Ideal Gas Law

By Levi Clancy for Student Reader on
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Ideal Gas Law

The Ideal Gas Law (aka the Equation of State for an ideal gas) relates pressure, volume, mass and temperature. The word ideal refers to the fact that in reality, gases do not follow it perfectly, particularly at high density or near the liquefaction/boiling point. The Ideal Gas Law is derived from Boyle's, Charles' and Gay-Lussac's Laws (shown below).

Based on the number of moles of substance:

P V = n R T

P = Pressure (in N m-2)
V = Volume (in m3)
n = Number of moles
T = Temperature (in °K)
R = Ideal Gas Constant = 8.314 J mol-1 K-1

Based on the number of molecules of substance:

P V = N k T

N = Number of molecules of substance
k = Boltzmann Constant = 1.38 × 10-23 J K-1

Ideal Gas Constant (R)8.314 J mol-1 K-1
Boltzmann Constant (k)1.38 × 10-23 J K-1
Standard Conditions

When dealing with thermodynamics, it is often stated that a system is under standard conditions (aka standard temperature and pressure, or STP). This refers to:

The temperature is 273 K (0°C)
The pressure is 1.00 atm = 101.3 kPa

Simplifications

Sometimes not enough data on changes in a closed system is provided in order to use the Ideal Gas Law straightforward. In that case, use the below equation and cancel out what is in common between the before and the after of the closed system.

( P1 × V1 ) / ( n1 × T1 ) = ( P2 × V2 ) / ( n2 × T2 )

For example, consider a situation where the temperature changes, but the quantity and volume remains constant. You want to find the new pressure. The equation simplifies to:

P1 / T1 = P2 / T2

Boyle's Law

Pressure × Volume = constant

If you increase pressure then the volume (V) will decrease; and vice-versa, such that the product of pressure and volume is always constant.
Charles' Law

Volume ∝ Temperature

Volume is proportional to temperature; they decrease and increase in tandem.

Gay-Lussac's Law

Pressure ∝ Temperature

Pressure is proportional to temperature; they decrease and increase in tandem.

Practice Problems

If 10.00 moles of helium gas is at 20.0°C and a gauge pressure of 0.300 atm, calculate the volume of the helium gas under these conditions. Calculate the temperature if the gas is compressed to precisely half the volume at a gauge pressure of 1.00 atm.

P V = n R T

The lowest pressure attainable using the best available vacuum techniques is about 10-12 N m-2. At such a pressure, how many molecules are there per cubic centimeter at 0°C?

P V = N k T

A cubic box of volume 5.12 × 10−2 m3 is filled with air at atmospheric pressure at 10°C. The box is closed and heated to 100°C. What is the net force on each side of the box?

You must find out the pressure at 100°C.
Then, the force may be deduced using Pressure = Force / Area.
This is a great example of an Ideal Gas Law problem where simplification must be performed.
Consider setting up two equations for the system at 10°C and at 100°C

P1 V1 = n1 R T1P2 V2 = n2 R T2

However, n1 = n2 since there is no loss of air when the box is heated.
Also, V1 = V2 since the volume is kept constant.
Further, the Ideal Gas Constant is of course the same.
Thus, solve both equations for V, n and R on one side.

P1 / T1 = n R / V
P2 / T2 = n R / V
P1 / T1 = P2 / T2

P1 = 1 atm = 101325 Pa
T1 = 10°C = 283.15°K
P2 = ?
T2 = 100°C = 373.15°K

101325 / 283.15 = P2357.849197 = P2133531.428 = P2 → The pressure at 100°C is 133531.428 Pa

The volume of the box is 5.12 × 10-2 m3Volume = ( Length of Side )3 → The cube root of 5.12 × 10-2 m3 is .8 m
Area of Side = ( Length of Side )2 → The surface area of each side is .64 m2There are 133531.428 Pa of pressure exerted on each side.
Pressure = Force / Area → Force = Pressure × Area
Thus, the internal force on each side is 133531.428 × .64 = 85460.1139 N

However, the question is asking for the net force on each side.
There is the force of the pressure inside the box, pushing outwards, which was calculated above.
There is also the force of the atmospheric pressure against the outside of the box, pushing inwards.
External Force = Atmospheric Pressure × Area → External Force = 101325 × .64 = 64848 N

The internal and external pressures exert force in opposite directions.
Thus, the net force is the difference between the two.
85460.1139 - 64848 = 20612.1139 N → There is a net force of 20600 N pushing outwards on each side.

A standard tank of oxygen used in a hospital has a gauge pressure of 2000 PSI and a volume of 5 liters at a temperature of 20°C. How long will the cylinder last if the flow rate, measured at atmospheric pressure, is constant at 2½ liters per minute?

2½ liter of atmospheric pressure air exit the tank every minute.
How many liters of air at atmospheric pressure are actually in the tank?
P V = n R T

P1 = 2000 PSI → 2000 × 6894.75729 Pa = 13789514.6 Pa
P2 = 1 atm → 101325 Pa
V1 = 5 L = .005 m3V2 = ?
T1 = T2 = 20°C = 293.15°K
n1 = n2 = ?

P1 V1 = n R T
P2 V2 = n R T
P1 V1 = P2 V213789514.6 · .005 = 101325 · V20.68045964 m3 = V2

The tank contains a total of .68045964 m3 of air at atmospheric pressure.
The tank loses .0025 m3 of air at atmospheric pressure every minute.
Thus, it will last .68045964 / .0025 → 272.183856 minutes → The tank will last 4½ hours.