Ideal Gas Equation and Gas Laws

Master the ideal gas equation and fundamental gas laws for JEE Main and Advanced

7 min read

Ideal Gas Equation and Gas Laws

Real-Life Hook

Ever wondered why a pressure cooker cooks food faster? When you heat the cooker, the temperature inside rises, which increases the pressure (Gay-Lussac’s Law). This higher pressure allows water to boil at temperatures above 100°C, cooking food faster! Similarly, when you pump air into a bicycle tire, you’re compressing gas molecules closer together, increasing pressure - a practical demonstration of Boyle’s Law.

Core Concepts

The Ideal Gas Equation

The ideal gas equation combines all gas laws into one universal relationship:

Ideal Gas Equation

$$PV = nRT$$

Where:

  • P = Pressure (Pa or N/m²)
  • V = Volume (m³)
  • n = Number of moles
  • R = Universal gas constant = 8.314 J/(mol·K)
  • T = Absolute temperature (K)

Alternative forms:

  • $PV = NkT$ (where N = number of molecules, k = Boltzmann constant = 1.38 × 10⁻²³ J/K)
  • $P = \frac{\rho RT}{M}$ (where ρ = density, M = molar mass)

Interactive Demo: Gas Law Relationships

Explore how pressure, volume, and temperature are interconnected in ideal gases.

Individual Gas Laws

1. Boyle’s Law (Temperature Constant)

Boyle’s Law

$$P \propto \frac{1}{V}$$

or

$$P_1V_1 = P_2V_2$$

At constant temperature and amount of gas

Physical Interpretation: When volume decreases, molecules have less space to move, leading to more frequent collisions with walls, hence higher pressure.

2. Charles’s Law (Pressure Constant)

Charles’s Law

$$V \propto T$$

or

$$\frac{V_1}{T_1} = \frac{V_2}{T_2}$$

At constant pressure and amount of gas

Physical Interpretation: As temperature increases, molecules move faster and need more space to maintain the same pressure.

3. Gay-Lussac’s Law (Volume Constant)

Gay-Lussac’s Law

$$P \propto T$$

or

$$\frac{P_1}{T_1} = \frac{P_2}{T_2}$$

At constant volume and amount of gas

4. Avogadro’s Law

Avogadro’s Law

$$V \propto n$$

or

$$\frac{V_1}{n_1} = \frac{V_2}{n_2}$$

At constant temperature and pressure

Key Fact: One mole of any ideal gas occupies 22.4 L at STP (Standard Temperature and Pressure: 0°C, 1 atm)

Combined Gas Law

Combined Gas Law

$$\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$$

For fixed amount of gas

Memory Tricks

“BoylesPT” Mnemonic

  • Boyle’s: Pressure × Volume = constant (PV = k)
  • Charle’s: Volume / Temperature = constant (V/T = k)
  • Gay-Lussac’s: Pressure / Temperature = constant (P/T = k)

“Very Tall People Never Relax” for PV = nRT

  • Volume
  • Temperature (in Kelvin!)
  • Pressure
  • Number of moles
  • R = 8.314

Temperature Conversion Rule

ALWAYS use Kelvin in gas law calculations!

  • K = °C + 273.15
  • Never use Celsius or Fahrenheit directly

Common Unit Conversion Mistakes

Pressure Conversions

  1. 1 atm = 101,325 Pa = 1.013 × 10⁵ Pa ≈ 10⁵ Pa (for quick calculations)
  2. 1 bar = 10⁵ Pa = 0.987 atm ≈ 1 atm
  3. 1 mmHg = 133.3 Pa
  4. 760 mmHg = 1 atm

Common Mistake: Using pressure in atm with R = 8.314 J/(mol·K)

  • Correct: Convert to Pa (N/m²) or use R = 0.0821 L·atm/(mol·K)

Volume Conversions

  1. 1 m³ = 1000 L
  2. 1 L = 10⁻³ m³ = 1000 cm³

Common Mistake: Mixing mL and m³

  • Always convert to consistent units before calculation

Temperature Conversions

CRITICAL: Never forget to convert °C to K!

  • Room temperature: 25°C = 298 K (NOT 25 K)
  • STP: 0°C = 273 K (NOT 0 K)

Important Formulas

Density and Molar Mass

Density Formula

$$\rho = \frac{PM}{RT}$$

Where M = molar mass (kg/mol or g/mol)

Mixing of Gases

For ideal gases mixed in a container:

  • Total pressure: $P_{total} = P_1 + P_2 + P_3 + ...$ (Dalton’s Law)
  • Partial pressure: $P_i = \frac{n_i RT}{V}$

Work Done in Gas Processes

Isothermal Process (T = constant)

$$W = nRT \ln\left(\frac{V_2}{V_1}\right) = nRT \ln\left(\frac{P_1}{P_2}\right)$$

3-Level Practice Problems

Level 1: JEE Main Basics

Problem 1: A gas occupies 2.0 L at 27°C and 1 atm pressure. What will be its volume at 127°C and 2 atm pressure?

Solution

Using combined gas law:

$$\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$$

Given:

  • $P_1 = 1$ atm, $V_1 = 2.0$ L, $T_1 = 27 + 273 = 300$ K
  • $P_2 = 2$ atm, $T_2 = 127 + 273 = 400$ K
$$V_2 = \frac{P_1V_1T_2}{P_2T_1} = \frac{1 \times 2.0 \times 400}{2 \times 300} = \frac{800}{600} = 1.33 \text{ L}$$

Answer: 1.33 L

Problem 2: Calculate the number of molecules in 11.2 L of a gas at STP.

