5. Molecular PhysicsSavchenko Formulas, chapter 5 of 14, 56 formulas

S

area cut out on the sphere

L

radius of the sphere, distance to the patch

A solid angle is the area a cone cuts out on the unit sphere. The whole sphere subtends $4\pi$, a half space $2\pi$. The fraction of molecules heading for a distant target is its solid angle over $4\pi$.

$v_{\text{кв}}$

root of the mean square speed

$\mu$

molar mass

R

gas constant

The thermal speed of a molecule follows from $m v^2/2 = 3kT/2$. Through the molar mass the same formula reads $\sqrt{3RT/\mu}$, so light gases move faster at the same temperature. For estimates $v \sim \sqrt{kT/m}$ is enough.

N₁, N₂

numbers of molecules passing per second

$\mu_1, \mu_2$

molar masses

T₁, T₂

temperatures on the two sides

Through a hole small against the mean free path molecules leave one by one, and the flux is proportional to $n v$, that is to $p/\sqrt{mT}$. So a light gas leaks faster than a heavy one, and two vessels at different temperatures settle at $p_1/\sqrt{T_1} = p_2/\sqrt{T_2}$, not at equal pressures.

k

Boltzmann's constant

T

absolute temperature

m

mass of a molecule

Temperature measures the mean kinetic energy of random motion. Each translational degree of freedom carries $kT/2$, three of them $3kT/2$. The same holds for a Brownian particle and any body in thermal equilibrium, so $kT$ estimates the thermal swing of a pendulum or a torsion fibre.

$f(v)$

probability density of the speed

$dN$

number of molecules with speeds from $v$ to $v + dv$

N

total number of molecules

The fraction of molecules within a speed interval $dv$ is $f(v)\,dv$, and the whole area under the graph is one. Averages and fluxes are computed through $f$. If every speed is multiplied by $l$, the distribution stretches and normalization asks for a factor $1/l$.

U

potential energy of a molecule

n₀

concentration where $U = 0$

h

height

In a force field the concentration of molecules falls with their potential energy as $e^{-U/kT}$. In gravity this is the barometric formula, and pressure falls with height as the concentration does. The ratio of concentrations at two points depends only on the energy difference.

F

force on the molecule

$\tau$

flight time

l

path against the force

While a molecule crosses a region with a field, the force changes its momentum by $F\tau$ and the field's work changes its kinetic energy. This is how a molecular beam is deflected and sorted by speed, including in a rotating selector where the speed is tied to the pitch and the angular velocity.

v_x

velocity component along an axis

m

mass of a molecule

T

temperature of the gas

A velocity component is Gaussian with variance $kT/m$. The fraction of molecules with $v_x$ above a threshold $b$ is an error integral, falling as $e^{-b^2}$ for a high threshold. For the speed the Gaussian gets an extra factor $4\pi v^2$.

n

number density of molecules

$\sigma$

collision cross-section

d

effective diameter of a molecule

A molecule flying through a gas hits every molecule inside a cylinder of cross-section $\sigma = \pi d^2$, so the path between collisions is about $1/(n\sigma)$. The factor $\sqrt{2}$ accounts for the other molecules moving too, with a relative speed $\sqrt{2}$ times larger on average. The free path is inversely proportional to pressure.

r₁, r₂

radii of the colliding molecules

Z

collisions of one molecule per second

$\langle v\rangle$

mean speed

Two molecules collide when their centres come within the sum of the radii, hence the cross-section $\pi(r_1+r_2)^2$. In a second a molecule covers $v$ and hits $n\sigma v$ neighbours, with the $\sqrt{2}$ correction for their motion. The time between collisions is $\lambda/v$.

q

heat flux density

$\kappa$

thermal conductivity

c₁

heat capacity of one molecule, $\frac{i}{2}k$

Molecules from the hot side bring more energy to a surface than those from the cold side take away, the difference building up over one free path. The product $n\lambda$ does not depend on pressure, so a gas conducts heat independently of pressure as long as the free path is shorter than the vessel. Light gases with small molecules conduct best.

D

diffusion coefficient

$\lambda$

mean free path

$d\rho/dx$

density gradient of the carried gas

The flux of matter is proportional to the concentration gradient, and the coefficient is $\lambda v/3$ in order of magnitude. Molecules arrive at a surface from one free path on either side, carrying the density difference. Through $\lambda \propto 1/p$ the diffusion coefficient is inversely proportional to pressure.

