Ideal Gas Law

The Ideal Gas Law is an equation of state for an ideal gas, combining Boyle’s, Charles’s, and Avogadro’s Laws into a single relationship:

\( PV = nRT \)

\( P \): Pressure (in pascals, Pa)
\( V \): Volume (in cubic meters, m³)
\( n \): Number of moles of gas
\( R \): Universal gas constant \( = 8.314 \, \text{J mol}^{-1} \text{K}^{-1} \)
\( T \): Temperature (in kelvins, K)

Conditions for Ideal Gas Behavior

Low pressure
High temperature
Negligible volume of gas particles
No intermolecular forces between gas molecules

Conversions to Use in the Ideal Gas Law

Temperature must be in kelvins: \( T = \text{°C} + 273.15 \)
Pressure often needs to be converted to pascals: \( 1 \, \text{atm} = 1.013 \times 10^5 \, \text{Pa} \)
Volume should be in cubic meters: \( 1 \, \text{cm}^3 = 1 \times 10^{-6} \, \text{m}^3 \), \( 1 \, \text{L} = 1 \times 10^{-3} \, \text{m}^3 \)

Alternate Forms and Applications

\( \frac{PV}{T} = \text{constant} \) for a fixed amount of gas
Combining states: \( \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \)
Useful in calculating molar mass from density: \( M = \frac{mRT}{PV} \)

Kinetic Interpretation

Relates macroscopic behavior to microscopic motion of molecules
Average kinetic energy of a molecule: \( \frac{3}{2}kT \)
Total internal energy of an ideal monatomic gas: \( U = \frac{3}{2}nRT \)
Explains pressure as the result of molecular collisions with container walls

A 10 L container holds oxygen gas at a pressure of \( 2 \times 10^5 \, \text{Pa} \) and a temperature of \( 300 \, \text{K} \). Find the number of moles of gas in the container.

Solution:

Convert volume to m³: \( V = 10 \, \text{L} = 0.01 \, \text{m}^3 \)
Use \( PV = nRT \)
\[ n = \frac{PV}{RT} = \frac{(2 \times 10^5)(0.01)}{8.314 \times 300} \approx 0.802 \, \text{mol} \]

0.5 moles of an ideal gas is kept in a 5 L container at 400 K. Calculate the pressure of the gas.

Solution:

Convert volume to m³: \( V = 5 \, \text{L} = 0.005 \, \text{m}^3 \)
Use \( PV = nRT \)
\[ P = \frac{nRT}{V} = \frac{(0.5)(8.314)(400)}{0.005} = 332560 \, \text{Pa} = 3.33 \times 10^5 \, \text{Pa} \]

An ideal gas is compressed from 1.0 L to 0.5 L at constant temperature of \( 273 \, \text{K} \). What will happen to the pressure?

Solution:

Since temperature and number of moles are constant, we can use Boyle's Law:
\[ P_1 V_1 = P_2 V_2 \Rightarrow P_2 = \frac{P_1 V_1}{V_2} = \frac{P_1 \times 1.0}{0.5} = 2P_1 \] So, pressure doubles.

Written by Thenura Dilruk