An ideal gas at a temperature of 39°C, a pressure of 2.00 atmospheres and a volume of 2.10 liters. What is the temperature if the pressure is halved and the volume is tripled?

To solve this problem, we can use the ideal gas law equation, which states that the product of the pressure and volume of a gas is directly proportional to the temperature of the gas. This can be written as:

P1V1/T1 = P2V2/T2

Where: P1 = initial pressure = 2.00 atm V1 = initial volume = 2.10 L T1 = initial temperature = 39°C = 273 + 39 = 312 K (Kelvin)

P2 = final pressure = 2.00 atm / 2 = 1.00 atm V2 = final volume = 2.10 L * 3 = 6.30 L T2 = final temperature = ?

We can substitute these values into the equation and solve for T2:

(2.00 atm * 2.10 L) / 312 K = (1.00 atm * 6.30 L) / T2

Solving for T2 gives:

T2 = (1.00 atm * 6.30 L) * 312 K / (2.00 atm * 2.10 L) = 936 K

So, the final temperature of the gas, when the pressure is halved and the volume is tripled, is 936 K.