Properties of Pure Substances Questions and Answers

Answer: D

The temperature is 35.2 °C.

With a constant enthalpy of vaporization, the Clausius-Clapeyron equation gives

\[ \begin{aligned} \ln\frac{P_{2}}{P_{1}} &= \frac{\Delta h_{v}}{R} \\ &\quad \times \left(\frac{1}{T_{1}} - \frac{1}{T_{2}}\right) \end{aligned} \]

Here \(\ln(13.5/2.7) = \ln 5 = 1.609\), \(\Delta h_{v}/R = 42700/8.314 = 5136\) K and \(T_{1} = 281.15\) K:

\[ \begin{aligned} \frac{1}{T_{2}} &= \frac{1}{281.15} - \frac{1.609}{5136} \\ &= 3.557\times10^{-3} - 3.13\times10^{-4} \\ &= 3.243\times10^{-3}\ \text{K}^{-1} \end{aligned} \]

so \(T_{2} = 308.3\) K = 35.2 °C. The other values, 44.3, 51.4 and 58.7 °C, would correspond to vapour pressures well above 13.5 kPa on this curve.

Answer: D

The triple point is at 195.2 K, 0.59 atm.

At the triple point the solid and liquid vapour pressures are equal, so the two equations hold together:

\[ \begin{aligned} 15.16 - \frac{3063}{T} &= 18.7 - \frac{3754}{T} \\ \frac{691}{T} &= 3.54 \Rightarrow T = 195.2\ \text{K} \\ \ln p &= 15.16 - \frac{3063}{195.2} = -0.532 \\ p &= e^{-0.532} = 0.59\ \text{atm} \end{aligned} \]

295.2 K is 100 K too high and makes the two curves disagree. 0.69 atm is a slip in the exponential; the liquid and solid equations both give 0.59 atm at 195.2 K.

Answer: C

The latent heat of vaporisation is about 1498 kJ/kg.

The liquid's vapour-pressure equation has the Clausius-Clapeyron form \(\ln p = A - L/(\bar{R}T)\) with L per kmol, so the coefficient 3063 K equals \(L/\bar{R}\):

\[ \begin{aligned} L &= 3063 \times 8.314 = 25466\ \text{kJ/kmol} \\ &= \frac{25466}{17} = 1498\ \text{kJ/kg} \end{aligned} \]

Using the solid's coefficient, 3754, would give the latent heat of sublimation, about 1836 kJ/kg; their difference, about 338 kJ/kg, is the latent heat of fusion. The other options do not follow from either coefficient.

Answer: D

At the triple point solid, liquid and vapour phases co-exist in equilibrium.

The triple point is the single combination of temperature and pressure at which the three phase boundaries meet on the phase diagram. For water it is 0.01°C and 611.7 Pa. By the Gibbs phase rule, F = C - P + 2 = 1 - 3 + 2 = 0, so it is an invariant point.

Two phases alone co-exist along the lines of the phase diagram: solid-liquid on the melting curve, liquid-vapour on the vaporisation curve and solid-vapour on the sublimation curve. Each of the 'only two phases' options describes one of those curves, not the triple point.

Answer: A

Quality is \(x = \dfrac{m_{vapour}}{m_{liquid} + m_{vapour}}\).

Quality, or dryness fraction, is the mass fraction of vapour in a saturated liquid-vapour mixture. It runs from 0 for saturated liquid to 1 for saturated vapour, and gives mixture properties such as \(v = v_{f} + x\,v_{fg}\).

The liquid fraction \(m_{liquid}/(m_{liquid} + m_{vapour})\) is \(1 - x\), sometimes called the wetness fraction. The ratios of vapour to liquid mass and liquid to vapour mass are not bounded between 0 and 1 and are not the quality.

Answer: D

The slope of a constant-pressure line on the Mollier diagram is the temperature.

The Mollier diagram plots enthalpy h against entropy s. From the Gibbs relation \(dh = T\,ds + v\,dp\), at constant pressure \(dp = 0\), so

\[ \left(\frac{\partial h}{\partial s}\right)_{p} = T \]

That is why isobars in the wet region are straight lines (the temperature is constant there) and curve upward in the superheated region as the temperature rises.

Enthalpy and entropy are the axes of the diagram, not a slope, and internal energy does not appear in this derivative.

Answer: A

For a superheated vapour, its pressure is less than the saturation pressure at the given temperature.

A vapour is superheated when it is hotter than the boiling point at its pressure or, equivalently, at a lower pressure than the saturation pressure for its temperature. Either way it lies to the right of the saturated vapour line.

The other statements are reversed. Its temperature is higher, not lower, than the saturation temperature at the given pressure. Its specific volume is larger than that of saturated vapour at the same temperature, and its enthalpy is higher than that of saturated vapour at the same pressure, because heat has been added beyond saturation.

Answer: D

The heat transfer is about 77470 kJ.

The tank is rigid, so no work is done and \(Q = U_{2} - U_{1}\). Find the masses first:

\[ \begin{aligned} m_{f} &= \frac{0.1}{0.001043} = 95.88\ \text{kg} \\ m_{g} &= \frac{1.9}{1.694} = 1.12\ \text{kg} \\ m &= 97.00\ \text{kg} \end{aligned} \]

\[ \begin{aligned} U_{1} &= 95.88(417.33) + 1.12(2506.06) \\ &\approx 42823\ \text{kJ} \end{aligned} \]

The specific volume stays \(v = 2/97.00 = 0.02062\) m³/kg. At 2 MPa:

\[ \begin{aligned} x_{2} &= \frac{0.02062 - 0.001177}{0.09963 - 0.001177} = 0.1975 \\ u_{2} &= 906.42 + 0.1975(1693.84) = 1240.9 \\ U_{2} &= 97.00 \times 1240.9 \approx 120367\ \text{kJ} \end{aligned} \]

So \(Q \approx 120367 - 42823 \approx 77500\) kJ, which matches the keyed 77470 kJ within the rounding of the table values. The other options are well short of the energy needed to take nearly 97 kg of mostly liquid water from 100 kPa to a wet mixture at 2 MPa, as the working shows.

Answer: D

The degrees of freedom are 4.

Gibbs' phase rule for a non-reacting system is F = C - P + 2. There are four independent components (CO₂, N₂, O₂ and H₂O) and two phases (a gas phase and a liquid phase), so:

\[ F = 4 - 2 + 2 = 4 \]

Each gas dissolves to some extent in the water and water vapour enters the gas, so all four components are present in both phases, but that does not change C. Counting only the three gases as components gives 3; treating the system as a single phase gives 5, not offered.

Answer: B

The direct change from ice to vapour is called sublimation.

Sublimation is a phase change from solid straight to vapour without passing through the liquid state. For water it occurs at pressures below the triple point pressure, about 0.611 kPa, which is why snow and ice slowly vanish in cold, dry air.

Vaporization is liquid to vapour, saturation describes the condition at which a phase change takes place, and the triple point is the single state at which all three phases coexist.