Ideal Gas Equation
A rigid container of volume 0.0224 m^3 holds an ideal gas at 273 K and 1.0 \times 10^5 Pa; estimate the number of moles present.
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Solution
0.99 mol
Relating the macroscopic state variables of a gas, the ideal gas equation PV=nRT allows the number of moles n to be found from pressure, volume, and temperature. It combines Boyle's law, Charles's law, and Avogadro's principle into a single relation that is valid for dilute gases. Rearranging gives n=RTPV. Substituting P=1.0×105 Pa, V=0.0224 m\textsuperscript{3}, T=273 K, and R=8.314 J/mol\cdotK: n=8.314×273(1.0×105)(0.0224)=2269.72240≈0.99 mol. This is the well-known result that one mole of ideal gas occupies 22.4 litres at standard temperature and pressure. The value 1.98 mol doubles the answer by mistakenly halving the temperature. The value 0.50 mol arises from using twice the correct volume in the denominator. The value 2.0 mol ignores the gas-constant scaling entirely. This problem is a direct application of the NCERT ideal gas law, which holds best at low pressure and high temperature where intermolecular forces are negligible. As a plausibility check, the conditions given are essentially STP, where one mole is expected, confirming n≈1 mol.
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About This Question
- Subject
- physics
- Chapter
- kinetic theory of gases
- Topic
- ideal gas equation
- Difficulty
- Easy
- Year
- 2025
Solution
Correct Answer:
0.99 mol
Relating the macroscopic state variables of a gas, the ideal gas equation PV=nRT allows the number of moles n to be found from pressure, volume, and temperature. It combines Boyle's law, Charles's law, and Avogadro's principle into a single relation that is valid for dilute gases. Rearranging gives n=RTPV. Substituting P=1.0×105 Pa, V=0.0224 m\textsuperscript{3}, T=273 K, and R=8.314 J/mol\cdotK: n=8.314×273(1.0×105)(0.0224)=2269.72240≈0.99 mol. This is the well-known result that one mole of ideal gas occupies 22.4 litres at standard temperature and pressure. The value 1.98 mol doubles the answer by mistakenly halving the temperature. The value 0.50 mol arises from using twice the correct volume in the denominator. The value 2.0 mol ignores the gas-constant scaling entirely. This problem is a direct application of the NCERT ideal gas law, which holds best at low pressure and high temperature where intermolecular forces are negligible. As a plausibility check, the conditions given are essentially STP, where one mole is expected, confirming n≈1 mol.
This easy difficulty physics question is from the chapter kinetic theory of gases, covering the topic of ideal gas equation. It appeared in the 2025 exam.
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