An ideal gas `(1` mol, monatomic) is in the intial state `P` (see Fig.) on an isothermal `A` at temperature `T_(0)`. It is brought under a constant volume `(2V_(0))` process to `Q` which lies on an adiabatic `B` intersecting the isothermal `A` at `(P_(0),V_(0),T_(0))`. The change in the internal energy of the gas during the process is (in terms of `T_(0)) (2^(2//3) = 1.587)`

Options

A. (a)`2.3 T_(0)`
B. (b)`-4.6T_(0)`
C. (c)`-2.3T_(0)`
D. (d)`4.6 T_(0)`

Answer

B

Explanation

b. Temperature at state `P = T_(0)`. Since `P` lies on the isothermal meal of temperature `T_(0)`. If `T` be the temperature at `Q`, then for the adiabatic process `B`, we have, `T_(0) V_(0)^(gamma - 1) = T (2V_(0))^(gamma - 1)`
`T = (T)/(2^(gamma - 1)) = (T)/(2^(gamma - 1))`
change in the internal energy of the gas is
`Delta U = C_(V) (T - T_(0)) = ((R )/(gamma - 1)) ((T_(0))/ (2^(2//3)) - T_(0))`
`= (3 RT_(0) (1 - 2^(2//3)))/(2 xx 2^(2//3)) = - 4.6 T_(0)`