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Ideal Gas Equation

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Estimated time: 10 minutes
  • Introduction
  • Definition: Ideal Gas Equation
  • Deriving the Ideal Gas Equation
  • Formula Combines Gas Law
  • Example
  • Key Points: Ideal Gas Equation
Maharashtra State Board: Class 11

Introduction

Gases are everywhere — in the air we breathe, the tyres of our vehicles, the pressure cookers in our kitchens, and even the hot air balloons that float across the sky. Unlike solids and liquids, gases can be compressed, expanded, and heated in ways that dramatically change their behaviour.

Three measurable properties define the state of any gas:

Property Symbol What It Measures SI Unit
Pressure P Force exerted by gas molecules per unit area on container walls Pascal (Pa)
Volume V Space occupied by the gas
Temperature T Measure of the average kinetic energy of molecules Kelvin (K)
Maharashtra State Board: Class 11

Definition: Ideal Gas Equation

“The relation between three properties of a gas, i.e., pressure, volume and temperature, is called the ideal gas equation.”

OR

The relation between the three properties of a gas - pressure (P), volume (V), and temperature (T) - expressed as PV = nRT, is called the ideal gas equation.

Maharashtra State Board: Class 11

Deriving the Ideal Gas Equation

Step-by-Step Derivation

  1. From Boyle's Law (at constant T): V ∝ \[\frac {1}{P}\]  ...(i)
  2. From Charles' Law (at constant P): V ∝ T  ...(ii)
  3. Combining (i) and (ii): V ∝ \[\frac {T}{P}\]
  4. Introducing the proportionality constant R (for 1 mole): V = \[\frac {RT}{P}\]
  5. Rearranging: PV = RT (for 1 mole)  ...(4)
  6. For n moles of gas: PV = nRT
    where:
    p = pressure (Pa)
    V = volume (m³)
    n = number of moles of gas
    R = universal gas constant = 8.31JK-1mol-1
    T = absolute temperature (K)

This is the Ideal Gas Equation — also known as the equation of state of an ideal gas.

Maharashtra State Board: Class 11

Formula: Combined Gas Law

\[\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}\]

Maharashtra State Board: Class 11

Example

Problem: The pressure reading in a thermometer at the steam point is 1.367 × 103 Pa. What is pressure reading at the triple point, knowing the linear relationship between temperature and pressure?

Solution:
Step 1: Convert steam point to Kelvin: T = 273.15 + 100 = 373.15 K

Step 2: Using Gay-Lussac's Law (P ∝ T):
\[\frac {P_{triple}}{T_{triple}}\] = \[\frac {P}{T}\]

Step 3: Solve:
Ptriple = \[\frac {273.16}{373.15}\] × 1.367 × 103

Answer: Ptriple = 1.000 × 103 Pa
Maharashtra State Board: Class 11

Key Points: Ideal Gas Equation

  • An ideal gas has point-mass molecules, no intermolecular forces, and perfectly elastic collisions.
  • The Ideal Gas Equation, PV = nRT, combines all three laws into a single universal relationship.
  • The Universal Gas Constant R = 8.314 J mol⁻¹ K⁻¹ is the same for all ideal gases.
  • Real gases approximate ideal behaviour at low pressure and high temperature.
  • Always use absolute temperature (Kelvin) in gas law calculations. T(K) = T(°C) + 273.15

Test Yourself

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