The ideal gas law

PV = NkBT is called the ideal gas law.  It is the equation of state, which relates the quantities V, P, and T.  Most real gases at ordinary temperatures and pressures obey the ideal gas law.  The ideal gas law can be rewritten as

PV = nNAkBT = nRT.

Here n is the number of moles of the gaseous substance, NA is Avogadro's number, NA = 6.022*1023 molecules/mol,  and R = NAkB is a constant, called the universal gas constant,  R = 8.31 J/(mol K). 
The number of moles n is given by n = m/M where m is the average mass of the gas particles in the volume,  and M is the molar mass of the gas.

In the 17th and 18th century experiment with gases at very low temperature and pressures revealed three relations that are generalized by the ideal gas law.

The ideal gas law holds well for real gases at low densities and pressures, such as atmospheric density and pressure.  If we use T = 0 oC = 273 K and P = 1 atm, then we find that one mole of gas occupies a volume of 22.4 liters.  One mole of gas contains NA gas particles.  For all low density gases, NA gas particles occupy 22.4 liters at T = 273K and P = 1 atm.

Nothing in the ideal gas law depends on the nature of the gas particles.  The value of PV/T depends only on the number of gas particles and a universal constant. 

Note:  In all gas laws T denote the absolute temperature, measured in Kelvin in SI units.

Problem:

The particle density of atmospheric air at 273.15 K at sea level is 2.687*1025/m3.  Calculate the pressure P.

Solution:

Problem:

If a helium-filled balloon initially at room temperature is placed in a freezer, will its volume increase, decrease, or remain the same?

Solution:

External link:  Balloons in Liquid Nitrogen  (Youtube)

Problem:

A tank having a volume of 0.1 m3 contains helium gas at 150 atm.  How many balloons can the tank blow up, if each filled balloon is a sphere 0.3 m in diameter at an absolute pressure of 1.2 atm?

Solution:

Problem:

The mass of a hot air balloon and its cargo (not including the air inside) is 200 kg.  The air outside is at 10 oC and 101 kPa.  The volume of the balloon is 400 m3.  To what temperature must the air in the balloon be heated before the balloon will lift off.  (Air density at 10 oC is 1.25 kg/m3.)

Solution:


imageThe earth atmosphere is a gas in the presence of gravity.  Gravity prevents the atmosphere from escaping.  Near the Earth's surface the acceleration due to gravity is approximately constant (g = 9.8 m/s2).  The atmosphere is a fluid, and the pressure in this fluid is a function of the altitude.  For a liquid, i.e.  an incompressible fluid, we found that P = P0 + ρgy, where y is the distance below the reference point.  But the atmosphere is a compressible gas.  Let us consider a column of gas containing ρparticle particles of mass m per unit volume near the Earth's surface.

The pressure times the area (i.e.  the force) at height y must exceed that at height y + dy by the weight of the intervening gas.  We need 
Py+dyA - PyA = -mρparticleVg = -mρparticlegAdy,  or 
dP = -mρparticlegdy. 

But from the ideal gas law we know that P = ρparticlekT, with k being the Boltzmann constant.  We may therefore write
dP = -P(mg/kT)dy, or 
dP/P = -(mg/kT)dy. 

If the temperature T is constant, then this integrates to 
P = P0e-mgy/(kT)

The pressure decreases exponentially with altitude.
Near Earth's surface the temperature T is not constant, it generally decreases with altitude y.  In the standard atmosphere T = T0 - ay, with a = 0.0065 K/m.  We then have 
dP/P = -mgdy/(k(T0 - ay)). 

This integrates to 
ln(P/P0) =( mg/ka)ln(1 - ay/T0), or 
(P/P0)(ka/mg) = (1 - ay/T0).

Since the Earth's atmosphere consists mostly of nitrogen molecules, we have 
(P/P0)0.19 = (1 - (0.0065 (K/m))y/T0).

T0 and P0 are the temperature (in Kelvin) and the pressure at sea level and the altitude y is measured in m.  The pressure, density, and temperature of the atmosphere decrease as the altitude increases.

External link:  Standard Atmosphere Calculator


How can we make a balloon rise?

We have to make its weight smaller than the weight of an equal shaped volume of surrounding air.  The skin of the balloon and the basket certainly are heavier than an equal volume of the surrounding air.  The inside of the skin therefore has to be filled with something appreciably lighter than the surrounding air.  Ideally we would empty it out completely.  But then the pressure inside would be zero.  The air pressure from the outside would collapse the skin.  To keep the skin stretched, the inside and outside pressures on each section of skin must be equal, but the weight of the inside material must be less than that of an equal volume of surrounding air.  The pressure is proportional to the particle density and the temperature, while the weight is proportional to the particle density and the weight of each particle.

P = kρparticleT,

weight = ρparticle * volume * weight per particle.

To get the weight down and keep the pressure the same we have two options.