Ideal Gas Law: Units, Conversions, and Practice Problems for Physics

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21 Terms

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Ideal gas equation

PV = nRT

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Pressure (P)

Pressure in pascals (Pa)

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Volume (V)

Volume in cubic meters (m³)

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Number of moles (n)

Number of moles in moles (mol)

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Gas constant (R)

Gas constant equal to 8.31 J K⁻¹ mol⁻¹

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Temperature (T)

Temperature in kelvins (K)

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Convert pressure from kPa to Pa

Multiply by 1000 (e.g. 2 kPa = 2000 Pa)

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Convert Celsius to Kelvin

Add 273 (e.g. 25 °C = 298 K)

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Convert cm³ to m³

Multiply by 10⁻⁶

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Convert dm³ to m³

Multiply by 10⁻³

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Use of SI units in ideal gas equation

Because R (8.31 J K⁻¹ mol⁻¹) is defined using SI units.

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Spotting ideal gas equation usage

If the gas constant (R = 8.31) is given in the question.

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Calculation for moles of oxygen gas

n = PV ÷ RT = (70,000 × 0.040) ÷ (8.31 × 350) = 0.96 mol

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Volume of CO₂ at given conditions

V = nRT ÷ P = (0.65 × 8.31 × 280) ÷ 100,000 = 0.0151 m³ (15.1 dm³)

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Moles of nitrogen gas at given conditions

Convert: 0.55 dm³ = 0.00055 m³; T = 35 + 273 = 308 K; n = PV ÷ RT = (90,000 × 0.00055) ÷ (8.31 × 308) = 0.019 mol

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Temperature of H₂ in °C

Convert: V = 1200 cm³ = 1.20 × 10⁻³ m³; T = PV ÷ nR = (110,000 × 0.00120) ÷ (0.0500 × 8.31) = 317 K → 44 °C

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Volume of He at given conditions

Convert: T = 295 K, P = 75,000 Pa; V = nRT ÷ P = (0.75 × 8.31 × 295) ÷ 75,000 = 0.0245 m³ (24.5 dm³)

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Finding Mr of gas

Convert: V = 1.5 × 10⁻³ m³; P = 80,000 Pa; n = PV ÷ RT = (80,000 × 0.0015) ÷ (8.31 × 300) = 0.048 mol; Mr = mass ÷ moles = 2.6 ÷ 0.048 = 54.2 g mol⁻¹

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Mass of Ne gas

Convert: T = 317 K, P = 100,000 Pa; n = PV ÷ RT = (100,000 × 0.00300) ÷ (8.31 × 317) = 0.115 mol; Mr(Ne) = 20.2 → mass = n × Mr = 0.115 × 20.2 = 2.32 g

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Finding Mr at given conditions

Convert: V = 1.1 × 10⁻³ m³, T = 333 K, P = 250,000 Pa; n = PV ÷ RT = (250,000 × 0.0011) ÷ (8.31 × 333) = 0.0993 mol; Mr = mass ÷ moles = 1.60 ÷ 0.0993 = 16.1 g mol⁻¹

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Common mistakes with ideal gas equation

Forgetting to convert cm³/dm³ to m³; forgetting to convert °C to K; using the wrong unit for pressure (kPa instead of Pa).