One mole of an ideal gas undergoes a cyclic process ABCDA as shown in the P-V diagram, The net work done in the process is
`(1 atm`=`10^(6)` dyne `cm^(-2))`

Options

A. 500 J
B. 700 J
C. 800 J
D. 900 J

Answer

C

Explanation

As the loop is traced in clockwise direction the work done is positive
Work done in the process = area of rectangle ABCD
`=ABxxAD`
Here AB=4-2=2 litre =`2xx10^(3)cm^(3)`
AD =8-4 = 4atm = `4xx10^(6)` dyne `cm^(-2)`
Work done , =`ABxxAD=2xx10^(3)xx4xx10^(6)`
=`8xx10^(9)erg=800 J`