Skip to content
Physics 1

Thermodynamics in depth: kinetic theory, Kelvin and Clausius, Carnot's theorem, Otto and Diesel cycles

Pressure and temperature from molecular motion, equipartition and specific heats, equivalence of the statements of the second law, Carnot's theorem, efficiency of the Otto and Diesel cycles, entropy in mixing and free expansion, table of processes.

9 min read 0 comments

This page goes deeper into the lessons on the first and second laws of thermodynamics. It derives the pressure of a gas from the motion of its molecules and the microscopic meaning of temperature, explains specific heats with the equipartition of energy, proves the equivalence of the Kelvin and Clausius statements and Carnot’s theorem, computes the efficiency of the Otto and Diesel cycles of combustion engines, and shows with two examples that the entropy of the universe increases in real processes. At the end is the table to remember of ideal-gas processes.

Pressure from molecular motion

Theorem. A gas of N molecules of mass m in a volume V exerts on the walls the pressure

    \[ p = \frac{1}{3}\,\frac{N}{V}\,m\,\langle v^2\rangle \]

where \langle v^2\rangle is the mean of the square of the speed.

Proof. Take a cubic container of side L and a wall of area A = L^2 perpendicular to the x axis. A molecule with component v_x hits the wall elastically: its momentum along x goes from mv_x to -mv_x, and the wall receives the impulse 2mv_x. The molecule returns to the same wall after travelling 2L back and forth, i.e. every 2L/v_x seconds. The average force it exerts is impulse over time: \frac{2mv_x}{2L/v_x} = \frac{mv_x^2}{L}.

Summing over all molecules: F = \frac{m}{L}\sum v_x^2 = \frac{mN\langle v_x^2\rangle}{L}, and p = \frac{F}{L^2} = \frac{Nm\langle v_x^2\rangle}{V}. The directions are equivalent, so \langle v_x^2\rangle = \langle v_y^2\rangle = \langle v_z^2\rangle = \frac{1}{3}\langle v^2\rangle. ∎

A molecule hitting the wall of a container
The blue molecule reaches the yellow wall of area A with momentum m v x along x and leaves with minus m v x: the wall receives the impulse 2 m v x. The time between two hits on the same wall is the time to travel twice the side L.

The meaning of temperature. Comparing with the equation of state pV = Nk_BT (with k_B = R/N_A = 1.38\cdot 10^{-23} J/K):

    \[ \frac{1}{2}m\langle v^2\rangle = \frac{3}{2}k_BT \]

Absolute temperature measures the mean translational kinetic energy of the molecules. The root-mean-square speed is v_{rms} = \sqrt{3k_BT/m} = \sqrt{3RT/M}: for nitrogen at 300 K about 517 m/s, for hydrogen 1930 m/s.

Equipartition and specific heats

The previous result says that each translational direction carries on average \frac{1}{2}k_BT. The equipartition theorem (from statistical mechanics) generalises this: every quadratic term in the energy of a molecule (a degree of freedom) has on average \frac{1}{2}k_BT.

  • Monatomic gas: only 3 translations. U = \frac{3}{2}nRT, c_V = \frac{3}{2}R, c_p = \frac{5}{2}R, \gamma = \frac{5}{3}.
  • Diatomic gas at room temperature: 3 translations and 2 rotations (rotation about the molecular axis does not count). c_V = \frac{5}{2}R, c_p = \frac{7}{2}R, \gamma = 1.4. This is the value for air.
  • At high temperatures (above about 1000 K for N_2) vibrations “unfreeze”, with two quadratic terms (kinetic and elastic): c_V tends to \frac{7}{2}R.

Why vibrations are “frozen” at room temperature is explained by quantum mechanics: vibrational energy can only grow in jumps, and if the jump is much larger than k_BT collisions cannot excite it. It is one of the first cracks in classical physics, noticed by Maxwell as early as 1860.

Kelvin and Clausius say the same thing

Kelvin-Planck statement: there is no cyclic machine that converts into work all the heat taken from a single reservoir. Clausius statement: there is no cyclic machine whose only effect is to transfer heat from a cold body to a hot one.

Theorem. The two statements are equivalent: if one is false, so is the other.