N

number of steps, collisions

$\lambda$

length of one step, the free path

D

diffusion coefficient

With random steps in independent directions it is the squared displacements that add, so after $N$ steps a molecule has strayed $\lambda\sqrt{N}$. Hence the diffusion time over a distance $L$ grows as $L^2/D$, with $D \sim \lambda v$. This estimates how long a smell takes to cross a room.

p

gas pressure

S

cross-section of the body

v

speed of the body

$\sqrt{RT/\mu}$

thermal speed of the molecules

When the free path exceeds the body, molecules hit it independently. More of them arrive at the front, and with more momentum, than at the back, and the pressure difference is $p$ times $v/v_{\text{th}}$ in order of magnitude. The force grows linearly with speed while the body is slower than the molecules.

q

heat flux per unit area

n

number density

$\Delta T$

temperature difference of the walls

i

degrees of freedom

When the free path exceeds the gap, a molecule flies wall to wall without collisions and carries $\frac{i}{2}k\,\Delta T$ per crossing. The heat flux is proportional to the number density and hence to pressure, unlike in a dense gas. The Pirani gauge and the Dewar flask rest on this.

p₁, p₂

pressures in the vessels

T₁, T₂

temperatures of the vessels

F

force on the plate

When vessels join through a hole smaller than the free path, equilibrium means equal opposite fluxes $n\sqrt{T}$, not equal pressures. Molecules leaving the hot side of a plate carry more momentum, and in a rarefied gas the plate feels a force that measures the pressure.

r₁, r₂

radii of the inner and outer cylinders

$\omega, \omega'$

angular velocities of the cylinders

A molecule leaving a rotating wall carries its tangential speed and keeps its angular momentum on the way to the other wall. The momentum the molecules bring balances only when $\omega r^2$ agrees for both cylinders, and a rarefied gas transmits rotation this way rather than by viscosity.

p

pressure

V

volume

$\nu$

number of moles

$\mu$

molar mass

R

gas constant, $8.31$ J/(mol K)

The equation of state ties the pressure, volume and temperature of any amount of ideal gas. It holds while molecules rarely interact, that is for ordinary gases at moderate densities. All the isoprocesses follow from it, and for a mixture it is written for each gas separately.

n

number density, molecules per unit volume

k

Boltzmann's constant, $1.38\cdot10^{-23}$ J/K

The equation of state per molecule. The pressure of a mixture is the sum of the partial pressures, since each kind of molecule pushes on the wall independently. Through $n$ one computes the free path, fluxes and the number of molecules in a volume.

p₀

atmospheric pressure

M

mass of the piston or load

S

area of the piston

h

height of the liquid column

The gas pressure under a piston or a liquid column follows from the equilibrium of the piston or the column, and the volume and temperature then follow from the equation of state. A column of mercury or water in a tube adds $\rho g h$ to the atmosphere or takes it away, depending on which side the gas is.

$V = \text{const}$

isochore, gas sealed in a rigid vessel

$p = \text{const}$

isobar, a free piston or the atmosphere sets the pressure

In a rigid vessel pressure is proportional to absolute temperature, under a free piston the volume is. Both lines pass through the origin on the Kelvin scale, which makes converting states easy.

$\rho$

density of the gas

$\mu$

molar mass

From $pV = \frac{m}{\mu}RT$ the density is proportional to pressure and molar mass and inversely proportional to temperature. At one pressure and temperature the densities of gases are as their molar masses. This gives the lift of a balloon and the mass of gas in a cylinder.

p

pressure

V

volume

At constant temperature the product of pressure and volume of a given mass of gas stays the same. Air trapped in a tube by a column of mercury or water behaves so when moved slowly enough for heat to escape.

$\mu_{\text{возд}}$

air

$\mu_{\text{H}_2}$

hydrogen

$\mu_{\text{He}}$

helium

$\mu_{\text{H}_2\text{O}}$

water vapour

The molar mass in grams per mole equals the relative molecular mass numerically. Air has 29, nitrogen 28, oxygen 32, carbon dioxide 44. SI formulas take kilograms per mole.

$\rho_{\text{возд}}$

density of the surrounding air

$\rho_{\text{газа}}$

density of the gas in the envelope

V

volume of the balloon

A balloon is lifted by the buoyant force less the weight of the gas inside, both expressed through densities at one pressure, that is through molar masses. The envelope adds its own weight, so for a small balloon it wins and there is a smallest radius. Hot air is lighter than cold air at the same pressure.

$\nu$

number of moles

$\mu$

molar mass

N_A

Avogadro's number, $6.02\cdot10^{23}$ per mole

A mole holds $N_A$ molecules and weighs $\mu$. The gas constant and Boltzmann's constant are tied by $R = kN_A$. A mixture's molar mass is its total mass over its total number of moles.

p₁, V₁, T₁

parameters in the first state

p₂, V₂, T₂

in the second

For a fixed mass of gas $pV/T$ is the same in every state, because the right side of the equation of state is $\nu R$. The isoprocesses are the cases with one parameter fixed.

A

work done by the gas on its surroundings

p

gas pressure

$dV$

change of volume

The work of a gas is the area under the process on the $pV$ diagram, positive on expansion. On an isobar it is $p\Delta V = \nu R\Delta T$, on a straight-line process the area of a trapezoid, over a cycle the area inside the loop. The work of outside forces on the gas is equal and opposite.

