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The Particulate Nature of Matter - Gas Laws

Grade 11IBPhysics

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

🔑Concepts

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An Ideal Gas is a theoretical model that assumes: 1. Molecules are identical point masses with negligible volume. 2. There are no intermolecular forces except during collisions. 3. Collisions are perfectly elastic. 4. Molecules move in random motion.

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The state of a gas is defined by its pressure (PP), volume (VV), and absolute temperature (TT in Kelvin). The Kelvin scale is defined as T(K)=t(∘C)+273T(K) = t(^{\circ}C) + 273.

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Boyle's Law states that for a fixed mass of gas at constant temperature, P∝1VP \propto \frac{1}{V}.

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Charles's Law states that for a fixed mass of gas at constant pressure, V∝TV \propto T.

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Gay-Lussac's (Pressure) Law states that for a fixed mass of gas at constant volume, P∝TP \propto T.

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The mole (nn) represents the amount of substance. One mole contains NA=6.02×1023N_A = 6.02 \times 10^{23} particles (Avogadro's constant). n=NNA=mMn = \frac{N}{N_A} = \frac{m}{M}, where MM is the molar mass.

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The average random kinetic energy of the molecules in an ideal gas is directly proportional to the absolute temperature: Ek∝TE_k \propto T.

📐Formulae

PV=nRTPV = nRT

PV=NkBTPV = Nk_B T

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

P=13ρc2ˉP = \frac{1}{3} \rho \bar{c^2}

Ekˉ=32kBT=32RNAT\bar{E_k} = \frac{3}{2} k_B T = \frac{3}{2} \frac{R}{N_A} T

vrms=3kBTmv_{rms} = \sqrt{\frac{3k_B T}{m}}

💡Examples

Problem 1:

A cylinder contains 0.20 mol0.20\text{ mol} of an ideal gas at a temperature of 27∘C27^{\circ}C and a pressure of 1.5×105 Pa1.5 \times 10^5\text{ Pa}. Calculate the volume of the gas.

Solution:

V=nRTPV = \frac{nRT}{P} V=0.20×8.31×(27+273)1.5×105V = \frac{0.20 \times 8.31 \times (27 + 273)}{1.5 \times 10^5} V=0.20×8.31×3001.5×105V = \frac{0.20 \times 8.31 \times 300}{1.5 \times 10^5} V≈3.32×10−3 m3V \approx 3.32 \times 10^{-3}\text{ m}^3

Explanation:

We use the ideal gas law PV=nRTPV = nRT. Temperature must be converted from Celsius to Kelvin by adding 273273. RR is the universal gas constant 8.31 J K−1 mol−18.31\text{ J K}^{-1}\text{ mol}^{-1}.

Problem 2:

Calculate the average kinetic energy of a gas molecule at 100∘C100^{\circ}C.

Solution:

Ekˉ=32kBT\bar{E_k} = \frac{3}{2} k_B T Ekˉ=32×(1.38×10−23)×(100+273)\bar{E_k} = \frac{3}{2} \times (1.38 \times 10^{-23}) \times (100 + 273) Ekˉ=1.5×1.38×10−23×373\bar{E_k} = 1.5 \times 1.38 \times 10^{-23} \times 373 Ekˉ≈7.72×10−21 J\bar{E_k} \approx 7.72 \times 10^{-21}\text{ J}

Explanation:

The average kinetic energy depends only on the absolute temperature. We use the Boltzmann constant kB=1.38×10−23 J K−1k_B = 1.38 \times 10^{-23}\text{ J K}^{-1} and ensure the temperature is in Kelvin.

Problem 3:

A gas at pressure P1P_1 and volume V1V_1 is compressed to half its volume while the absolute temperature is doubled. Find the new pressure P2P_2 in terms of P1P_1.

Solution:

P1V1T1=P2V2T2\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} V2=0.5V1,T2=2T1V_2 = 0.5 V_1, \quad T_2 = 2T_1 P1V1T1=P2(0.5V1)2T1\frac{P_1 V_1}{T_1} = \frac{P_2 (0.5 V_1)}{2 T_1} P1=0.5P22P_1 = \frac{0.5 P_2}{2} P1=0.25P2  ⟹  P2=4P1P_1 = 0.25 P_2 \implies P_2 = 4 P_1

Explanation:

Using the combined gas law, we substitute the ratios for volume and temperature to find that the pressure increases fourfold.

Gas Laws Grade 11 Notes & Examples | IB Physics