Question 2 of 7: Reading a Two-Component $p$–$T$ Diagram

Question 2: Reading a Two-Component $p$–$T$ Diagram (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Vapour-pressure data read off the chart. Component 1 (triangles): approximately $(-50\,{}^{\circ}\text{F},20\text{ psia})$, $(0,190)$, $(50,450)$, $(100,850)$, $(150,1600)$, with its trend curve leaving the top of the chart (2000 psia) near $165\,{}^{\circ}\text{F}$. Component 2 (dots): approximately $(-100,0)$, $(0,100)$, $(50,230)$, $(100,430)$, $(150,800)$, $(200,1500)$, with its trend curve leaving the top of the chart near $230\,{}^{\circ}\text{F}$. Neither curve ends (reaches its critical point) inside the plotted range.

Find. (a) The heavier component; (b)–(d) the phase state at the three stated $(p,T)$ points.

Approach. For a pure component at fixed $T$ (below its critical temperature), a point above its vapour-pressure curve ($p>p_{sat}(T)$) is liquid, a point below it ($p<p_{sat}(T)$) is vapour, and a point on the curve is saturated, with liquid and vapour coexisting. So read or interpolate $p_{sat}$ at the query temperature and compare it with the query pressure.

[Figure not reproduced: Chart as read from the source: vapour-pressure curves for Component 1 and Component 2, with the query points of parts (b)–(d). The (b)/(c) point lies below Component 1's curve (vapour) but above Component 2's (liquid); the (d) point sits on Component 2's curve. See the official exam paper.]

1. Part (a) — heavier component. Across the chart Component 2 has the lower vapour pressure; only at the very bottom, below about 50 psia, do the two sets of markers merge within reading precision. At $100\,{}^{\circ}\text{F}$, for example, $p_{sat,1}\approx850$ psia while $p_{sat,2}\approx430$ psia, and at $150\,{}^{\circ}\text{F}$ the values are $\approx1600$ vs. $\approx800$ psia. Equivalently, Component 2 needs a higher temperature to reach any given pressure. A lower vapour pressure means a less volatile substance with a higher boiling point and critical temperature, which for hydrocarbons goes with higher molecular weight. So $\boxed{\text{Component 2 is heavier}}$.
2. Part (b) — Component 1 at 1000 psia, 140°F. Component 1 still has a plotted saturation point at $150\,{}^{\circ}\text{F}$, so $140\,{}^{\circ}\text{F}$ is below its critical temperature and $p_{sat,1}$ is defined. Interpolating between $(100,850)$ and $(150,1600)$: $p_{sat,1}(140)\approx850+(1600-850)\times\frac{40}{50}=1450$ psia (about 1410 psia if interpolated along the exponential-looking trend curve instead). The query pressure $1000<p_{sat,1}$, so the point lies below Component 1's curve: $\boxed{\text{Component 1 is a vapour (superheated gas)}}$ at $(1000\text{ psia},140\,{}^{\circ}\text{F})$.
3. Part (c) — Component 2 at 1000 psia, 140°F. Interpolating between $(100,430)$ and $(150,800)$: $p_{sat,2}(140)\approx430+(800-430)\times\frac{40}{50}=726$ psia (about 707 psia along the trend curve). The query pressure $1000>p_{sat,2}$, so the point lies above Component 2's curve: $\boxed{\text{Component 2 is a liquid (compressed/subcooled)}}$ at $(1000\text{ psia},140\,{}^{\circ}\text{F})$. At the same $(p,T)$ the lighter component is a gas and the heavier one a liquid, which is part (a)'s volatility difference seen directly.
4. Part (d) — Component 2 at 1500 psia, 200°F. The chart has a Component 2 data marker at exactly $(200\,{}^{\circ}\text{F},1500\text{ psia})$, i.e. $p_{sat,2}(200)\approx1500$ psia, equal to the query pressure. The point lies on the vapour-pressure curve, so $\boxed{\text{Component 2 is saturated: liquid and vapour coexist in equilibrium (two-phase)}}$. For a pure component, $p$ and $T$ on the curve do not fix how much of each phase is present; that needs one more property, such as the specific volume or quality.