# Ideal Gas Law

By Levi Clancy for Student Reader on *updated *

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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 Based on the number of moles of substance: P V = n R T P = Pressure (in N m Based on the number of molecules of substance: P V = N k T N = Number of molecules of substance | |
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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) | |

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. ( P 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: P |

###### 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

P V = n R T

^{-12}N m

^{-2}. At such a pressure, how many molecules are there per cubic centimeter at 0°C?

P V = N k T

^{−2}m

^{3}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

P_{1} V_{1} = n_{1} R T_{1}P_{2} V_{2} = n_{2} R T_{2}

However, n_{1} = n_{2} since there is no loss of air when the box is heated.

Also, V_{1} = V_{2} 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.

P_{1} / T_{1} = n R / V

P_{2} / T_{2} = n R / V

P_{1} / T_{1} = P_{2} / T_{2}

P_{1} = 1 atm = 101325 Pa

T_{1} = 10°C = 283.15°K

P_{2} = ?

T_{2} = 100°C = 373.15°K

101325 / 283.15 = P_{2357.849197 = P2133531.428 = P2 → The pressure at 100°C is 133531.428 Pa}

The volume of the box is 5.12 × 10^{-2} m^{3}Volume = ( Length of Side )^{3} → The cube root of 5.12 × 10^{-2} m^{3} is .8 m

Area of Side = ( Length of Side )^{2} → The surface area of each side is .64 m^{2}There 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.

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

P_{1} = 2000 PSI → 2000 × 6894.75729 Pa = 13789514.6 Pa

P_{2} = 1 atm → 101325 Pa

V_{1} = 5 L = .005 m^{3}V_{2} = ?

T_{1} = T_{2} = 20°C = 293.15°K

n_{1} = n_{2} = ?

P_{1} V_{1} = n R T

P_{2} V_{2} = n R T

P_{1} V_{1} = P_{2} V_{2}13789514.6 · .005 = 101325 · V_{2}0.68045964 m^{3} = V_{2}

The tank contains a total of .68045964 m^{3} of air at atmospheric pressure.

The tank loses .0025 m^{3} of air at atmospheric pressure every minute.

Thus, it will last .68045964 / .0025 → 272.183856 minutes → The tank will last 4½ hours.