Ideal Gas Equation Explained: PV = nRT, Units, Forms, and JEE Tips [2025 Guide]

 

📌 Ideal Gas Equation: A Complete Guide for JEE & NEET Aspirants

The Ideal Gas Equation, one of the most essential and frequently used equations in physics and chemistry, connects pressure, volume, temperature, and moles of gas into one compact relationship:

👉 The Equation:

PV = nRT

Where:

• P = Pressure

• V = Volume

• n = Number of moles

• R = Universal gas constant

• T = Temperature (in Kelvin)

Let’s dive deep into this equation, understand each part, know when to use which form, and crack tricky numerical problems easily in JEE, NEET, NDA, or Class 11-12 Boards.

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🔍 Understanding Each Term in PV = nRT

✅ P — Pressure

• Unit: atm or Pascal (Pa)

• 1 atm = 1.013 × 10⁵ Pa

• Represents the force gas molecules exert on the container walls.

✅ V — Volume

• Unit: Litres (L) or cubic meters (m³)

• 1 L = 10⁻³ m³

• The space occupied by gas.

✅ n — Number of Moles

• n = mass / molar mass

• Represents how many moles (or molecules) of gas are present.

✅ R — Gas Constant

• Value depends on units used:

  • R = 0.0821 L·atm/mol·K

  • R = 8.314 J/mol·K (when P is in Pa, V in m³)

✅ T — Temperature

• Always in Kelvin (K)

• K = °C + 273.15

• Higher the temperature, faster the gas particles.

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🔁 Rearranged Forms of PV = nRT

In numerical problems or derivations, you’ll often need to rearrange the formula:

• To find number of moles (n):

  $$n=\frac{PV}{R\mathrm{T}}$$

• To find pressure (P):

  $$P=\frac{nRT}{\mathrm{V}}$$

• To find volume (V):

  $$V=\frac{nRT}{\mathrm{P}}$$

• To find temperature (T):

  $$T=\frac{PV}{n\mathrm{R}}$$

These forms are crucial when solving problems from JEE Main, NEET, or even simple board questions.

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🔢 How to Choose the Right Value of R?

Situation	Use This Value
P in atm, V in litres	R = 0.0821 L·atm/mol·K
P in pascal (Pa), V in m³	R = 8.314 J/mol·K

Always convert all units to match the version of R you're using.

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📌 Example Problem (JEE Level):

Q. A gas occupies 2 L at 1 atm and 300 K. Find the number of moles.
(Given: R = 0.0821 L·atm/mol·K)

Solution:

$$n = \frac{PV}{RT} = \frac{1 \times 2}{0.0821 \times 300} ≈ 0.081 mol$$

🧠 JEE/NEET Tips to Ace Gas Law Problems

1. ✅ Always convert Celsius to Kelvin.

2. ✅ Use correct value of R based on units of P & V.

3. ✅ Combine with other gas laws like Boyle’s, Charles’, or Dalton’s when needed.

4. ✅ For density-based problems, use:

   $$M = \frac{dRT}{P}$$

   where M is molar mass, d is density.

5. ✅ If mass is given:
   Use n = mass / molar mass first, then plug into PV = nRT.

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🧮 Derivation (In Brief): From Kinetic Theory

Using kinetic theory, the pressure exerted by an ideal gas is:

$$P=\frac{1}{3}\frac{Nm\overline{v^2}}{\mathrm{V}}$$

This eventually simplifies to the ideal gas law when combined with molecular definitions and Avogadro’s hypothesis.

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📘 When Not to Use PV = nRT?

• ❌ Real Gases at high pressure or low temperature — instead, use van der Waals equation.

• ❌ Non-ideal conditions (strong intermolecular forces, high densities).

• ❌ Mixture of gases — use partial pressures or Dalton’s Law instead.

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⚙️ Applications of Ideal Gas Equation

1. 🚆 Thermodynamics Problems (heat, work done by gas)

2. 🔥 Combustion reactions involving gases

3. 🎈 Balloon problems (expansion, compression)

4. 💨 Rate of diffusion when combined with Graham’s law

5. 📐 Measuring molar mass using density method

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📚 Practice Questions

1. A 5L container holds 2 moles of oxygen at 300 K. Find pressure.

2. Calculate the volume occupied by 3.5 mol of nitrogen at 2 atm and 273 K.

3. What is the temperature of 1 mole of gas exerting 101.3 kPa pressure in 1 m³?

(Don’t forget: Practice = Perfection!)

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🧠 Final Revision Tips:

• PV = nRT works best under ideal gas conditions.

• Understand unit systems before solving.

• Focus on interlinked concepts like moles, temperature, and pressure.

• Don’t blindly memorize — visualize gas behavior mentally.

• Make a unit conversion table for exam time!

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