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States of Matter: Gases and Liquids - Ideal Gas Equation

Grade 11ICSEChemistry

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

🔑Concepts

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The Ideal Gas Equation is a state equation that describes the relationship between the four variables of a gas sample: Pressure (PP), Volume (VV), Absolute Temperature (TT), and the number of moles (nn).

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It is derived by combining Boyle's Law (V∝1PV \propto \frac{1}{P}), Charles's Law (V∝TV \propto T), and Avogadro's Law (V∝nV \propto n).

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An 'Ideal Gas' is a theoretical gas that strictly follows the gas laws at all temperatures and pressures. Real gases behave ideally only at high temperatures and low pressures.

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The Universal Gas Constant (RR) has different values depending on the units used: R=0.0821 L atm K−1 mol−1R = 0.0821 \text{ L atm K}^{-1} \text{ mol}^{-1}, R=8.314 J K−1 mol−1R = 8.314 \text{ J K}^{-1} \text{ mol}^{-1}, or R=0.08314 L bar K−1 mol−1R = 0.08314 \text{ L bar K}^{-1} \text{ mol}^{-1}.

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Standard Temperature and Pressure (STP) is defined as 273.15 K273.15 \text{ K} (0∘C0^\circ\text{C}) and 1 bar1 \text{ bar} pressure. At STP, the molar volume of an ideal gas is 22.7 L mol−122.7 \text{ L mol}^{-1} (Older STP definition used 1 atm1 \text{ atm}, resulting in 22.4 L22.4 \text{ L}).

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The Ideal Gas Equation can be rearranged to calculate the density (dd) or molar mass (MM) of a gas using the formula PM=dRTPM = dRT.

📐Formulae

PV=nRTPV = nRT

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

n=mMn = \frac{m}{M}

M=mRTPVM = \frac{mRT}{PV}

d=PMRTd = \frac{PM}{RT}

💡Examples

Problem 1:

Calculate the volume occupied by 8.8 g8.8 \text{ g} of CO2CO_2 at 31.1∘C31.1^\circ\text{C} and 1 bar1 \text{ bar} pressure. (Given: R=0.083 L bar K−1 mol−1R = 0.083 \text{ L bar K}^{-1} \text{ mol}^{-1}, Atomic masses: C=12,O=16C = 12, O = 16)

Solution:

  1. Calculate moles of CO2CO_2: n=massmolar mass=8.8 g44 g/mol=0.2 moln = \frac{\text{mass}}{\text{molar mass}} = \frac{8.8 \text{ g}}{44 \text{ g/mol}} = 0.2 \text{ mol}.
  2. Convert temperature to Kelvin: T=31.1+273.15=304.25 KT = 31.1 + 273.15 = 304.25 \text{ K}.
  3. Use Ideal Gas Equation: V=nRTPV = \frac{nRT}{P}.
  4. Substitute values: V=0.2×0.083×304.251V = \frac{0.2 \times 0.083 \times 304.25}{1}.
  5. V≈5.05 LV \approx 5.05 \text{ L}.

Explanation:

The number of moles was first determined using the given mass and the molar mass of CO2CO_2. Then, the temperature was converted to the absolute scale (Kelvin) before applying the Ideal Gas Equation PV=nRTPV = nRT.

Problem 2:

A gas at 25∘C25^\circ\text{C} and 760 mm Hg760 \text{ mm Hg} pressure occupies a volume of 600 mL600 \text{ mL}. What will be its pressure at a height where temperature is 10∘C10^\circ\text{C} and volume is 640 mL640 \text{ mL}?

Solution:

  1. Identify given values: P1=760 mm HgP_1 = 760 \text{ mm Hg}, V1=600 mLV_1 = 600 \text{ mL}, T1=25+273=298 KT_1 = 25 + 273 = 298 \text{ K}.
  2. New conditions: V2=640 mLV_2 = 640 \text{ mL}, T2=10+273=283 KT_2 = 10 + 273 = 283 \text{ K}.
  3. Apply Combined Gas Law: P1V1T1=P2V2T2\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}.
  4. Rearrange for P2P_2: P2=P1V1T2T1V2=760×600×283298×640P_2 = \frac{P_1 V_1 T_2}{T_1 V_2} = \frac{760 \times 600 \times 283}{298 \times 640}.
  5. P2≈676.6 mm HgP_2 \approx 676.6 \text{ mm Hg}.

Explanation:

Since the amount of gas (nn) remains constant, the combined gas law is used. Temperature must always be in Kelvin for gas law calculations.

Ideal Gas Equation Class 11 Notes & Examples