Ideal gases and constant-parameter processes

A rigid sealed bottle is heated. No particles have been added, yet the pressure rises. What changed in the molecular impacts? Follow how this invisible picture leads to a measurable relationship between pressure, volume and temperature.

A molecule strikes a wall and changes momentum; the wall receives an opposite impulse. In a short time, an enormous number of impacts combine to produce what appears to be a continuous pressure force. Increasing particle count per unit volume at unchanged motion produces more impacts. Increasing speeds at unchanged number density changes both collision frequency and the impulse per impact. A single mean speed is therefore insufficient for a quantitative description.

Let particle mass be m_0, number density n, and mean square speed \langle v^2\rangle. Square each speed first, then average; the unit is m²/s². For two particles with speeds v and 3v, \langle v^2\rangle=5v^2, whereas the square of their mean speed is 4v^2. An equilibrium gas has no preferred direction: the mean squares of the three velocity components are equal.

A faster particle transfers more energy when brought to rest. One particle's translational kinetic energy is m_0v^2/2, so averaging gives \langle E\rangle=m_0\langle v^2\rangle/2 in joules. Doubling every speed quadruples the mean energy. This is not the energy of every molecule: speeds differ and continually change in collisions. The mean describes the stable state of a large collection of particles.

Adding the impulses transferred to a wall by an equilibrium ideal gas gives p=m_0n\langle v^2\rangle/3=2n\langle E\rangle/3. The factor of one third appears because the motion component normal to this wall matters, and all three directions are equivalent. Pressure p is measured in pascals. Doubling number density at fixed mean energy doubles pressure; doubling energy at fixed number density does the same. In this model, removing the particles, n\to0, gives p\to0.

Temperature measures the intensity of thermal motion without tracking individual molecules. For a classical ideal gas, mean translational energy is \langle E\rangle=3kT/2, where T is absolute temperature in kelvins and k\approx1{,}38\cdot10^{-23}\,\text{Дж/К} is the Boltzmann constant. Doubling T doubles this energy. J/K multiplied by K gives J; temperature measures the mean energy scale here.

If light and heavy gases have the same temperature, must their particles move equally fast? The formula gives equal mean translational energies, not equal speeds. In m_0\langle v^2\rangle/2, a smaller mass must be balanced by a larger mean square speed. Lighter particles are faster in this sense. Rotation and vibration of polyatomic molecules are not included in 3kT/2. A classical model cannot be extended to very low temperatures without checking its validity.

Gas laws relate pressure to energy, so the temperature scale's zero matters. If t is the numerical temperature in °C, then T=(t+273{,}15)\,\text{К}. For example, 0 °C means 273.15 K: ordinary water does not lose thermal motion when it freezes. Use kelvins in temperature ratios. Dividing by 0 °C would otherwise imply an infinite pressure ratio where no such effect occurs.

The two scales have shifted zeros but equal division sizes. A change of 1 °C therefore equals a change of 1 K: subtracting temperatures cancels the 273.15 offset. Division does not cancel it. Heating from 10 to 20 °C means going from 283.15 to 293.15 K, not doubling absolute temperature. This distinction between a difference and a ratio explains why heating calculations can sometimes use Celsius differences while gas laws require kelvins.

To express the mean square using familiar m/s, define v_{\text{кв}}=\sqrt{\langle v^2\rangle}, the root-mean-square speed. It does not mean every molecule moves at that speed. To move from a particle to a mole, use molar mass M in kg/mol and R=N_Ak\approx8{,}31\,\text{Дж/(моль}\cdot\text{К)}. Multiplying k by the particle count per mole gives the same energy scale per mole.

Equate m_0\langle v^2\rangle/2 with 3kT/2, then use M=N_Am_0 and R=N_Ak. This gives v_{\text{кв}}=\sqrt{3RT/M}. The square root matters: quadrupling T only doubles the speed, since energy depends on speed squared. At the same T, quadrupling M halves the speed. Units inside the square root give m²/s²; entering M in g/mol instead of kg/mol would invalidate the calculation.

Substitute mean energy 3kT/2 into pressure 2n\langle E\rangle/3: the factors cancel, leaving p=nkT. This is the equilibrium ideal-gas equation of state in terms of number density. At fixed volume, heating increases mean impact energy and p; at fixed temperature, adding particles increases n and p. If the gas can expand during heating, however, number density can fall. A rise in T alone does not guarantee a rise in pressure.

For a whole sample, substitute n=N/V and express particle count in moles. The ideal-gas equation, also called the Mendeleev–Clapeyron equation, is pV=\nu RT=(m/M)RT, where \nu=m/M. Use pressure in Pa, volume in m³, mass m in kg, M in kg/mol and T in K. At fixed T and V, doubling amount of substance doubles p; at fixed p and T, doubling the amount requires twice the V. These parameters cannot be chosen independently.

If an instrument reports density, why find mass and volume separately? A unit volume contains n particles of mass m_0, so \rho=m/V=m_0n in kg/m³. Dividing the equation of state by V gives pM=\rho RT. At fixed T and composition, doubling density doubles pressure. At constant pressure, heating lowers density: the same mass occupies more volume. This describes the same ideal gas, not a new law.

Compare the initial and final equilibrium states of the same sample. If amount of substance is unchanged, the right side of pV/T=\nu R is constant: pV/T=\text{const}. Identify another parameter that stays constant and cancel it. Leakage changes the number of moles, so equal temperature and volume no longer guarantee equal pressure. Check that the vessel is sealed and no gas is added before choosing a simplified law.

Slowly compress the gas while allowing heat transfer that keeps T constant: an isothermal process. For the same sample, pV=\text{const}, Boyle's law (Boyle–Mariotte law). Halving volume doubles number density while mean energy stays the same, so pressure doubles. Expansion gives the reverse result. Strong compression can make a real gas depart from the ideal model; do not extend the mathematical hyperbola to zero volume without checking validity.

Return to the rigid sealed bottle: V and particle count are constant, and only the intensity of motion changes. An isochoric process gives p/T=\text{const}, the pressure law. The Russian naming used in the master calls it Charles's law; English texts commonly call it Gay-Lussac's or Amontons's law. Increasing absolute temperature by a factor of one and a half increases pressure by the same factor; cooling lowers it. The ideal p(T) graph is straight, but extending it to 0 K does not mean a real gas remains gaseous down to that temperature.

Replace the rigid wall with a freely moving piston under a constant load. Heating makes the gas expand at unchanged pressure: an isobaric process. For the same sample, V/T=\text{const}, commonly called Charles's law in English; the Russian convention used in the master calls it Gay-Lussac's law. Doubling T requires twice the V: lower number density compensates for greater particle energy. Again, use kelvins and equilibrium states; piston friction or a changing load could prevent constant pressure.

An ideal gas in a rigid sealed vessel is heated from 300 K to 450 K. Its initial pressure is 100 kPa. Find the final pressure.

The rigid vessel keeps volume constant. Sealing keeps the amount in moles unchanged.

Choose the constant-volume law: p₁/T₁=p₂/T₂.

p₂=p₁T₂/T₁=100×450/300=150 кПа.

Check: temperature and pressure both increased by a factor of 1.5.

The final pressure is 150 kPa.

Pressure results from momentum transfer to the walls. The relation p = nkT connects particles per unit volume with the mean energy scale.

Temperature ratios require kelvins; equal scale divisions allow equal numerical temperature differences in K and °C.

Choose a gas law after checking that amount of substance and the specified parameter remain constant. Limiting values must stay within the ideal-gas model's validity.

Next, include work done by the gas and energy transfer.