Gases MCQs
MDCAT · Chemistry · 6 questions
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Revision summary
PMDC MDCAT syllabus: States of matter (gases)
The gas laws relate pressure P, volume V, temperature T (always in kelvin) and amount n. Boyle: PV is constant at fixed T, so V ∝ 1/P. Charles: V/T is constant at fixed P, so doubling T in kelvin doubles V. Avogadro: equal volumes of gases at the same T and P hold equal numbers of molecules. Together they give the ideal gas equation PV = nRT, with R = 8.314 J K⁻¹ mol⁻¹ (or 0.0821 dm³ atm K⁻¹ mol⁻¹). One mole of an ideal gas at 273 K and 1 atm occupies 22.414 dm³. Graham's law: the rate of diffusion or effusion is inversely proportional to the square root of molar mass, so hydrogen spreads fastest. Real gases follow the ideal laws only approximately; they deviate most at high pressure (molecular volume matters) and low temperature (attractions matter), which the van der Waals equation corrects. In numerical questions, convert °C to K first.
Questions with answers
Written by Parcha from the syllabus. Samples until a subject reviewer checks them; report anything wrong from the practice screen.
Q1 · Parcha original · Sample
Boyle's law states that, at constant temperature, the volume of a fixed mass of gas is:
- A. Directly proportional to pressure
- B. Inversely proportional to pressure
- C. Directly proportional to temperature
- D. Independent of pressure
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Answer: B. Inversely proportional to pressure
Boyle's law: PV = constant at constant temperature and amount of gas, so V ∝ 1/P. Doubling the pressure halves the volume.
Open this question on its own pageQ2 · Parcha original · Sample
A gas occupies 2 dm³ at 300 K. At constant pressure, its volume at 600 K will be:
- A. 1 dm³
- B. 2 dm³
- C. 6 dm³
- D. 4 dm³
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Answer: D. 4 dm³
Charles's law: V/T is constant at constant pressure (T in kelvin). Doubling T from 300 K to 600 K doubles V from 2 dm³ to 4 dm³.
Q3 · Parcha original · Sample
The value of the general gas constant R in SI units is:
- A. 8.314 J K⁻¹ mol⁻¹
- B. 0.0821 J K⁻¹ mol⁻¹
- C. 1.987 J K⁻¹ mol⁻¹
- D. 22.4 J K⁻¹ mol⁻¹
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Answer: A. 8.314 J K⁻¹ mol⁻¹
R = 8.314 J K⁻¹ mol⁻¹. The same constant is 0.0821 dm³ atm K⁻¹ mol⁻¹ and about 1.987 cal K⁻¹ mol⁻¹; the units decide which number to use.
Q4 · Parcha original · Sample
One mole of an ideal gas at 0 °C and 1 atm occupies about:
- A. 11.2 dm³
- B. 24.0 dm³
- C. 22.4 dm³
- D. 44.8 dm³
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Answer: C. 22.4 dm³
At 273 K and 1 atm (the STP used in Pakistani textbooks), the molar volume of an ideal gas is 22.414 dm³. At room temperature (about 25 °C) it is about 24 dm³.
Q5 · Parcha original · Sample
Under the same conditions, which gas diffuses fastest?
- A. O₂
- B. CO₂
- C. He
- D. H₂
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Answer: D. H₂
Graham's law: rate of diffusion ∝ 1/√M. Hydrogen has the smallest molar mass (2 g/mol), so it diffuses fastest; helium (4) is next.
Q6 · Parcha original · Sample
Real gases deviate most from ideal behaviour at:
- A. High pressure and low temperature
- B. Low pressure and high temperature
- C. Low pressure and low temperature
- D. High temperature only
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Answer: A. High pressure and low temperature
At high pressure, the volume of the molecules themselves is no longer negligible; at low temperature, attractive forces between molecules matter. Both break the ideal-gas assumptions.