One mole of oxygen gas is at a pressure of 6.50 atm and a temperature of 28.5°C.(a) If the gas is heated at constant volume until the pressure triples, what is the final temperature? °C(b) If the gas is heated so that both the pressure and volume are doubled, what is the final temperature? °C

(a) We can use the ideal gas law for this problem, which states that the pressure and temperature of a gas are directly proportional at constant volume. This can be expressed as P1/T1 = P2/T2, where P1 and T1 are the initial pressure and temperature, and P2 and T2 are the final pressure and temperature.

First, we need to convert the initial temperature from Celsius to Kelvin. The conversion formula is K = °C + 273.15, so T1 = 28.5°C + 273.15 = 301.65 K.

The final pressure P2 is three times the initial pressure, so P2 = 3 * 6.50 atm = 19.5 atm.

We can now solve for T2: T2 = P2/P1 * T1 = 19.5 atm / 6.50 atm * 301.65 K = 903.95 K.

To convert this back to Celsius, we subtract 273.15: T2 = 903.95 K - 273.15 = 630.8°C.

(b) If both the pressure and volume are doubled, we can use the combined gas law, which states that P1V1/T1 = P2V2/T2.

The final pressure P2 is twice the initial pressure, so P2 = 2 * 6.50 atm = 13 atm. The final volume V2 is also twice the initial volume V1, so V2 = 2 * V1.

We can now solve for T2: T2 = P2V2/P1V1 * T1 = 13 atm * 2V1 / 6.50 atm * V1 * 301.65 K = 602.3 K.

To convert this back to Celsius, we subtract 273.15: T2 = 602.3 K - 273.15 = 329.15°C.