligned} \]

Only the initially liquid fraction, \(1-x=0.87\), must evaporate. Afterwards, the entire mole of vapor is heated from \(T_0\) to \(T_1\). At constant pressure,

\[ Q=n(1-x)\Delta H_{\mathrm{vap}}(T_0) +nC_{p,\mathrm{gas}}(T_1-T_0). \]

Therefore,

\[ \begin{aligned} Q &=(0.87\ \mathrm{mol})(22.334\ \mathrm{kJ\,mol^{-1}})\\ &\quad+ (1\ \mathrm{mol})(0.035014\ \mathrm{kJ\,mol^{-1}\,K^{-1}}) (275-263.992)\ \mathrm{K}\\ &=19.816\ \mathrm{kJ}. \end{aligned} \]

Thus,

\[ \boxed{Q\approx19.82\ \mathrm{kJ}.} \]

As an independent energy-balance check,

\[ \Delta U_{0\to1}=17.815\ \mathrm{kJ}, \qquad p_1\Delta V=2.001\ \mathrm{kJ}, \]

so \(Q=\Delta U_{0\to1}+p_1\Delta V=19.816\ \mathrm{kJ}\).

1.9

For the adiabatic compression,

\[ W=\Delta U=nC_{V,\mathrm{gas}}(T_2-T_1). \]

Thus,

\[ W=(1\ \mathrm{mol})(26.7\ \mathrm{J\,mol^{-1}\,K^{-1}}) (393-275)\ \mathrm{K} =3150.6\ \mathrm{J}. \]

Therefore,

\[ \boxed{W\approx3.15\ \mathrm{kJ},} \]

\[ \boxed{\mathrm{COP}=\frac{19.816}{3.1506}\approx6.29.} \]

Using the supplied fallback \(Q=20.15\ \mathrm{kJ}\) would instead give \(\mathrm{COP}\approx6.40\).

1.10

For the specified return step \(3\to0\), both heat and work are zero. Therefore,

\[ U_0-U_3=q_{3\to0}+w_{3\to0}=0, \]

so

\[ U_3=U_0. \]

At point 3 the refrigerant is liquid. At point 0 it contains a vapor fraction \(x\), so its internal energy is

\[ U_0=nU_{m,\mathrm{liq}}(T_0) +nx\Delta U_{\mathrm{vap}}(T_0). \]

Consequently,

\[ nC_{V,\mathrm{liq}}(T_3-T_0) =nx\Delta U_{\mathrm{vap}}(T_0). \]

Neglecting liquid volume and treating the vapor as ideal,

\[ \Delta U_{\mathrm{vap}}(T_0) =\Delta H_{\mathrm{vap}}(T_0)-RT_0. \]

Thus,

\[ \Delta U_{\mathrm{vap}}(T_0) =22.334-\frac{8.314(263.992)}{1000} =20.139\ \mathrm{kJ\,mol^{-1}}. \]

It follows that

\[ T_3=T_0+ \frac{x\Delta U_{\mathrm{vap}}(T_0)}{C_{V,\mathrm{liq}}}, \]

\[ T_3 =263.992\ \mathrm{K} +\frac{0.13(20139\ \mathrm{J\,mol^{-1}})} {77\ \mathrm{J\,mol^{-1}\,K^{-1}}} =297.994\ \mathrm{K}. \]

Therefore,

\[ \boxed{T_3\approx298\ \mathrm{K}.} \]

1.11

The correct choice is c), decrease.

At fixed \(T_0\), the energy balance for the return step gives

\[ x= \frac{C_{V,\mathrm{liq}}(T_3-T_0)} {\Delta U_{\mathrm{vap}}(T_0)}. \]

Increasing \(T_3\) therefore increases the initial vapor fraction \(x\). This fraction is no longer fixed at \(0.13\).

The heat absorbed is

\[ Q=n(1-x)\Delta H_{\mathrm{vap}}(T_0) +nC_{p,\mathrm{gas}}(T_1-T_0). \]

With \(T_0\) and \(T_1\) fixed, increasing \(x\) reduces \(Q\), because less liquid remains available to evaporate. Meanwhile,

\[ W=nC_{V,\mathrm{gas}}(T_2-T_1) \]

remains constant because \(T_1\) and \(T_2\) are unchanged. Hence,

\[ \boxed{\mathrm{COP}=Q/W\ \text{decreases}.} \]

The solution was checked against the official solution.