One mole of an ideal mono atomic gas is at a temperature `T_(0) = 1000 K` and its pressure is `P_(0)`. The gas is adiabatically cooled so that its pressure becomes `(2)/(3) P_(0)`. Thereafter, the gas is cooled at constant volume to reduce its pressure to `(P_(0))/(3)` . Calculate the total heat absorbed by the gas during the process.
Take `R = (25)/(3)J mol^(-1)K^(-1)` and `((2)/(3))^(2//5) = 0.85`

Answer

`5312.5 J`