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. 17 dm3
- A. 100 dm3
- B. 75 dm3
- C. 17 dm3
- D. 51 dm3
Explanation
To find the volume of the gas at STP (Standard Temperature and Pressure), we can use the combined gas law equation: (P1 * V1) / T1 = (P2 * V2) / T2 Where: P1 = Initial pressure of the gas (in N/m^2) = 1.05 × 10^5 N/m^2 V1 = Initial volume of the gas (in dm^3) = 60 dm^3 T1 = Initial temperature of the gas (in Kelvin) = 600°C + 273.15 (conversion from Celsius to Kelvin) P2 = Final pressure of the gas at STP (in N/m^2) = 1.01 × 10^5 N/m^2 (STP pressure) T2 = Final temperature of the gas at STP (in Kelvin) = 273 K (STP temperature) V2 = Final volume of the gas at STP (unknown) First, let's convert the initial temperature to Kelvin: T1 = 600°C + 273.15 = 873.15 K Now, we can solve for V2: (1.05 × 10^5 N/m^2 * 60 dm^3) / 873.15 K = (1.01 × 10^5 N/m^2 * V2) / 273 K Now, let's solve for V2: V2 = (1.05 × 10^5 N/m^2 * 60 dm^3 * 273 K) / (1.01 × 10^5 N/m^2 * 873.15 K) V2 ≈ 17.22 dm^3 Therefore, the volume of the gas at STP would be approximately 17.22 dm^3.
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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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