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)?

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.