Water, the commonest substance around us, is an excellent system to understand many concepts of ther — Physical Chemistry — Thermodynamics Chemistry Question
Water
Water, the commonest substance around us, is an excellent system to understand many concepts of thermodynamics. It exists in three different phases: solid (ice), liquid and vapour. [At high pressures, different solid phases of ice exist, but we do not consider them here.] The phase diagram for water, which gives the pressure versus temperature curves for its different phases in equilibrium, is shown below :
Phase diagram:
Phase diagram of water (not to scale)
At what temperature and pressure do all the three phases of water coexist in equilibrium?
Model Answer
The three phases of water coexist in equilibrium at a unique temperature and pressure (called the triple point):
Ttr = 273.16 K = 0.01 °C ptr = 6.11×10–3 bar
What is the effect of decrease of pressure on boiling point of water and melting point of ice, as seen from the phase diagram?
Model Answer
If pressure decreases, boiling point decreases, but melting point increases (slightly).
The liquid-vapour coexistence curve ends at the point pc = 223 bar and Tc = 374 °C. What is the significance of this point?
Model Answer
Beyond this point, there is no distinction between liquid and vapour phases of water. Put alternatively, it is possible to have liquid to vapour transition by a continuous path going around the critical point. (In contrast, solid-liquid transition is discontinuous.)
What is the phase of water at T = 300 K, p = 12.0 bar; T = 270 K, p = 1.00 bar?
Model Answer
T = 300 K, p = 12.0 bar: liquid phase
T = 270 K, p = 1.00 bar: solid phase
Below what value of pressure will ice, when heated isobarically, sublimate to vapour?
Model Answer
Below p = 6.11×10–3 bar, ice heated isobarically will sublimate to vapour.
At a certain temperature and pressure on the liquid-vapour co-existence line, the molar volumes of water in the two phases are
Vliq = 3.15×10–5 m3 Vvap = 15.8×10–5 m3
Determine the volume fractions in liquid and vapour phases for 1.00 mol of water in a 0.100 dm3 vessel at this temperature and pressure,
Model Answer
If xl and xv are the mole fractions of water in liquid and vapour phases, respectively.
V = xl lV + xv vV = xl lV + (1 – xl) vV
...
Vl / V = 0.140
Vv / V = 0.860
B Clausius – Clapeyron equation
Explain your answer to part 1.1 ii. above on the basis of the Clapeyron equation.
Model Answer
dp/dT = ΔH / (T ΔV)
ΔH = molar enthalpy change in phase transition
ΔV = molar change in volume in phase transition
For ice – liquid water transition:
ΔH > 0 ΔV < 0 since ice is less dense than water.
dp/dT < 0
Since ΔV is not large, the p – T curve for this transition is steep with a negative slope. Thus, decrease of pressure increases the melting point slightly.
For liquid water - vapour transition:
ΔH > 0 ΔV < 0
dp/dT > 0
Decrease of pressure decreases the boiling point.
Autoclaves used for medical sterilisation need to have a temperature of 120°C of boiling water to kill most bacteria. Estimate the pressure required for the purpose. The molar enthalpy change of vaporisation of water is 40.66 kJ mol–1 at the normal boiling point. Indicate the assumptions made in your estimate.
Model Answer
Clausius - Clapeyron equation for (solid) liquid – vapour transition is
...
This equation follows from the Clapeyron equation under the assumptions:
1. Vapour follows ideal gas law.
2. Molar volume of the condensed phase is negligible compared to molar volume of vapour phase.
3. If further vapΔH is assumed to be constant (no variation with T), the eq. is integrated to give
...
Here p1 = 1.01 bar , T1 = 373.15 K
T2 = 393.15 K vapΔH = 40.66 kJ mol -1
R = 8.31 J K -1 mol -1
Then: p2 = 2.01 bar
The estimate is based on the assumptions 1, 2 and 3.
The molar enthalpy change of fusion at normal freezing point (273.15 K) is 6008 J mol–1. Estimate the pressure at which water and ice are in equilibrium at –0.200 °C. Density of ice is equal to 917 kg m–3 and that of water is 1000 kg m–3. Indicate the assumptions made in your estimate.
Model Answer
For ice - liquid water equilibrium, use Clapeyron equation
At T1 = 273.15 K, p1 = 1.01 bar
Assume that for a small change in T, ΔH / ΔV is constant.
Integrating the Clapeyron equation above
...
T2 = 272.95 K ΔH(fusion) = -16008 J mol-1
...
p2 – p1 = 27.0 bar
p2 = 28.0 bar
The estimate is based on assumption 1.
C Irreversible condensation
Consider 28.5 g of supercooled (liquid) water at –12.0 °C and 1.00 bar. Does this state lie on the p – T plane of the phase diagram?
Model Answer
On the p -T plane, this equilibrium state is a solid phase (ice). Water in liquid phase at this temperature and pressure is not an equilibrium state - it is a supercooled state that does not lie on the given p -T plane.
This metastable state suddenly freezes to ice at the same temperature and pressure. Treat the metastable state as an equilibrium state and calculate the heat released in the process. Molar heat capacities, assumed constant, are :
Cp(ice) = 76.1 J K–1 mol–1
Cp(liquid water) = 37.15 J K–1 mol–1
ΔH(fusion) = – 333.5 J g–1
Model Answer
Treating the metastable state as equilibrium state, we can go from the supercooled liquid state to the solid state at the same temperature and pressure by a sequence of 3 reversible steps.
1. Supercooled liquid at -12.0 °C to liquid at 0 °C
q1 = number of moles × pC (liquid water) × change of temperature
2. Liquid at 0 °C to ice at 0 °C
q2 = 28.5 g × (–333.5) = 9505 J
3. Ice at 0 °C to ice at –12.0 °C
q3 = number of moles × pC (liquid water) × change of temperature
= 28.5 g / 18.015 g mol-1 × 37.15 J K-1 mol-1 × (-12.0 K) = -705.3 J
q = q1 + q2 + q3 = – 8765 J
Since all the steps are at the constant pressure of 1.00 bar,
q = ΔH
But ΔH is independent on the path, i. e. it depends only on the end points. Thus for the irreversible condensation of supercooled liquid to ice
q = ΔH = – 8765 J
Determine the total entropy change of the universe in the process and assure yourself that the answer is consistent with the Second Law of Thermodynamics. Take the surroundings to be at –12.0 °C.
Model Answer
The actual irreversible path between the two end states of the system is replaced by the sequence of three reversible steps, as above. ΔS can be calculated for each reversible step.
ΔS1 = 5.41 J K-1
ΔS2 = – 34.79 J K-1
ΔS3 = -2.64 J K-1
ΔSsystem = ΔS1 + ΔS2 + ΔS3 = – 32.02 J K–1
ΔSsur = –qsur / Tsur = 18765 / 261.15 = 33.56 J K-1
ΔSuniv = ΔSsystem + ΔSsur = 1.54 J K–1
The entropy of the universe increases in the irreversible process, as expected by the Second Law of Thermodynamics.