Chemistry • Year 11 • Module 2 • Lesson 4
Gases & Molar Volume
Apply V = n × Vm in both directions, select the correct molar volume for given conditions, and identify and correct common student errors.
1. Calculation drill — six problems in both directions
For each problem: (i) identify the correct Vm value and state why; (ii) show full working; (iii) give the answer with correct units. 12 marks (2 per problem)
1.1 Calculate the volume occupied by 2.00 mol of oxygen gas (O2) at SATP.
1.2 A 45.42 L sample of nitrogen gas (N2) is collected at STP. How many moles does it contain?
1.3 A balloon is filled with 4.20 g of helium gas (He) at SATP. Calculate the volume of the balloon. (He = 4.003 g mol−1)
1.4 A gas syringe at STP contains 680 mL of chlorine gas (Cl2). How many moles of Cl2 are present? (Remember to convert units first.)
1.5 A chemistry teacher needs to prepare 0.250 mol of carbon dioxide gas at SATP. What volume of gas (in litres) must be collected?
1.6 Calculate the number of moles in 11.36 L of ammonia gas (NH3) at SATP.
2. Interpret graph — molar volume versus temperature
The graph below shows how the molar volume of an ideal gas changes with temperature at constant pressure (100 kPa). 7 marks
Figure 2.1. Molar volume of an ideal gas at 100 kPa as temperature increases from 0 °C to 50 °C. Illustrative data.
2.1 Describe the relationship between temperature and molar volume shown in the graph. 2 marks
2.2 Use the graph to identify the molar volume values at STP and SATP and explain why the values differ. 2 marks
2.3 A student claims that at 50 °C and 100 kPa, one mole of O2 gas occupies the same volume as one mole of CO2 gas. Do the data in this graph support or refute this claim? Justify your answer with reference to Avogadro’s law. 3 marks
3. Compare: applying the formula in two directions
Complete the two-column table to compare calculating volume from moles versus calculating moles from volume. 6 marks (1 per cell)
| Feature | Finding volume (V) | Finding moles (n) |
|---|---|---|
| Formula to use | ||
| What you need to know | ||
| Worked example (use n = 3 mol and Vm = 24.8 L mol−1, or V = 74.4 L and Vm = 24.8 L mol−1) |
4. Predict and justify — natural gas at a NSW power station
The Eraring Power Station (NSW) is a large coal-fired power station, but smaller gas-fired peakers use natural gas (principally methane, CH4) to supplement the grid. A storage tank holds methane at SATP.
5 marks
4.1 A burst releases 1.60 kg of methane (CH4) into the atmosphere at SATP. Calculate the volume of gas released. (C = 12.011, H = 1.008) 3 marks
4.2 An engineer argues that 1.60 kg of methane released at STP would occupy a smaller volume than at SATP. Predict whether this claim is correct and justify your answer. 2 marks
5. Error hunt — find and fix the student’s mistakes
A Year 11 student wrote the following solutions to three problems. Each solution contains one error. Identify the error and write the correction. 6 marks (2 per problem: 1 identify, 1 correct)
Student’s Problem A.
“Find the volume of 2.50 mol of N2 gas at STP.
Vm at STP = 24.8 L mol−1
V = n × Vm = 2.50 × 24.8 = 62.0 L”
5.1 Error:
Correction:
Student’s Problem B.
“Find the moles in 620 mL of CO2 at SATP.
n = V ÷ Vm = 620 ÷ 24.8 = 25.0 mol”
5.2 Error:
Correction:
Student’s Problem C.
“Find the volume of 1.00 mol of liquid water (H2O) at SATP.
V = n × Vm = 1.00 × 24.8 = 24.8 L”
5.3 Error:
Correction:
Q1 — Calculation drill
1.1 SATP: Vm = 24.8 L mol−1. V = 2.00 × 24.8 = 49.6 L.
1.2 STP: Vm = 22.71 L mol−1. n = 45.42 ÷ 22.71 = 2.00 mol.
1.3 MM(He) = 4.003 g mol−1. n = 4.20 ÷ 4.003 = 1.049 mol. SATP: V = 1.049 × 24.8 = 26.0 L.
1.4 Convert: 680 mL = 0.680 L. STP: n = 0.680 ÷ 22.71 = 0.0299 mol.
1.5 SATP: V = 0.250 × 24.8 = 6.20 L.
1.6 SATP: n = 11.36 ÷ 24.8 = 0.458 mol.
Q2.1 — Graph trend (2 marks)
The molar volume increases linearly with temperature [1]. As temperature rises, gas particles move faster, collide with the container walls more forcefully, and therefore occupy a larger volume at constant pressure [1].
Q2.2 — STP and SATP values and reason (2 marks)
From the graph: STP (0 °C) → Vm = 22.71 L mol−1; SATP (25 °C) → Vm = 24.8 L mol−1 [1]. The values differ because SATP is at a higher temperature; at higher temperature the gas particles have more kinetic energy and the volume they occupy at the same pressure is larger [1].
Q2.3 — Student claim about 50 °C (3 marks)
The claim is supported by the data [1]. Avogadro’s law states that equal amounts (moles) of all ideal gases at the same temperature and pressure occupy the same volume [1]. Because both O2 and CO2 are treated as ideal gases at 50 °C and 100 kPa, one mole of each occupies the same molar volume (approximately 26.9 L mol−1 from the graph) regardless of their different molecular masses [1].
Q3 — Compare table
Formula: Finding V = n × Vm; Finding n = V ÷ Vm. What you need: V: know n (mol) and Vm (L mol−1); n: know V (L) and Vm (L mol−1). Worked example: V = 3 × 24.8 = 74.4 L; n = 74.4 ÷ 24.8 = 3.00 mol.
Q4.1 — Methane volume at SATP (3 marks)
MM(CH4) = 12.011 + 4(1.008) = 16.043 g mol−1 [1]. n = 1600 ÷ 16.043 = 99.7 mol [1]. V = 99.7 × 24.8 = 2473 L (2.47 × 103 L) [1].
Q4.2 — STP vs SATP volume (2 marks)
The engineer is correct [1]. The same number of moles of methane (99.7 mol) at the lower temperature of STP (0 °C) would have V = 99.7 × 22.71 = 2264 L, which is smaller than 2473 L at SATP. At a lower temperature, gas particles have less kinetic energy, so the same number of moles exerts the same pressure in a smaller volume [1].
Q5 — Error hunt
5.1 Error: The student used Vm = 24.8 L mol−1 (SATP) when the question specified STP. Correction: At STP, Vm = 22.71 L mol−1. V = 2.50 × 22.71 = 56.8 L.
5.2 Error: The student substituted 620 mL directly without converting to litres. Correction: V = 620 mL = 0.620 L. n = 0.620 ÷ 24.8 = 0.0250 mol.
5.3 Error: The student applied V = n × Vm to a liquid. The molar volume formula only applies to gases. Correction: 1.00 mol of liquid water (M = 18.015 g mol−1) has a mass of 18.015 g; at a density of 1.00 g mL−1 it occupies approximately 18.0 mL (0.0180 L).