Problem Statement
Estimate the molar volume of saturated water vapour at $100°\text{C}$ using the ideal gas approximation and compare to the actual value.
Given Information
- See problem statement for all given quantities.
Physical Concepts & Formulas
This problem applies fundamental physics principles to the scenario described. The solution requires identifying the relevant conservation laws and equations of motion, then solving systematically with careful attention to units and sign conventions.
- See the step-by-step solution for the specific equations applied.
- All quantities are in SI units unless otherwise stated.
Step-by-Step Solution
Step 1 — Identify given quantities and set up the problem: At the boiling point ($T=373\ \text{K}$, $p=1.013\times10^5\ \text{Pa}$):
Step 2 — Apply the relevant physical law or equation: $$V_{m,ideal} = \frac{RT}{p} = \frac{8.314\times373}{1.013\times10^5} = \frac{3101}{1.013\times10^5} = 30.6\ \text{L/mol}$$
Step 3 — Solve algebraically for the unknown: Actual value: $30.14\ \text{L/mol}$ (steam tables). The ideal gas approximation is excellent here.
Step 4 — Substitute numerical values with units: Specific volume: $v = V_m/M = 30.6\times10^{-3}/0.018 \approx 1.70\ \text{m}^3/\text{kg}$ (actual: 1.673 m³/kg). ✓
Step 5 — Compute and check the result: Result: $V_m \approx 30.6\ \text{L/mol}$, within 1.5% of the actual value.
Worked Calculation
$$V_{m,ideal} = \frac{RT}{p} = \frac{8.314\times373}{1.013\times10^5} = \frac{3101}{1.013\times10^5} = 30.6\ \text{L/mol}$$
$$\text{Numerical result} = \text{given expression substituted with values}$$
$$\boxed{V_{m,ideal} = \frac{RT}{p} = \frac{8.314\times373}{1.013\times10^5} = \frac{3101}{1.013\times10^5} = 30.6\ \text{L/mol}}$$
At the boiling point ($T=373\ \text{K}$, $p=1.013\times10^5\ \text{Pa}$):
$$V_{m,ideal} = \frac{RT}{p} = \frac{8.314\times373}{1.013\times10^5} = \frac{3101}{1.013\times10^5} = 30.6\ \text{L/mol}$$
Actual value: $30.14\ \text{L/mol}$ (steam tables). The ideal gas approximation is excellent here.
Specific volume: $v = V_m/M = 30.6\times10^{-3}/0.018 \approx 1.70\ \text{m}^3/\text{kg}$ (actual: 1.673 m³/kg). ✓
Result: $V_m \approx 30.6\ \text{L/mol}$, within 1.5% of the actual value.
Answer
$$\boxed{V_{m,ideal} = \frac{RT}{p} = \frac{8.314\times373}{1.013\times10^5} = \frac{3101}{1.013\times10^5} = 30.6\ \text{L/mol}}$$
Physical Interpretation
The numerical answer is physically reasonable — matching expected orders of magnitude and dimensional analysis. The result confirms the theoretical prediction and provides quantitative insight into the system’s behaviour.
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