Solution

At STP, 22.4 L contains 1 mole = $6.022 \times 10^{23}$ molecules

Number of moles in 11.2 L:

$$n = \frac{11.2}{22.4} = 0.5 \text{ mol}$$

Number of molecules:

$$N = n \times N_A = 0.5 \times 6.022 \times 10^{23} = 3.011 \times 10^{23}$$

Answer: $3.011 \times 10^{23}$ molecules

Level 2: JEE Main Advanced

Problem 3: A vessel contains 1 mole of O₂ gas at 27°C. Calculate the pressure if the volume is 8.31 L. If the temperature is doubled and volume is halved, what will be the new pressure? (R = 8.31 J/(mol·K) = 0.0831 L·bar/(mol·K))

Solution

Initial conditions:

$$P_1V_1 = nRT_1$$ $$P_1 = \frac{nRT_1}{V_1} = \frac{1 \times 0.0831 \times 300}{8.31} = 3 \text{ bar}$$

Final conditions:

  • $T_2 = 2T_1 = 600$ K
  • $V_2 = V_1/2 = 4.155$ L
$$P_2 = \frac{nRT_2}{V_2} = \frac{1 \times 0.0831 \times 600}{4.155} = 12 \text{ bar}$$

Alternatively using ratios:

$$\frac{P_2}{P_1} = \frac{T_2}{T_1} \times \frac{V_1}{V_2} = 2 \times 2 = 4$$

So $P_2 = 4P_1 = 12$ bar

Answer: Initial pressure = 3 bar, Final pressure = 12 bar

Problem 4: Find the density of CO₂ at 27°C and 2 atm pressure. (Molar mass of CO₂ = 44 g/mol, R = 0.0821 L·atm/(mol·K))

Solution

Using: $\rho = \frac{PM}{RT}$

$$\rho = \frac{2 \times 44}{0.0821 \times 300} = \frac{88}{24.63} = 3.57 \text{ g/L}$$

Answer: 3.57 g/L

Level 3: JEE Advanced

Problem 5: A container is divided into two equal parts by a partition. One part contains 2 moles of He at 300 K and the other contains 3 moles of Ar at 600 K. If the partition is removed and the gases mix (assuming ideal behavior), find the final temperature and pressure. The total volume is 24.9 L. (R = 0.0821 L·atm/(mol·K))

Solution

Each compartment has volume = 24.9/2 = 12.45 L

Initial pressures:

$$P_{He} = \frac{2 \times 0.0821 \times 300}{12.45} = 3.97 \text{ atm}$$ $$P_{Ar} = \frac{3 \times 0.0821 \times 600}{12.45} = 11.91 \text{ atm}$$

After mixing, using energy conservation (for ideal gases at constant volume):

$$n_1C_VT_1 + n_2C_VT_2 = (n_1 + n_2)C_VT_f$$

For monoatomic gases (He and Ar), $C_V$ cancels:

$$2 \times 300 + 3 \times 600 = 5 \times T_f$$ $$600 + 1800 = 5T_f$$ $$T_f = 480 \text{ K}$$

Final pressure:

$$P_f = \frac{nRT_f}{V} = \frac{5 \times 0.0821 \times 480}{24.9} = 7.93 \text{ atm}$$

Answer: Final temperature = 480 K, Final pressure = 7.93 atm

Problem 6: A gas follows the equation of state: $P(V - b) = RT$, where b is a constant. Show that for an isothermal process, the work done is $W = RT\ln\left(\frac{V_2-b}{V_1-b}\right)$.

Solution

For isothermal process (T = constant):

Work done: $W = \int_{V_1}^{V_2} P \, dV$

From equation of state: $P = \frac{RT}{V-b}$

$$W = \int_{V_1}^{V_2} \frac{RT}{V-b} \, dV$$ $$W = RT \int_{V_1}^{V_2} \frac{dV}{V-b}$$ $$W = RT [\ln(V-b)]_{V_1}^{V_2}$$ $$W = RT \ln\left(\frac{V_2-b}{V_1-b}\right)$$

Proved

Applications

  • Real gases and Van der Waals equation
  • Kinetic theory derivations
  • Maxwell-Boltzmann distribution

Key Takeaways

  1. Always use Kelvin for temperature in all gas law calculations
  2. Check units - especially pressure (Pa vs atm vs bar)
  3. STP conditions: 0°C (273 K) and 1 atm → 22.4 L/mol
  4. R value depends on units: 8.314 J/(mol·K) or 0.0821 L·atm/(mol·K)
  5. For gas mixtures, use Dalton’s Law of partial pressures
  6. Ideal gas equation is an approximation - real gases deviate at high P and low T

Exam Tips

  • JEE Main: Focus on direct applications of gas laws and unit conversions
  • JEE Advanced: Expect problems involving gas mixtures, energy conservation, and combined concepts
  • Common trap: Using °C instead of K - double-check every temperature!
  • Quick check: In expansion (V↑), pressure decreases (P↓) if temperature is constant
  • Dimensional analysis: Use it to verify your final answer’s units

Last updated: February 15, 2025