Evaluating Properties

The question

At the base camp of Mount Everest, a climber's tea water boils furiously — at about 70 °C, too cool to brew a decent cup. In a pressure cooker the same water won't boil until 120 °C. And in the boiler of a power station it stays liquid at 300 °C. Why does water boil at 100 °C in Paris but 70 °C on Everest?

Two properties fix a state

The first law tells us how energy moves; to use it, we need numbers for $u$ , $h$ and $v$ of real substances. Happily, we need surprisingly few measurements to pin them down. The state principle says that for a simple compressible substance — one where the only way to do work is by changing volume, which covers water, air and refrigerants — two independent intensive properties fix the state. Choose temperature and pressure, and every other property, from density to entropy, is determined.

The catch is the word independent. During boiling, as you will see in a moment, temperature and pressure are not independent.

The vapour dome

Heat water steadily at constant pressure and watch what happens on the temperature–volume diagram:

ExploreBoiling water, wherever you are

Pressure1.013 bar

Heat added576 kJ/kg

Temperature100.0 °Csaturated mixture

Quality x10.7% vapour

Specific volume0.180 m³/kg

Enthalpy660 kJ/kg

Challenges0/3

Boil water below 72 °C.
Boil water at 119 °C or hotter.
Turn liquid into vapour without it ever boiling.

One kilogram of water heated at constant pressure, plotted on the temperature–specific volume diagram (volume on a log scale). The dome encloses the states where liquid and vapour coexist. Change the pressure — by going up a mountain, or into a boiler — and the boiling temperature moves with it.

Three regions appear, separated by a bell-shaped curve, the vapour dome:

To the left, compressed liquid: water below its boiling point. Its volume barely changes.
Under the dome, a two-phase mixture of liquid and saturated vapour. Here temperature and pressure are locked together: at 1.013 bar water boils at 100 °C, at 0.314 bar at 70 °C, at 120 bar at 325 °C. Adding heat converts liquid into vapour at constant temperature, and the volume grows enormously.
To the right, superheated vapour: steam hotter than its boiling point.

The top of the dome is the , 374 °C and 220.6 bar for water. Above it, liquid and vapour are indistinguishable: there is no boiling, just a smooth change from dense to light fluid. Modern "supercritical" power stations run their boilers up there, at 250–300 bar, precisely to avoid the constant-temperature boiling step.

Inside the dome, a mixture is described by its , the mass fraction that is vapour:

$x = \frac{m_\text{vapour}}{m_\text{liquid} + m_\text{vapour}}, \qquad v = v_f + x\,(v_g - v_f).$

The subscripts are the steam-table convention: $f$ for saturated liquid (from the German flüssig), $g$ for saturated vapour (gas). The same lever rule works for $u$ , $h$ and $s$ .

Reading a steam table

For a century engineers looked properties up in steam tables: saturated tables indexed by temperature or pressure, and superheated tables giving $v$ , $u$ , $h$ and $s$ on a grid of $(p, T)$ , with linear interpolation between grid points. Software has replaced the paper, but the logic is the same: give two independent properties, get the rest.

The temperature–entropy diagram below is a steam table you can drag. (Entropy, $s$ , arrives properly in Lesson 6; for now treat it as a coordinate that spreads the dome out nicely.)

Enthalpy: the property that flows

In the table you will find a combination that turns up so often it gets its own name, :

$h = u + pv.$

Its usefulness will become obvious in Lesson 4: when fluid flows into a device, it brings its internal energy and the work needed to push it in, $pv$ . Enthalpy bundles the two. For now, notice one thing: across the dome at constant pressure, the heat added per kilogram is exactly the change in enthalpy. The latent heat of vaporisation, $h_{fg} = h_g - h_f$ , is 2257 kJ/kg at 1 atm — over five times the 419 kJ/kg it takes to warm water from 0 to 100 °C.

Estimate first

How much does the volume of water increase when it boils at 1 atm (vapour volume ÷ liquid volume)?

1101001,00010,000100,0001,000,000

Drag along the line to place your guess — each tick is ten times the last.

Liquids and solids: the incompressible model

For liquids, the steam tables barely change with pressure. Compressing water from 1 to 100 bar shrinks it by only 0.5%. It is often good enough to treat liquids and solids as incompressible: $v$ constant, and $u$ depends only on temperature, $\Delta u = c\,\Delta T$ , with a single specific heat $c$ (4.18 kJ/kg·K for water). For the enthalpy, keep the pressure term:

$h(T,p) \approx h_f(T) + v_f\,\big(p - p_\text{sat}(T)\big).$

That second term is tiny for most purposes — but, as the pump in Lesson 4 will show, it is the whole story when you pressurise a liquid.

Gases: when is the ideal-gas law good enough?

