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Kinetic Theory - Equation of State of a Perfect Gas

Grade 11CBSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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An ideal or perfect gas is a theoretical gas that strictly follows the gas laws (Boyle′sBoyle's, Charles′sCharles's, and Avogadro′sAvogadro's) at all pressures and temperatures.

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The Equation of State for an ideal gas relates the macroscopic variables Pressure (PP), Volume (VV), and Absolute Temperature (TT): PV=nRTPV = nRT.

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Universal Gas Constant (RR): It is the same for all gases, with a value of approximately 8.314 J mol−1 K−18.314 \text{ J mol}^{-1} \text{ K}^{-1}.

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Boltzmann Constant (kBk_B): It relates the average relative kinetic energy of particles in a gas with the thermodynamic temperature. It is defined as kB=RNAk_B = \frac{R}{N_A}, where NAN_A is Avogadro's number (6.022×1023 mol−16.022 \times 10^{23} \text{ mol}^{-1}).

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Boyle's Law: For a fixed mass of gas at constant temperature, PV=constantPV = \text{constant} or P1V1=P2V2P_1V_1 = P_2V_2.

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Charles's Law: For a fixed mass of gas at constant pressure, VT=constant\frac{V}{T} = \text{constant} or V1T1=V2T2\frac{V_1}{T_1} = \frac{V_2}{T_2}.

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Dalton's Law of Partial Pressures: The total pressure exerted by a mixture of non-reactive ideal gases is equal to the sum of the partial pressures of individual gases: P=P1+P2+...+PnP = P_1 + P_2 + ... + P_n.

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Real gases approach ideal gas behavior at conditions of low pressure and high temperature.

📐Formulae

PV=nRTPV = nRT

PV=NkBTPV = N k_B T

n=mM=NNAn = \frac{m}{M} = \frac{N}{N_A}

kB=RNA≈1.38×10−23 J K−1k_B = \frac{R}{N_A} \approx 1.38 \times 10^{-23} \text{ J K}^{-1}

ρ=PMRT\rho = \frac{P M}{R T}

P1V1T1=P2V2T2\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}

💡Examples

Problem 1:

A vessel contains 8.0 g8.0 \text{ g} of Oxygen (O2O_2) at a temperature of 27∘C27^\circ\text{C} and pressure of 2 atm2 \text{ atm}. Find the volume of the vessel. (Given R=0.0821 L atm mol−1 K−1R = 0.0821 \text{ L atm mol}^{-1} \text{ K}^{-1}, Molar mass of O2=32 g mol−1O_2 = 32 \text{ g mol}^{-1})

Solution:

  1. Convert temperature to Kelvin: T=27+273=300 KT = 27 + 273 = 300 \text{ K}.
  2. Calculate number of moles: n=massmolar mass=8.032=0.25 moln = \frac{\text{mass}}{\text{molar mass}} = \frac{8.0}{32} = 0.25 \text{ mol}.
  3. Use ideal gas equation: V=nRTP=0.25×0.0821×3002V = \frac{nRT}{P} = \frac{0.25 \times 0.0821 \times 300}{2}.
  4. V=6.15752=3.07875 LV = \frac{6.1575}{2} = 3.07875 \text{ L}.

Explanation:

The volume is calculated by rearranging the equation of state PV=nRTPV = nRT to solve for VV. Note that temperature must always be in Kelvin for gas law calculations.

Problem 2:

If the volume of an ideal gas is reduced to half and its absolute temperature is doubled, what happens to the pressure?

Solution:

Let initial states be P1,V1,T1P_1, V_1, T_1 and final states be P2,V2,T2P_2, V_2, T_2. Given: V2=V12V_2 = \frac{V_1}{2} and T2=2T1T_2 = 2T_1. Using the combined gas law: P1V1T1=P2V2T2\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} P2=P1V1T2T1V2=P1×V1×2T1T1×(V1/2)P_2 = \frac{P_1 V_1 T_2}{T_1 V_2} = \frac{P_1 \times V_1 \times 2T_1}{T_1 \times (V_1/2)} P2=P1×2×2=4P1P_2 = P_1 \times 2 \times 2 = 4P_1.

Explanation:

The pressure becomes four times the initial pressure because pressure is inversely proportional to volume and directly proportional to absolute temperature.