Eduqas A-Level Chemistry A410QS
Formulae and Equations
Basic Ideas About Atoms
Chemical Calculations
Bonding
Solid Structures
The Periodic Table
Simple Equilibria and Acid-Base Reactions
Thermochemistry
Rates of Reaction
The Wider Impact of Chemistry
Organic Compounds
Hydrocarbons
Halogenoalkanes
Alcohols and Carboxylic Acids
Instrumental Analysis
Redox and Standard Electrode Potential
Redox Reactions
Chemistry of the P-Block
Chemistry of the D-Block Transition Metals
Chemical Kinetics
Enthalpy Changes for Solids and Solutions
Entropy and Feasibility of Reactions
Equilibrium Constants
Acid-Base Equilibria
Stereoisomerism
Aromaticity
Alcohols and Phenols
Aldehydes and Ketones
Carboxylic Acids and Their Derivatives
Amines
Amino Acids, Peptides and Proteins
Organic Synthesis and Analysis
Specified Practical Work
Practical Skills and Techniques
Mathematical Skills in Chemistry
Chemical Calculations
The ideal gas equation
- 3 things to know
- Components 1 and 2
Overview
Ideal gas equation and say what each symbol represents and assumptions made about the molecules of an ideal gas.
What you need to know
Answer each one before you reveal it.
In which units must each quantity be expressed when using the ideal gas equation?
Convert a volume of 1 dm³ into cubic metres.
Questions from this subtopic
Each one with the model answer underneath.
- 1
State the ideal gas equation and say what each symbol represents.
Model answer
The equation is pV = nRT, where p is pressure, V is volume, n is the amount of gas in moles, R is the gas constant and T is the absolute temperature.
- 2
In which units must each quantity be expressed when using the ideal gas equation?
Model answer
Pressure in pascals, volume in cubic metres, amount in moles and temperature in kelvin, with R equal to 8.31 J K⁻¹ mol⁻¹.
- 3
Convert a volume of 1 dm³ into cubic metres.
Model answer
1 dm³ is 1 × 10⁻³ m³.
- 4
Convert a volume of 250 cm³ into cubic metres.
Model answer
250 cm³ is 2.50 × 10⁻⁴ m³.
- 5
Calculate the volume occupied by 0.500 mol of a gas at 298 K and 1.00 × 10⁵ Pa.
Model answer
V = nRT/p = (0.500 × 8.31 × 298) ÷ 1.00 × 10⁵ = 1.24 × 10⁻² m³, which is 12.4 dm³.
- 6
A 0.440 g sample of a gas occupies 250 cm³ at 300 K and 1.00 × 10⁵ Pa. Calculate its molar mass.
Model answer
n = pV/RT = (1.00 × 10⁵ × 2.50 × 10⁻⁴) ÷ (8.31 × 300) = 1.00 × 10⁻² mol, so the molar mass is 0.440 ÷ 1.00 × 10⁻² = 44 g mol⁻¹.
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