Systems, states, properties and equations of state

Choose a system and describe its state

A system is the matter or region chosen for analysis. A closed system follows a fixed amount of matter; energy may cross its boundary. An open system, often described through a control volume, permits matter to cross. An isolated system exchanges neither matter nor energy. These labels describe the chosen boundary, not whether a container looks closed.

A state specifies properties sufficient for the description being used. At equilibrium, pressure p, temperature T and composition can be assigned consistently without continuing internal changes. A rapidly changing, uneven sample may need local values rather than one pressure and temperature. A process connects states; its route can affect heat and work even when endpoints are unchanged.

Count entities, then convert amount to mass

Amount of substance n is a count-based quantity. Its unit, the mole (mol), specifies a fixed number of entities: one mole corresponds to exactly 6.02214076×10²³ of them. Name the entities being counted, such as nitrogen molecules or carbon atoms. Equal amounts in moles mean equal numbers of those specified entities; they need not have equal masses.

Molar mass M is mass per mole. A sample of mass m has amount n=m/M; conversely m=nM. Match units: grams divided by grams per mole gives moles, as do kilograms divided by kilograms per mole. Converting 30 g with M=0.010 kg/mol requires first writing the mass as 0.030 kg. Dividing 30 by 0.010 without reconciling those units introduces a factor of one thousand.

Separate size from condition

Mass m and total internal energy U are extensive properties: doubling an otherwise identical sample doubles them. Temperature and pressure are intensive: copying the sample does not double either. Specific volume v=V/m and specific internal energy u=U/m are properties per unit mass. The capital V means total volume, not voltage in this chapter.

An equation of state links properties for a stated material and range. For an ideal gas, pV=nRT, where n is amount in moles and R is the gas constant. Here R is rounded to 8.314 J/(mol·K). Use absolute pressure in pascals, volume in cubic metres and absolute temperature in kelvin. One pascal times one cubic metre is one joule.

Check the model before using the equation

Kelvin temperature is Celsius temperature plus 273.15. Temperature differences have the same numerical size in kelvin and degrees Celsius, but ratios require absolute temperature. A change from 20 °C to 40 °C is not a doubling of absolute temperature. Gauge pressure likewise cannot replace absolute pressure in the ideal-gas equation.

The plotted isotherms are fixed-temperature states calculated from this model. Ideal-gas behavior is an approximation most useful far from condensation and at sufficiently low density. Liquids and two-phase mixtures need other property models or tables. In liquid–vapor coexistence for a pure substance, pressure and temperature are linked; specifying both may still leave the vapor mass fraction unknown.

Calculate pressure with a declared model

Take one mole of hypothetical ideal gas at 300 K in 0.024942 m³, or 24.942 L. The equation gives p=(1×8.314×300)/0.024942=100000 Pa=100 kPa absolute. This is a constructed state, not a container specification.

At the same amount and volume, 600 K gives 200 kPa absolute. If the surrounding pressure were 90 kPa, the first state would be 10 kPa gauge. That gauge reading alone is insufficient for the equation unless the surrounding pressure is also known.

For a separate count-to-mass example, assign hypothetical gas A a molar mass of 10 g/mol and gas B 30 g/mol. A 30 g sample of A contains 30/(10)=3 mol. The same amount of B has mass 3×30=90 g. Under the ideal-gas approximation, equal 3 mol amounts at the same temperature and volume have equal pressure because M does not appear in pV=nRT. Their masses and real-gas applicability can still differ; this example supplies molar masses and does not identify real gases.

Try the reasoning

1. Does a closed system exchange no energy?

   Show the answer and reasoning

   It exchanges no matter. Heat or work can still cross its boundary.

2. Why cannot 20 °C and 40 °C be used as a two-to-one temperature ratio?

   Show the answer and reasoning

   Their kelvin values are 293.15 and 313.15; thermodynamic ratios use the absolute scale.

3. Does the ideal-gas equation identify the phase of every substance?

   Show the answer and reasoning

   No. Its applicability must first be justified; condensation and liquid–vapor mixtures require appropriate property information.

4. A hypothetical gas has M=25 g/mol. A 100 g sample and a second gas with M=40 g/mol are to have the same amount. Find that amount, the second mass and the ideal-model pressure comparison at equal temperature and volume.

   Show the answer and reasoning

   The first amount is 100/(25)=4 mol. The second mass must be 4×40=160 g. Equal amounts give equal ideal-model pressures at the same T and V; equal masses would generally give different amounts.

5. A learner obtains 3000 mol by dividing 30 g by 0.010 kg/mol. Diagnose the calculation.

   Show the answer and reasoning

   The mass units differ. Use 0.030 kg/(0.010 kg/mol)=3 mol, or convert the molar mass to 10 g/mol and divide 30 g by it. The mole counts specified entities rather than kilograms.

What to carry forward

Describe the boundary, phase and state before selecting a property equation. Units and reference scales are part of the model.