Isothermal, isobaric, isochoric and adiabatic processes

For n moles of an ideal gas pV = nRT leaves only two of the three variables free, so a state is a point in the p-V plane and a quasi-static process is a curve through it. That curve is what the work integral W = ∫p dV needs before it can be evaluated at all, because the integral wants p as a known function of V. Each name supplies one. Isochoric: V fixed, a vertical line, and dV = 0 kills the integral. Isobaric: p fixed, a horizontal line, and p comes outside the integral. Isothermal: T fixed, so p = nRT/V, a hyperbola. Adiabatic: Q = 0, and the curve p = p₁(V₁/V)^γ falls away faster than the hyperbola because the gas cools as it expands. Draw all four through one starting point and the whole behaviour of the topic is visible before a single number is substituted.

Hold the volume fixed and the gas can do no work, so the first law reduces to Q = ΔU and the measured heat gives the molar heat capacity directly: C_V = (1/n)(ΔU/ΔT). What happens next decides half the problems in this topic. For an ideal gas the internal energy is a function of temperature alone — kinetic theory gives U = (f/2)nRT for f active degrees of freedom — so once C_V has been identified as the temperature coefficient of U, that identification travels with U. Take 2.00 mol of argon from 300 K to 450 K and ΔU = n C_V ΔT = 2.00 × 12.47 × 150 = 3.74 kJ, whether the gas was heated in a rigid bottle, allowed to expand at constant pressure, or compressed by a piston with the heat blocked. Only the split between Q and W changes with the path. ΔU does not, because U is a state function and temperature is the only state variable it answers to.

Let the gas expand against a fixed external pressure and both terms of the first law are alive. The work is W = pΔV, and pV = nRT turns that into W = nRΔT — a result that never mentions p, because a higher pressure forces a proportionally smaller volume change. The heat needed is therefore Q = ΔU + W = n C_V ΔT + nRΔT = n(C_V + R)ΔT, which is where Cₚ = C_V + R comes from. Numbers make the split concrete. Heat 1.00 mol of nitrogen from 300 K to 400 K at 1.013 × 10⁵ Pa: the gas expands by nRΔT/p = 8.21 × 10⁻³ m³, does 831 J of work, gains 2079 J of internal energy, and so needs 2910 J of heat. The fraction that leaves through the piston is R/Cₚ = 8.314/29.10 = 0.286, so 28.6% of every joule supplied to a diatomic gas at constant pressure never stays in the gas.

Clamp the temperature — a thin-walled cylinder in a large stirred bath, moved slowly — and ΔU = n C_V ΔT = 0 for the whole process. The first law then reads Q = W exactly: every joule the gas delivers to the piston is a joule the reservoir has just delivered to the gas, and the gas is only the conduit. The work follows from p = nRT/V with T held constant, so W = ∫nRT dV/V from V₁ to V₂, which is W = nRT ln(V₂/V₁). Expand 1.00 mol at 300 K from 10.0 L to 25.0 L and W = 8.314 × 300 × ln 2.50 = 2.29 kJ out, matched by 2.29 kJ of heat in. Compress instead and the logarithm turns negative: work goes in, and the same energy must be dumped to the bath or the temperature will not hold. Note what the formula needs — only the volume ratio, so litres and cubic metres give the same answer provided both volumes use one unit.

Insulate the cylinder, or move the piston faster than heat can leak through it, and Q = 0. The first law becomes dU = −dW, that is n C_V dT = −p dV: the work is now financed entirely out of internal energy, so an expanding gas must cool. Substituting p = nRT/V separates the variables into C_V dT/T = −R dV/V, and integrating with R = Cₚ − C_V gives T V^(γ−1) = constant, which pV = nRT converts into p V^γ = constant. Differentiate that last form and dp/dV = −γp/V, against the isotherm's −p/V: at any shared point the adiabat is steeper by the factor γ. The work is easiest from the first law itself, W = −ΔU = n C_V (T₁ − T₂). Compress 1.00 mol of air, γ = 1.40, from 20.0 L to 5.00 L starting at 300 K: T₂ = 300 × 4.00^0.40 = 522 K, so W = 20.79 × (300 − 522) = −4.62 kJ. That is 4.62 kJ done on the gas, all of it still inside as a 222 K rise — which is why a bicycle pump gets hot.

Choosing the name is a physical judgement about two times: how long the process takes, against how long heat needs to cross the walls. A gas compressed over a minute in a thin copper cylinder standing in a water bath holds the bath's temperature and is isothermal. The same compression done in 30 ms in an insulated cylinder is adiabatic to within a per cent. Between those extremes no named process applies and the work has to be read off a measured p-V curve instead. The second requirement is easier to forget: all four relations assume quasi-static change, so that one pressure and one temperature describe the whole sample at every instant. Let a gas burst through a valve into an evacuated chamber and it is thermally isolated, Q = 0, but it pushes nothing, so W = 0 and ΔU = 0. The temperature is unchanged, the gas has not cooled at all, and p V^γ = constant is simply false along that path.