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Behavior of Perfect Gas and Kinetic Theory - Equation of State of a Perfect Gas

Grade 11ICSEPhysics

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 composed of many randomly moving point particles that are not subject to interparticle interactions.

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The state of a gas is defined by four variables: Pressure (PP), Volume (VV), Temperature (TT in Kelvin), and Amount of substance (nn moles).

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Boyle's Law: For a fixed mass of gas at constant temperature, the volume is inversely proportional to its pressure: V∝1PV \propto \frac{1}{P} or PV=constantPV = \text{constant}.

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Charles's Law: For a fixed mass of gas at constant pressure, the volume is directly proportional to its absolute temperature: V∝TV \propto T or VT=constant\frac{V}{T} = \text{constant}.

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Gay-Lussac's Law: For a fixed mass of gas at constant volume, the pressure is directly proportional to its absolute temperature: P∝TP \propto T or PT=constant\frac{P}{T} = \text{constant}.

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Avogadro's Law: Under the same conditions of temperature and pressure, equal volumes of all gases contain an equal number of molecules: V∝nV \propto n.

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The Equation of State of a Perfect Gas (Ideal Gas Equation) combines these laws: PV=nRTPV = nRT.

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The Universal Gas Constant (RR) has a value of approximately 8.314 J mol−1K−18.314 \text{ J mol}^{-1} \text{K}^{-1}.

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Boltzmann Constant (kBk_B): It is the gas constant per molecule, defined as kB=RNAk_B = \frac{R}{N_A}, where NAN_A is Avogadro's number (6.023×1023 mol−16.023 \times 10^{23} \text{ mol}^{-1}).

📐Formulae

PV=nRTPV = nRT

PV=NkBTPV = N k_B T

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

R=NAkBR = N_A k_B

P=ρRTMP = \frac{\rho RT}{M}

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

💡Examples

Problem 1:

A sample of an ideal gas occupies a volume of 2.0 L2.0 \text{ L} at a pressure of 1.5 atm1.5 \text{ atm} and a temperature of 300 K300 \text{ K}. If the volume is compressed to 1.0 L1.0 \text{ L} and the temperature is raised to 400 K400 \text{ K}, what will be the final pressure of the gas?

Solution:

Given: P1=1.5 atmP_1 = 1.5 \text{ atm} V1=2.0 LV_1 = 2.0 \text{ L} T1=300 KT_1 = 300 \text{ K} V2=1.0 LV_2 = 1.0 \text{ L} T2=400 KT_2 = 400 \text{ K} Using the combined gas law: P1V1T1=P2V2T2\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} P2=P1V1T2V2T1P_2 = \frac{P_1 V_1 T_2}{V_2 T_1} P2=1.5×2.0×4001.0×300P_2 = \frac{1.5 \times 2.0 \times 400}{1.0 \times 300} P2=1200300=4.0 atmP_2 = \frac{1200}{300} = 4.0 \text{ atm}

Explanation:

The combined gas law is derived from the equation of state for a constant number of moles nn. By rearranging the formula to solve for the unknown final pressure P2P_2, we substitute the initial and final states of the system.

Problem 2:

Calculate the number of moles of an ideal gas present in a 0.5 m30.5 \text{ m}^3 container at a pressure of 2.0×105 Pa2.0 \times 10^5 \text{ Pa} and a temperature of 27∘C27^\circ\text{C}. (Use R=8.314 J mol−1K−1R = 8.314 \text{ J mol}^{-1} \text{K}^{-1})

Solution:

Convert temperature to Kelvin: T=27+273.15=300.15 KT = 27 + 273.15 = 300.15 \text{ K}. Using PV=nRTPV = nRT: n=PVRTn = \frac{PV}{RT} n=(2.0×105 Pa)×(0.5 m3)(8.314 J mol−1K−1)×(300.15 K)n = \frac{(2.0 \times 10^5 \text{ Pa}) \times (0.5 \text{ m}^3)}{(8.314 \text{ J mol}^{-1} \text{K}^{-1}) \times (300.15 \text{ K})} n=1000002495.45≈40.07 molesn = \frac{100000}{2495.45} \approx 40.07 \text{ moles}

Explanation:

To use the ideal gas equation, all units must be in SI. Pressure is in Pascals, volume in cubic meters, and temperature must be converted from Celsius to Kelvin by adding 273.15273.15.

Equation of State of a Perfect Gas - Revision Notes & Key Formulas | ICSE Class 11 Physics