One mole of an ideal gas at 300 K is heated at constant volume `(V_(1)` ) until its temperature is doubled, then it is expanded isothermally till it reaches the original pressure. Finally, the gas is cooled at the constant pressure till system reached to the half of original volume `(V_(2)).` Determine the total work done(`V _(f)//2)` in calories: [Use in `2-0.70, K = 2 cal K ^(-1) mol ^(-1)]`

Options

A. 60
B. 40
C. 20
D. 10

Explanation

The temperature, pressure and volume conditions at each of the points A,B,C,D as show in are `A ( p _(1) , V _(1) , T _(1)) , B ( 2 p _(1) , V _(1), 2 T _(1)) C ( p _(1) , 2 V_(1) , 2 T _(1)) , D (2 p _(1) + (V_(1))/(2) ,(T_(1))/(2))`
The total work done is given by `W _("total") = W _(AB) + W _(BC) + W _(CD),` For isochoric change `(AB) , W _(AB) = 0.`
For isothermal expansion `(BC), W _(BC) =- 1 xx R xx 2 T _(1) ln (( 2 V)/( V _(1))) =- 2 2 xx 300 xx 0. 7 =- 840 cal`
For isobaric change `(CD), W _(CD) =0 p _(1) Delta V =- R Delta V =- R [ (T _(1))/( 2) - 2 T _(1) ] = (3)/(2) RT _(1) = (3)/(2) xx 2 xx 300 = 900 cal `
Therefore , `W _("total") = - 840 + 900 = 60 cal`