The gaseous state: ideal and real gases and pV = nRT — practice questions

Step 1: List the variables and convert units. p = 150 kPa = 150 × 10³ Pa = 150,000 Pa V = ? n = 0.500 mol T = 25 °C = 25 + 273 = 298 K R = 8.31 J mol⁻¹ K⁻¹

Step 2: Rearrange the ideal gas equation to solve for V. pV = nRT => V = nRT / p

Step 3: Substitute the values and calculate. V = (0.500 mol × 8.31 J mol⁻¹ K⁻¹ × 298 K) / 150,000 Pa V = 1238.79 / 150,000 V = 0.0082586 m³

Step 4: Give the answer to an appropriate number of significant figures and in sensible units. The data is given to 3 significant figures, so the answer should be too. V = 0.00826 m³ Alternatively, converting to dm³: 0.00826 m³ × 1000 = 8.26 dm³.

Step 1: Since the amount of gas (n) is constant, we can use the combined gas law, which is a rearrangement of the ideal gas equation: p₁V₁/T₁ = p₂V₂/T₂.

Step 2: List the initial and final conditions. Note that for this type of 'two-state' problem, volume units can be kept consistent (e.g., both cm³) as they will cancel out. Temperature must always be in Kelvin. Initial state (1): p₁ = 1.00 × 10⁵ Pa V₁ = 250 cm³ T₁ = 300 K

Final state (2): p₂ = ? V₂ = 100 cm³ T₂ = 350 K

Step 3: Rearrange the equation to solve for p₂. p₂ = (p₁V₁T₂) / (T₁V₂)

Step 4: Substitute the values and calculate. p₂ = (1.00 × 10⁵ Pa × 250 cm³ × 350 K) / (300 K × 100 cm³) p₂ = 8,750,000,000 / 30,000 p₂ = 291,666.67 Pa

Step 5: Give the answer to an appropriate number of significant figures. The data is given to 3 significant figures. p₂ = 2.92 × 10⁵ Pa