What amount of heat will be released during isobaric cooling of 80 g of helium from 200 degrees to 100?

Data: m (mass of helium) = 80 g (0.08 kg); T1 (initial temperature of helium) = 200 ºС; T2 (temperature after cooling) = 100 ° C.

Constants: i (number of degrees of freedom) = 3; R (universal gas constant) = 8.314 J / (mol * K); M (molar mass of helium, monatomic gas) = 4 * 10 ^ -3 kg / mol.

The amount of heat that will be released when helium is cooled: ΔU (decrease in internal energy) = U1 – U2 = i / 2 * (m / M) * R * (T1 – T2).

Calculation: ΔU = 3/2 * (0.08 / (4 * 10 ^ -3)) * 8.314 * (200 – 100) = 24 942 J.