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States of Matter: Gases and Liquids - Gas Laws

Grade 11ICSEChemistry

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

🔑Concepts

Measurable properties of gases include Pressure (PP), Volume (VV), Temperature (TT in Kelvin), and Amount (nn in moles). Temperature must always be converted using T(K)=t(C)+273.15T(K) = t(^{\circ}C) + 273.15.

Boyle's Law: For a fixed mass of gas at constant temperature, volume is inversely proportional to pressure (V1PV \propto \frac{1}{P}), meaning PV=constantPV = \text{constant}.

Charles's Law: For a fixed mass of gas at constant pressure, volume is directly proportional to absolute temperature (VTV \propto T).

Gay-Lussac's Law: For a fixed mass of gas at constant volume, pressure is directly proportional to absolute temperature (PTP \propto T).

Avogadro's Law: Equal volumes of all gases under the same conditions of temperature and pressure contain equal number of molecules (VnV \propto n).

Ideal Gas Equation: Combining Boyle's, Charles's, and Avogadro's laws gives PV=nRTPV = nRT, where RR is the Universal Gas Constant (0.0821 L atm K1mol10.0821 \text{ L atm K}^{-1}\text{mol}^{-1} or 8.314 J K1mol18.314 \text{ J K}^{-1}\text{mol}^{-1}).

Dalton's Law of Partial Pressures: The total pressure of a mixture of non-reacting gases is the sum of the partial pressures of the individual gases: Ptotal=P1+P2+P3+P_{\text{total}} = P_1 + P_2 + P_3 + \dots.

Graham's Law of Diffusion/Effusion: The rate of diffusion (rr) of a gas is inversely proportional to the square root of its molar mass (MM) or density (dd): r1Mr \propto \frac{1}{\sqrt{M}}.

Kinetic Molecular Theory (KMT): Postulates that gas particles are in constant random motion, have negligible volume compared to the container, and undergo perfectly elastic collisions.

Real Gases and Deviation: Real gases deviate from ideal behavior at high pressure and low temperature. The van der Waals equation accounts for intermolecular forces (aa) and molecular volume (bb): (P+an2V2)(Vnb)=nRT(P + \frac{an^2}{V^2})(V - nb) = nRT.

📐Formulae

P1V1=P2V2 (Boyle’s Law)P_1V_1 = P_2V_2 \text{ (Boyle's Law)}

V1T1=V2T2 (Charles’s Law)\frac{V_1}{T_1} = \frac{V_2}{T_2} \text{ (Charles's Law)}

P1T1=P2T2 (Gay-Lussac’s Law)\frac{P_1}{T_1} = \frac{P_2}{T_2} \text{ (Gay-Lussac's Law)}

P1V1T1=P2V2T2 (Combined Gas Law)\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} \text{ (Combined Gas Law)}

PV=nRT (Ideal Gas Equation)PV = nRT \text{ (Ideal Gas Equation)}

Pi=XiPtotal (Partial Pressure where Xi is mole fraction)P_i = X_i \cdot P_{\text{total}} \text{ (Partial Pressure where } X_i \text{ is mole fraction)}

r1r2=M2M1=d2d1 (Graham’s Law)\frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}} = \sqrt{\frac{d_2}{d_1}} \text{ (Graham's Law)}

(P+an2V2)(Vnb)=nRT (van der Waals Equation)\left(P + \frac{an^2}{V^2}\right)(V - nb) = nRT \text{ (van der Waals Equation)}

💡Examples

Problem 1:

A sample of neon gas (NeNe) occupies 4.0 L4.0 \text{ L} at 0.8 atm0.8 \text{ atm} and 27C27^{\circ}C. Calculate the volume it will occupy at STPSTP (Standard Temperature and Pressure).

Solution:

  1. Identify given values: V1=4.0 LV_1 = 4.0 \text{ L}, P1=0.8 atmP_1 = 0.8 \text{ atm}, T1=27+273=300 KT_1 = 27 + 273 = 300 \text{ K}.
  2. Standard conditions (STPSTP): P2=1.0 atmP_2 = 1.0 \text{ atm}, T2=273 KT_2 = 273 \text{ K}.
  3. Use Combined Gas Law: P1V1T1=P2V2T2\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}.
  4. Solve for V2V_2: V2=P1V1T2P2T1=0.8×4.0×2731.0×300=2.912 LV_2 = \frac{P_1 V_1 T_2}{P_2 T_1} = \frac{0.8 \times 4.0 \times 273}{1.0 \times 300} = 2.912 \text{ L}.

Explanation:

The combined gas law allows us to calculate changes in volume when both pressure and temperature vary. Temperature must be in Kelvin.

Problem 2:

Determine the relative rate of diffusion of Helium (HeHe, M=4M = 4) compared to Methane (CH4CH_4, M=16M = 16).

Solution:

  1. Use Graham's Law: rHerCH4=MCH4MHe\frac{r_{He}}{r_{CH_4}} = \sqrt{\frac{M_{CH_4}}{M_{He}}}.
  2. Substitute molar masses: rHerCH4=164=4=2\frac{r_{He}}{r_{CH_4}} = \sqrt{\frac{16}{4}} = \sqrt{4} = 2.

Explanation:

Helium atoms are lighter than methane molecules; therefore, Helium diffuses 22 times faster than Methane under the same conditions.

Gas Laws - Revision Notes & Key Formulas | ICSE Class 11 Chemistry