An ideal monoatomic gas is carried along the ABCDA as shown in the figure. The total heat absorbed during this process is

Options

A. `10.5 P_(0)V_(0)`
B. `7.5 p_(0)V_(0)`
C. `2.5 P_(0)V_(0)`
D. `1.5P_(0)V_(0)`

Answer

A

Explanation

Heat absorbed means heat is extracted from source i.e.Q must be positive.
This occurs along path
`A to B to C `.
`:. ` Heat added = `DeltaQ_("ABC") = DeltaU_(ABC) + Delta W _(ABC) `
`= nC_(V) ( T_(C)-T_(A)) + 3P_(0)(2V_(0)-V_(0))`
`= n (3)/(2) R(T_(C)-T_(A))+ 3p_(0)V_(0)= (3)/(2)( nRT_(C) - nRT_(A)) + 3p_(0)V_(0)`
`= (3)/(2)( p_(C)V_(C)-P_(A)V_(A)) + 3p_(0)V_(0)`
`=(3)/(2)(3p_(0)2V_(0)-p_(0)V_(0)) + 2p_(0) V_(0)`
`= (3)/(2) xx 5p_(0)V_(0)+3p_(0)V_(0)= (21)/(2)P_(0)V_(0)= 105p_(0)V_(0)`