Proof. *If Clausius is false, Kelvin is false.* Suppose a refrigerator F exists that, without work, carries heat Q_2 from the cold reservoir T_2 to the hot one T_1. Couple F with an ordinary heat engine M that takes Q_1 from T_1, produces W = Q_1 - Q_2 and gives exactly Q_2 to T_2. Overall the cold reservoir does not change (it receives and gives Q_2), the hot reservoir gives Q_1 - Q_2, and the work W = Q_1 - Q_2 is obtained: a machine that turns entirely into work the heat of a single reservoir. Kelvin is violated.

*If Kelvin is false, Clausius is false.* Suppose a machine exists that turns into work W all the heat Q taken from T_1. Use W to run an ordinary refrigerator that takes Q_2 from T_2 and gives Q_2 + W to T_1. Overall the hot reservoir receives Q_2 + W - Q = Q_2, the cold one gives Q_2, and there is no external work: Clausius is violated. ∎

Proof of the equivalence of the Clausius and Kelvin statements
The refrigerator F, in the dashed box, carries the heat Q2 from the cold to the hot reservoir without work, violating Clausius. Together with the engine M, which gives exactly Q2 to the cold reservoir, it forms a system that takes heat from a single reservoir and turns it all into the work W.

Carnot’s theorem

Theorem. No engine working between two reservoirs T_1 \gt T_2 has an efficiency greater than a reversible engine between the same reservoirs; all reversible engines between T_1 and T_2 have the same efficiency, \eta_C = 1 - \frac{T_2}{T_1}.

Proof. Let X be an engine with efficiency \eta_X \gt \eta_R, where R is reversible. Adjust the two engines so that they produce the same work W: X takes Q_1^X = W/\eta_X from T_1, R would take Q_1^R = W/\eta_R \gt Q_1^X. Run R in reverse, as a refrigerator, driven by the work W of X: R gives Q_1^R to T_1 and takes Q_1^R - W from T_2. Overall no work is produced, the hot reservoir receives Q_1^R - Q_1^X \gt 0 and the cold one gives the same amount: heat from cold to hot with no other effect, against Clausius. So \eta_X \le \eta_R.

If X is also reversible, swapping the roles gives \eta_R \le \eta_X: the efficiencies are equal. It is then enough to compute it for the Carnot cycle of the ideal gas, as in the lesson. ∎

The Carnot efficiency depends neither on the substance nor on the machine, only on the temperatures: it is a limit of nature, not of technology. A power plant with steam at 840 K and a condenser at 300 K cannot exceed 64 %; real ones reach about 45 %.

The Carnot cycle in the temperature-entropy plane
In the T S plane the Carnot cycle is a rectangle: two horizontal isotherms and two vertical reversible adiabats. The area under the upper isotherm is the heat absorbed, the area under the lower isotherm the heat given out Q2, and the area of the rectangle is the work W.

Engine cycles: Otto and Diesel

The Otto cycle idealises the petrol engine: adiabatic compression 1→2, instantaneous combustion at constant volume 2→3, adiabatic expansion 3→4, exhaust at constant volume 4→1. The parameter is the compression ratio r = V_1/V_2.

Theorem. The efficiency of the ideal Otto cycle is \eta = 1 - \dfrac{1}{r^{\gamma - 1}}.

Proof. Heat enters only in the isochoric 2→3, Q_{in} = nc_V(T_3 - T_2), and leaves only in the isochoric 4→1, Q_{out} = nc_V(T_4 - T_1). On the adiabats TV^{\gamma - 1} is constant, and the volume ratios are the same: T_2 = T_1r^{\gamma - 1} and T_3 = T_4r^{\gamma - 1}. So T_3 - T_2 = (T_4 - T_1)r^{\gamma - 1} and

    \[ \eta = 1 - \frac{Q_{out}}{Q_{in}} = 1 - \frac{T_4 - T_1}{(T_4 - T_1)r^{\gamma - 1}} = 1 - \frac{1}{r^{\gamma - 1}} \]

∎

With r = 9 and \gamma = 1.4, \eta = 58.5 %. Increasing r increases the efficiency, but beyond about 10-12 the mixture ignites by itself before the spark (detonation, “engine knock”).

