The ideal gas law connects the pressure, volume, and temperature:

(1) for a fixed volume, a higher temperature leads to an increase in pressure, P ≈ T,

(2) at constant pressure, a higher temperature leads to an increase volume, V ≈ T, and

(3) at constant temperature, a higher pressure is associated with a smaller volume, P ≈ 1/V. This can be summarized as

**PV / T = const, **

where T is the absolute temperature [measured in Kelvin]. This law assume that the amount of gas [i.e., the mass] is kept constant.

**Example:** A glass column is filled with air. At room temperature, the pressue is 2.5-times the atmospheric pressure. What is the pressure inside the glass column when it immersed in boiling water?

**Solution:**

We have the absolute temperatures T_{0} = 298 K and T_{f} = 373 K.

Since the volume is constant **V _{0} = V_{1} = V, P_{0} V / T_{0} = P_{1} V / T,** so that,

**P _{1} = (T_{1}/T_{0}) P_{0} = (373K/298K) 2.5atm = 3.1 atm**

In an ideal gas, molecules are moving in random directions aat random speeds inside a container. The molecules bounce off the walls. During a collision, the wall of the container exerts a force on a molecule; thus, the molecule exerts a (reaction) force on the container wall: the summation of all forces from all collisions produces the macroscopic pressure.

A simplified derivation starts from a cubic container with volume V = L^{3} ; we we choose a coordinate system aligned with the cube. We assume that the molecule travels with speed v along the x -coordinate. The collision with the wall gives the change in momentum Δp = (-mv_{x}) – m(v_{x}) = – 2mv_{x}, . Since the time between consecutive collisons is t = 2L/v , the (average) force by one molecule follows F_{aye} = ( -2mv_{x})/ (2 L/v_{x}) = – mv_{x}^{2}/L . The particle moves in all directions so that v_{x}^{2} = (v^{2}) /3, where we introduced the mean square value (v^{2}) . Thus for the force due to N molecules F = Nm (v^{2}) /3L so that for the pressure P = F/L^{2} = Nm (v^{2}) /3V.

We find,

where we introduced the average kinetic energy of the molecule **(KE) = m (v ^{2}) /2**. Comparison with ideal gas law PV/T = const yields

**(KE) = (m/2) V ^{2}_{rms} = 3/2 kT,**

where k = 1.381 x 10^{-23} J/K is the Boltzmann constant.

The number of molecules in a gas is enormous N > > 1. Avogardo’s number is used to quantify large numbers,

N_{A} = 6.022 x 10^{23}

The number of moles is then given by **n = N/ N _{A}**

The product of the molecular mass and Avogadro’s number is the molar mass M = N_{A} m. For helium M_{he} ≈ 4 g; and air [80% N^{2} and 20% O_{2}]. M_{air} = 28.8 g. Avogadro’s number and the Boltzmann constant are the connection between microscopic and macroscopic quantities. The unit for masses of molecules is the atomic mass unit:

**1 u = (1 g / N _{A}) = 1.66 x 10^{-27} kg. N_{A}**