If the volume of a gas collected at a temperature of 600 °C and pressure of 1.05 x 10^5 Nm-2 is 60 dm3, what would be the volume of gas at STP (P=1.01 x 10^5 Nm-2, T = 273 K)?
Correct answer: C. 19.5 dm3
- A. 100 dm3
- B. 75 dm3
- C. 19.5 dm3
- D. 51 dm3
Explanation
To find the volume of the gas at STP (Standard Temperature and Pressure), we apply the combined gas law equation:(P1 * V1) / T1 = (P2 * V2) / T2Given:P1 = 1.05 × 105 N/m2V1 = 60 dm3T1 = 873.15 K (600°C converted to Kelvin)P2 = 1.01 × 105 N/m2T2 = 273 KSolving for V2:V2 = (1.05 × 105 N/m2 * 60 dm3 * 273 K) / (1.01 × 105 N/m2 * 873.15 K) ≈ 19.5 dm3Thus, the correct volume at STP is approximately 19.5 dm3, confirming that Option C is correct. Other options do not satisfy the equation when calculated under given conditions.
This question appeared on the UHS MDCAT 2015 paper, which you can sit online with every answer explained.
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The ideal gas equation, PV = nRT, relates pressure, volume, amount and absolute temperature when gas particles are assumed to have negligible volume and no intermolecular forces. Applications include finding molar mass, density or an unknown gas variable, with careful use of Kelvin temperature and consistent units, and comparison with real-gas behaviour.
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