The Otto cycle in the pressure-volume plane
Adiabatic compression from 1 to 2, heating at constant volume from 2 to 3 with heat input, adiabatic expansion from 3 to 4, cooling at constant volume from 4 to 1 with heat output. The enclosed area is the work.

The Diesel cycle differs in that combustion happens at constant pressure (2→3), because the diesel fuel is injected and burns while the piston moves down. With the compression ratio r = V_1/V_2 and the cut-off ratio r_c = V_3/V_2, the same reasoning (with Q_{in} = nc_p(T_3 - T_2)) gives

    \[ \eta = 1 - \frac{1}{r^{\gamma - 1}}\cdot\frac{r_c^{\gamma} - 1}{\gamma(r_c - 1)} \]

For the same r the Diesel is less efficient than the Otto (the second factor is greater than 1), but by compressing only air it can use r = 15–22 without detonation, and in the end it is more efficient.

The Diesel cycle in the pressure-volume plane
Compared with the Otto cycle, heat enters at constant pressure, along the horizontal stretch from 2 to 3; the rest of the cycle is the same: two adiabats and cooling at constant volume from 4 to 1.

The entropy of the universe increases: two examples

Mixing hot and cold water. Two equal masses m of water, at T_1 and T_2, reach T_f = \frac{T_1 + T_2}{2}. Heat flows irreversibly, but to compute the entropy one imagines a reversible heating between the same initial and final states: dS = \frac{mc\,dT}{T}, so

    \[ \Delta S = mc\ln\frac{T_f}{T_1} + mc\ln\frac{T_f}{T_2} = mc\ln\frac{T_f^2}{T_1T_2} = 2mc\ln\frac{(T_1 + T_2)/2}{\sqrt{T_1T_2}} \gt 0 \]

because the arithmetic mean is always greater than the geometric mean (unless T_1 = T_2). The cold body gains more entropy than the hot one loses, because it receives the same heat at a lower temperature.

Free expansion. An isolated ideal gas expands into a vacuum from V to 2V when a valve is opened. There is no work (the vacuum exerts no pressure) and no heat (isolated): \Delta U = 0, and for the ideal gas the temperature does not change. Yet the process is irreversible: the gas does not go back by itself into the starting half. The entropy is computed along a reversible isotherm between the same states: \Delta S = nR\ln 2 \gt 0, although no heat has flowed. Entropy is not just “heat over temperature”: it measures the number of accessible microscopic states, which is multiplied by 2^N when the volume doubles.

Free expansion of a gas into a vacuum
On the left the gas fills half of the container, separated from the vacuum by a wall. On the right the wall has been opened and the gas fills the whole volume, at the same temperature.

Table to remember

Reversible processes of n moles of ideal gas, with molar c_V and c_p:

ProcessLawW (done by the gas)Q\Delta U\Delta S
isochoricV constant0nc_V\Delta Tnc_V\Delta Tnc_V\ln\frac{T_2}{T_1}
isobaricp constantp\Delta V = nR\Delta Tnc_p\Delta Tnc_V\Delta Tnc_p\ln\frac{T_2}{T_1}
isothermalpV constantnRT\ln\frac{V_2}{V_1}= W0nR\ln\frac{V_2}{V_1}
adiabaticpV^\gamma constant-nc_V\Delta T = \frac{p_1V_1 - p_2V_2}{\gamma - 1}0nc_V\Delta T0

Other formulas to keep at hand: c_p - c_V = R (Mayer); \gamma = c_p/c_V; on the adiabat TV^{\gamma - 1} and T^\gamma p^{1 - \gamma} are also constant; \Delta S = nc_V\ln\frac{T_2}{T_1} + nR\ln\frac{V_2}{V_1} between any two states; efficiency \eta = W/Q_{in} = 1 - |Q_{out}|/Q_{in}; coefficients of performance COP_{fridge} = Q_2/W \le \frac{T_2}{T_1 - T_2} and COP_{hp} = Q_1/W \le \frac{T_1}{T_1 - T_2}; R = 8.314 J/(mol·K) = 0.082\,06 L·atm/(mol·K).

Exercises on processes, cycles, engines and entropy, from basic to exam level, are on the exercise page linked at the top.

Related articles

Subscribe
Notify of
guest
0 Comments
Oldest
Newest Most Voted
0
Would love your thoughts, please comment.x
()
x