Which is the temperature of a given mass of a gas initially at 0°C and 9 atm, if the pressure is reduced to 3 atm at constant volume?

We can solve this problem using the combined gas law, which relates the pressure, volume, and temperature of a gas: (P1 V1) / T1 = (P2 V2) / T2 where P1 and T1 are the initial pressure and temperature, P2 and T2 are the final pressure and temperature, and V1 and V2 are the initial and final volumes (which are held constant in this problem). We can plug in the values we know and solve for the unknown temperature: (9 atm)(V1) / 273 K = (3 atm)(V1) / T2 Cross-multiplying and simplifying gives: T2 = (3 atm / 9 atm) * 273 K = 91 K Therefore, the temperature of the gas is 91K. Answer is correct.