One mole of an ideal monoatomaic gas is taken from `A` to `C` along the path `ABC` . The temperature of the gas at `A` is `T_(0)` . For the process `ABC` :

Options

A. Work done by the gas is `RT_(0)`
B. Change in internal energy of the gas is `(11)/(2)RT_(0)`
C. Heat absorbed by the gas is `(11)/(2)RT_(0)`
D. Heat absorbed by the gas is `(13)/(2)RT_(0)`

Answer

A, C

Explanation

Work done = Area of `ABC` with `V`- axis
=`P_(0)(2V_(0)-V_(0)) + 0 = P_(0)V_(0) = nRT_(0) = RT_(0)`
Change in internal energy = `nC_(V)DeltaT`
` 1 xx (3)/(2)R xx (4T_(0)-T_(0)) = (9)/(2)RT_(0)`
`therefore` Heat absorbed
= `(9)/(2) RT_(0) + RT_(0) = (11)/(2)RT_(0)`