One mole of a monatomic gas is taken from a point `A` to another point `B` along the path `ACB`. The initial temperature at `A` is `T_(0)`. Calculate the heat abosrbed by the gas in the process `A rarr C rarr B`.

Explanation

If `T_(B)` be the temperature at `B`, then by gas law
`(P_(A) V_(A))/(T_(A)) = (P_(B) V_(B))/(T_(B))`
`:. T_(B) = (P_(B) V_(B))/(P_(A) V_(A)) T_(A) = ((2P_(0)) (2 V_(0)))/(P_(0) V_(0)) T_(0)`
The change in internal energy from `A` t o `B`
`Delta U = n C_(v) Delta T = 1 xx (3R)/(2) xx (4 T_(0) - T_(0))`
`= (9RT_(0))/(2)`
Work done in the process `A` to `C`
`W_(AC) = P Delta V = P_(0) (2 V_(0) - V_(0))`
`= P_(0) V_(0) = RT_(0)`
and `W_(CB) = 0`
`:.` Total work done form `A rarr C rarr B`
`W_(AC) + W_(CB) = RT_(0) + 0 = RT_(0)`
From the first law of themodynamics,
`Q = Delta U + W`
`= (9RT_(0))/(2) + RT_(0)`
`(11 RT_(0))/(2)`
Thus heat absorbed by the gas from `A rarr C rarr B` is
`(11 RT_(0))/(2)`