Q

heat supplied to the gas

$\Delta U$

change of internal energy

A

work done by the gas

Heat received by a system raises its internal energy and does work against outside forces. Signs matter, an ideal gas's $\Delta U$ depends only on temperature while the work depends on the path on the $pV$ diagram. In an adiabatic process $Q = 0$ and the work comes out of the internal energy.

i

degrees of freedom, 3 for a monatomic, 5 for a diatomic gas

$\nu$

number of moles

C_V

molar heat capacity at constant volume

The internal energy of an ideal gas is the sum of the molecules' kinetic energies, $kT/2$ per degree of freedom. It depends only on temperature, not on volume or pressure, so its change in any process is $\nu C_V\Delta T$. Through the equation of state $U = \frac{i}{2}pV$.

$\gamma$

adiabatic exponent

i

degrees of freedom, 3 or 5

The ratio of heat capacities enters the adiabatic equation and the speed of sound. A monatomic gas has $\gamma = 5/3$, a diatomic one $7/5$. Through $\gamma$ the internal energy reads $pV/(\gamma-1)$.

c

specific heat capacity

m

mass of the body

$\Delta T$

temperature change

The heat needed to warm a body is proportional to its mass and to the temperature change. In an insulated system the heat given up by some bodies equals the heat received by others, and this equation together with the latent heats solves every mixing problem. A heater of constant power delivers heat proportional to time.

V₁, V₂

initial and final volumes

T

temperature of the process

On an isotherm $p = \nu RT/V$, and the integral $\int p\,dV$ gives the logarithm of the volume ratio. The internal energy stays fixed, so all the heat supplied turns into work. On compression the gas's work is negative.

C_V

molar heat capacity at constant volume

C_p

at constant pressure

i

degrees of freedom

At constant volume all the heat goes into internal energy, at constant pressure also into the work $R\,\Delta T$ per mole, hence $C_p - C_V = R$. A monatomic gas has $C_V = \frac{3}{2}R$, a diatomic one $\frac{5}{2}R$. The specific heat is the molar one over $\mu$.

$\gamma$

adiabatic exponent $C_p/C_V$

Without heat exchange $\nu C_V\,dT = -p\,dV$, and with the equation of state this gives $pV^\gamma = \text{const}$. An adiabat is steeper than an isotherm, and adiabatic compression heats the gas. A process is adiabatic when fast or the vessel is insulated, and the work equals the drop in internal energy.

F

force on the piston, friction or a load

S

piston area

p₀

atmospheric pressure

While the piston stands, the gas heats at constant volume, and once the gas's force beats friction and the atmosphere it moves isobarically at $p_0 + F/S$. The heat is split between internal energy, work against the atmosphere, lifting the load and friction, and heat from the piston's friction may partly return to the gas.

$A_{\text{цикл}}$

work per cycle, the area inside the loop on the $pV$ diagram

$Q_{\text{подв}}$

heat taken from the heater

$Q_{\text{отд}}$

heat given to the cooler

Over a cycle the internal energy returns to its value, so the work equals the heat taken minus the heat given up and the area of the loop. The work is summed over the legs, the heat on each from the first law with its sign. In a cycle of two isobars and two isochores the corner temperatures obey $T_2^2 = T_1T_3$.

n

polytropic index

C

heat capacity of the gas in that process

Any process with constant heat capacity is a polytrope, and the converse holds. The heat capacity is found by writing the first law for a small step and dividing by $dT$. For a gas under a spring-loaded piston, where $p \propto V$, a monatomic gas has $2\nu R$.

z

height

$\mu$

molar mass of air

C_p

molar heat capacity at constant pressure

A rising parcel of air expands adiabatically, since heat exchange with its surroundings is slow, and cools. With $dp = -\rho g\,dz$ and the adiabat in the form $\nu C_p\,dT = V\,dp$ this gives a drop of about one degree per hundred metres. It estimates the temperature on a summit and the height of clouds.

T

temperature of the gas in the vessel

C_p

molar heat capacity at constant pressure

$\mu$

molar mass

The largest jet speed comes when the whole enthalpy of the gas in the vessel turns into kinetic energy, that is on discharge into vacuum. Discharging into a medium at pressure $p_2$ the factor $1 - (p_2/p_1)^{(\gamma-1)/\gamma}$ reduces it. A rocket's thrust is the mass flow times the exhaust speed.

$p, \rho$

pressure and density of the gas

v

flow speed

$\gamma$

adiabatic exponent

In steady flow without heat exchange the kinetic energy per unit mass plus its enthalpy $u + p/\rho$ is conserved, the flow work $p/\rho$ being added to the internal energy. For an ideal gas the enthalpy is $c_pT$, so heating of the jet and its acceleration trade one for the other.