To what temperature must a gas 273 K be heated in order to double both its volume and pressure?

According to the Ideal Gas Law, PV = nRT, where P is the pressure, V is the volume, n is the number of moles of gas, R is the gas constant, and T is the temperature in Kelvin. If we want to double both the volume and pressure of the gas while keeping the number of moles constant, then we can use the combined gas law, which states: (P1V1)/T1 = (P2V2)/T2 where P1 and V1 are the initial pressure and volume, and P2 and V2 are the final pressure and volume. Since we want to double both the pressure and volume, we can set: P2 = 2P1 V2 = 2V1 Substituting these values into the combined gas law, we get: (P1*2V1)/T1 = (2P1*2V1)/T2 Simplifying, we get: T2 = T1*4 So, the temperature must be quadrupled in order to double both the pressure and volume. Since the initial temperature is 273 K, we can multiply it by 4 to get the final temperature: T2 = 273 K * 4 = 1092 K Therefore, the answer is 1092 K.