For gases, the ideal-gas law $pv = RT$ is beautifully simple. But it ignores the size of molecules and the forces between them, and those matter when molecules are crowded — at high pressure or near condensation. The compressibility factor measures the error:

$Z = \frac{pv}{RT}.$

Remarkably, when pressure and temperature are scaled by their critical values — $p_R = p/p_c$ , $T_R = T/T_c$ — most gases fall on nearly the same :

ExploreWhen is a gas ideal?

Temperature300 K · T_R = 2.38

Pressure100 bar · p_R = 2.94

Compressibility Z0.989gas

Ideal-gas law error+1.2%in predicted density

Reduced state2.38, 2.94T_R, p_R

Challenges0/2

Find a gas state where the ideal-gas law is off by more than 20%.
Find a state above 50 bar where the ideal-gas law is still within 2%.

The generalised compressibility chart: Z = pv/RT against reduced pressure p/p_c, for several reduced temperatures T/T_c (curves from the Peng–Robinson equation of state). An ideal gas has Z = 1 everywhere. Pick a gas and a state: the dot shows where it sits.

The rule of thumb that falls out: a gas is close to ideal when its pressure is low compared with its critical pressure ( $p_R \ll 1$ ) or its temperature is well above critical ( $T_R \gtrsim 2$ ). Air at room conditions is both: $T_R \approx 2.3$ , $p_R \approx 0.03$ , $Z = 0.9997$ . Steam in a power-plant boiler is neither — which is why we need steam tables at all.

Check your intuition

Is the ideal-gas model reasonable for superheated steam at 0.1 bar and 100 °C, in a power-plant condenser?

Ideal-gas internal energy and enthalpy

For an ideal gas, molecules don't feel each other, so their internal energy can't depend on how far apart they are: $u$ and $h$ depend on temperature alone. Then

$du = c_v(T)\,dT, \qquad dh = c_p(T)\,dT, \qquad c_p - c_v = R.$

The difference $c_p - c_v = R$ is the extra energy it takes, at constant pressure, to push back the surroundings as the gas expands. The specific heats themselves come from the molecules — and they are not constants.

ExploreMolecules wake up as they warm

Temperature300 K

c_p1.039 kJ/(kg·K)3.50 R per mole

γ = c_p / c_v1.400

Challenges0/3

Find the gas whose γ is 5/3 at every temperature.
Heat nitrogen until its vibration is at least half awake.
Push γ of a gas below 1.15.

The molar heat capacity c_p of four gases, in units of the gas constant R, from a model in which molecules translate, rotate and vibrate as quantum oscillators. Each vibration only joins in once the gas is hot enough to excite it — so heat capacities climb with temperature, and γ = c_p/c_v falls.

The equipartition picture explains the numbers. Each way a molecule can store energy "freely" contributes $\tfrac12 R$ to $c_v$ : three directions of translation give every gas $\tfrac32 R$ ; a linear molecule adds two rotations, a bent one three. Vibrations are different: they are quantised, and a vibration with a characteristic temperature of 3390 K, like nitrogen's, simply cannot be excited at room temperature. As the gas heats up, one vibration after another "wakes up", and $c_p$ climbs.

That is why the hot gas in a gas turbine has $\gamma \approx 1.3$ instead of air's familiar 1.4 — and why accurate cycle calculations use ideal-gas tables of $h(T)$ instead of a constant $c_p$ . The course's models do the same.

Going deeper
Going deeper: the cold-air standard and when to use it

For quick estimates, Moran and Shapiro use the cold-air-standard assumption: air with constant specific heats evaluated at 300 K, $c_p = 1.005$ kJ/(kg·K), $c_v = 0.718$ , $k = 1.4$ . It turns every process into closed-form algebra — for example, an isentropic process obeys $T_2/T_1 = (p_2/p_1)^{(k-1)/k}$ .

It is accurate to a few percent below about 600 K. Above that, vibrations of N₂ and O₂ (and, in combustion products, of CO₂ and H₂O) raise $c_p$ by 10–25%, and the constant- $c_p$ model overpredicts temperatures and efficiencies. Variable-property calculations integrate $c_p(T)$ , which is what the tables of $h(T)$ and $s^\circ(T)$ do for you.

Recap

What to carry forward

State principle: two independent intensive properties fix the state of a simple compressible substance.
Under the vapour dome, temperature and pressure are tied together: lower the pressure and water boils cooler. The dome ends at the critical point (374 °C, 220.6 bar).
In the two-phase region, quality $x$ interpolates every property: $v = v_f + x v_{fg}$ .
Enthalpy $h = u + pv$ ; at constant pressure, the heat added equals the enthalpy rise.
Liquids are nearly incompressible. Gases are nearly ideal when $p \ll p_c$ or $T \gg T_c$ ; the compressibility factor $Z$ measures the error.
For ideal gases, $u$ and $h$ depend only on $T$ , and $c_p$ rises with temperature as molecular vibrations wake up.

Next: open the system up. How do you apply energy conservation to turbines, pumps and nozzles that fluid flows straight through?