Gases and thermal processes

Before calculating: what is the system?

Imagine a certain amount of gas in a cylinder with a piston. We will call system this gas; the piston, walls and outside air belong to the surroundings. This choice makes it possible to say whether energy enters or leaves. A closed system can exchange energy but not matter. An open system also has matter entering or leaving, and the balance needs additional terms that we will not use here.

An equilibrium state can be described by pressure p, volume V, temperature T and amount of substance n. Pressure is perpendicular force per area; volume measures occupied space; absolute temperature is expressed in kelvin. A change connects states but is not determined only by its endpoints: slow expansion with thermal exchange and expansion inside an insulated box can connect states through very different processes.

When we use a continuous curve on a p–V graph, we are usually idealising a sufficiently slow succession of near-equilibrium states. This is a quasi-static process. An explosion or free expansion does not necessarily have uniform gas pressure at each instant, so we should not invent an equilibrium curve to calculate work.

The microscopic model and the meaning of pressure

A gas consists of particles in irregular motion. When they collide with a wall, they change their momentum and exert a force on it. Macroscopic pressure results from the average effect of many collisions. If we keep volume and particle number fixed and increase thermal agitation, collisions transfer more momentum per unit time, increasing pressure.

In the ideal gas model, particle size is negligible compared with their separation, and intermolecular interactions are neglected outside collisions. Collisions are treated as elastic. This is a useful approximation for many dilute gases, far from condensation; it is not an exact description of any gas at any pressure.

For this model, the average translational kinetic energy of a particle is 3kT/2, where k is Boltzmann’s constant. The particles do not all have the same speed: there is a distribution. Temperature characterises an average of the assembly; we should not assign a macroscopic temperature to a single collision. Molecules can also store energy in rotation and vibration, so total internal energy is not always just translational energy.

Building and using pV = nRT

The number N of particles can be counted through the amount of substance n: N = nN_A, with N_A = 6,02214076 × 10²³ mol⁻¹. A mole contains this number of specified entities. Total mass is not n: if M is molar mass, n = m/M, using consistent units. A gas of heavier molecules can have fewer particles for the same mass.

In this lesson we use R ≈ 8,314 J/(mol·K), p in Pa, V in m³, n in mol and T in K. Since Pa·m³ = N·m = J, both sides have the unit of energy. Pressure must be absolute: if a gauge measures the difference from the atmosphere, local atmospheric pressure must be added before using the equation.

For a fixed amount of ideal gas, pV/T is constant. Keeping T constant gives constant pV; keeping p constant gives constant V/T; keeping V constant gives constant p/T. These proportional relationships cannot be combined without stating what remains fixed. If the container loses gas, n changes and p₁V₁/T₁ = p₂V₂/T₂ no longer works without correction. Nor do we use Celsius in temperature ratios: 600 K is twice 300 K; 60 °C is not twice 30 °C on the absolute scale.

The first law and signs of transfers

Internal energy U is associated with the system’s microscopic degrees of freedom. It is not heat that the gas “contains”. Heat Q and work W describe energy crossing the boundary during a process. U is a state function: between the same equilibrium states, ΔU is the same even if the individual transfers differ.

We use positive Q when energy enters the gas because of a temperature difference, and positive W when the gas does work on the surroundings. During compression, work done by the gas is negative: the surroundings supply energy to the system. Some books define work on the gas as positive and write ΔU = Q + W_on. The two conventions are equivalent because W_on = −W; the mistake is changing convention halfway through.

This form assumes that the relevant changes are in internal energy and not, for example, acceleration of the entire cylinder or its rising. When those macroscopic energies vary, they must enter the total balance. The first law conserves energy, but does not by itself determine a change’s spontaneous direction.

Why work appears as area on a p–V graph

On a piston of area A, an external pressure p_ext exerts a force p_extA. A small displacement Δx corresponds to a volume change ΔV = AΔx. Thus the work done by the gas against that pressure is approximately p_extΔV. With constant external pressure, W = p_ext(V₂ − V₁). In a quasi-static process in mechanical equilibrium, gas pressure approaches external pressure and we can use the p(V) curve.

If pressure varies, we divide the change into small intervals and add the pΔV contributions. The limit is W = ∫p dV for the indicated quasi-static path. The area under the curve measures signed work: expansion gives positive W, compression gives negative W. Two paths with the same initial and final volumes can have different areas. Knowing ΔV alone is not enough to know W.

