8 liters of an ideal gas is in an isolated container at 30 degrees Celsius. The container is heated at constant pressure until its volume is doubled. What is the new temperature of the gas?

Answer 1

For this problem we are looking at the ideal gas law, PV = nRT.

We can rewrite this equation as PV/T = nR, and since the number of molecules in the isolated container is constant, and R is a constant, we know that PV/T must be constant. So if we change one of those three variables, the others will change to maintain the constant.

In this particular example, our pressure is kept constant during the heating process, so we can also divide that out as a constant. Therefore, we get V/T = constant, or V₁/T₁ = V₂/T₂, where V₁ and T₁ are the volume and temperature initially, respectively, and V₂ and T₂ are the volume and temperature after the heating process.

We just need to plug the given values into this equation; however, we also need to convert Celsius to Kelvin for use in this equation:

8/(30+273.15) = 16/(T₂+273.15)

Doing the algebra, we get T₂ is equal to 333.15 degrees Celsius