1.

A thermodynamic system undergoes cyclic process ABCDA as shown in figure. The work done by the system in the cycle is

Answer»

`P_(0)V_(0)`
`2P_(0)V_(0)`
`(P_(0)V_(0))/(2)`
zero

Solution :In a CYCLIC process work done is EQUAL to the area under the cycle and is positive if the cycle is clockwise and negative if anticlockwise.
As is CLEAR from figure,
`W_(AEDA) = + "area of " DELTA AED = + (1)/(2)P_(0)V_(0)`
`W_(BCEB) = -"Area of "Delta BCE = -(1)/(2)P_(0)V_(0)`
The net work done by the system is
`W_("net") = W_(AEDA) + W_(BCEB) = + (1)/(2)P_(0)V_(0) - (1)/(2)P_(0)V_(0)` = zero


Discussion

No Comment Found