In free expansion into an empty region, external pressure is zero and there is no boundary work, even though volume increases. Using the gas’s initial pressure in W = pΔV would give an improper result: that was not the external pressure resisting motion.

Four changes we must not confuse

• Isochoric: constant volume. There is no p dV boundary work, so ΔU = Q if this is the only type of work. For a fixed amount of ideal gas, increasing T increases p.
• Isobaric: constant pressure along the path considered. If volume changes quasi-statically, W = pΔV. Energy received can be divided between an increase in U and work.
• Isothermal: constant temperature. For a fixed amount of ideal gas, U depends only on T, so ΔU = 0 and Q = W. This does not mean there is no heat exchange: it may be precisely that exchange that keeps temperature constant.
• Adiabatic: Q = 0. It does not mean constant T. If the gas expands and does work, internal energy decreases. Adiabatic compression can raise temperature.

For a reversible isothermal expansion of an ideal gas, we substitute p = nRT/V into the integral and obtain W = nRT ln(V₂/V₁). The logarithm’s argument is a dimensionless ratio. For a reversible adiabatic change of an ideal gas with constant heat capacities, pV^γ is constant, with γ = C_p/C_V. The word reversible matters: an adiabatic free expansion does not automatically obey this relationship.

How much of the energy received raises temperature?

The molar heat capacity at constant volume C_V relates ΔU and ΔT of an ideal gas over an interval where the capacity can be treated as constant: ΔU = nC_VΔT. For a monatomic ideal gas, considering only translational energy, C_V = 3R/2 and U = 3nRT/2. For molecules with other active degrees of freedom, the capacity differs and may vary with temperature.

At constant pressure the gas also does work. Using pΔV = nRΔT, the first law gives Q = nC_VΔT + nRΔT = nC_pΔT. Therefore C_p − C_V = R for an ideal gas. More energy per mole and kelvin is needed when some is delivered as expansion work.

A diatomic gas near temperatures where translation and rotation are active may have C_V approximately 5R/2. This is an approximation for a particular regime, not a universal temperature-independent property. Always check the model and heat capacity given before using memorised numbers.

Cycles, heat engines and the second law

In a cycle, the system returns to its initial state, so the total change in internal energy is zero. Net work equals net heat received. On a quasi-static p–V graph, it corresponds to the area enclosed by the cycle, with its sign depending on the direction of travel.

A cyclic heat engine receives energy Q_q from a hot source, delivers part as work and rejects an amount of energy Q_f to a cold source. Here Q_q and Q_f represent positive magnitudes: W = Q_q − Q_f and η = W/Q_q. Efficiency is a ratio; an engine with efficiency 0,40 delivers 40% of the energy received as net work.

The second law rules out a cyclic engine whose only effect is to take heat from a single source and convert all of it into work. Between reservoirs at T_q and T_f, the reversible Carnot limit is η_max = 1 − T_f/T_q, with absolute temperatures. Friction, finite temperature differences and other irreversible processes make actual performance lower than the ideal limit.

Entropy helps quantify this irreversibility. In a reversible process, the small entropy change is dS = δQ_rev/T. The total entropy of an isolated system does not decrease. It is not energy disappearing: energy is conserved, but the ability to obtain useful work may decrease. This introduction prepares the topic; calculating entropies for general processes requires careful definitions of states, boundaries and reversible reference paths.

Worked examples, without skipping steps

Follow the reasoning first. Then cover the solution and try to reconstruct it in your own words.

Now try it yourself

There is no timer. Write what you know, choose a relationship and justify each step. Hints and solutions are available whenever you need them.

Mistakes worth avoiding

Using Celsius in a temperature ratio.

Convert to kelvin. The scale origin changes ratios such as T₂/T₁, although intervals in °C and K have the same size.

Adiabatic means constant temperature.

Adiabatic means no heat transfer. Work can change internal energy and temperature.

ΔU always equals heat received.

Only in that simple form when the total work considered is zero. First write ΔU = Q − W with signs defined.

Any volume increase allows W = initial p × ΔV.

Work depends on the path and external pressure. That multiplication does not describe free expansion.

Conserving energy is enough for any engine to operate.

The second law and efficiency limits between reservoirs must also be respected.