The volume and pressure of a given mass of gas at 27oC are 76cm3 and 80cm of mercury respectively. Calculate the volume at s.t.p

To solve the problem, we need to use the combined gas law, which relates the pressure, volume, and temperature of a gas. The combined gas law states that: P1V1/T1 = P2V2/T2 where P1, V1, and T1 are the initial pressure, volume, and temperature, respectively, and P2, V2, and T2 are the final pressure, volume, and temperature, respectively. We are given the initial pressure, volume, and temperature of the gas as P1 = 80 cmHg, V1 = 76 cm^3, and T1 = 27°C. We want to find the volume of the gas at STP, which is a temperature of 0°C and a pressure of 1 atm (or 760 mmHg). To use the combined gas law, we need to convert the initial temperature to kelvin (K), since temperature must be in kelvin in this equation. We do this by adding 273 to the initial temperature: T1 = 27°C + 273 = 300 K We also need to convert the initial pressure from cmHg to atm, using the conversion factor: 1 atm = 760 mmHg = 101.325 kPa = 76 cmHg So, the initial pressure in atm is: P1 = 80 cmHg × (1 atm/76 cmHg) = 1.05 atm Now we can plug in the values into the combined gas law, with the final pressure P2 = 1 atm and the final temperature T2 = 0°C + 273 = 273 K: P1V1/T1 = P2V2/T2 (1.05 atm)(76 cm^3)/(300 K) = (1 atm)(V2)/(273 K) Solving for V2 gives: V2 = (1.05 atm)(76 cm^3)(273 K)/(300 K) = 72.8 cm^3 Therefore, the volume of the gas at STP is 72.8 cm^3, which is.