One mole of a monatomic ideal gas initially at a pressure of 2.00 bar and a temperature of 273 K is taken to a final pressure of 4.00 bar by a reversible path defined by `p//V`= constant. Taking `C_(V)` to be equal to 12.5 K `mol^(-1) K^(-1)`, the value of `(Delta U)/(w)` for this process is calculated to be

Options

A. `-3.0`
B. `-1.5`
C. `+1.5`
D. `+3.0`

Answer

A

Explanation

Given, `(P)/(V) = K` (constant) also, `(P_(1)V_(1))/(T_(1)) = (P_(2)V_(2))/(T_(2))`
`rArr T_(2) = T_(1) ((p_(2)V_(2))/(p_(1)V_(1))) = T_(1) ((p_(2))/(p_(1)))`......(i)
`rArr Delta U = C_(V) Delta t = 3C_(V) T_(1) rArr -dw = pdV = kVdV`
`rArr w = - (K)/(2) xx (3T_(1)R)/(K) = (3RT_(1))/(2)`
`rArr (DeltaU)/(w) = - (2C_(V))/(R) = -3`