The above p-v diagram represents the thermodynamic cycle of an engine, operating with an ideal monoatomic gas. The amount of heat, extracted from the source in a single cycle is

Options

A. `P_(0)V_(0)`
B. `(13/2)P_(0)V_(0)`
C. `(11/2)P_(0)V_(0)`
D. `4P_(0)V_(0)`

Answer

B

Explanation

In the (figure)shown, suppose temp. of monoatomic gas at A is `T_(0)`
In going A to B , volume is constant.
As pressure increases from `P_(0)` to `2P_(0)` temp. increases from `T_(0)` to `2T_(0)`
`:.` Heat extracted from source `= 1xx C_(v).dT`
`Q_(1)=3/2R(2T_(0)-T_(0))= 3/2RT_(0)= 3/2P_(0)V_(0)`
In going from B to C, pressure is constant. As volume increases from `V_(0) to 2 V_(0)`, temp.
increases from `2T_(0) to 4T_(0) ( :. T prop V)`
`:.` Heat extracted from source `=1xxc_(p).dT`
`Q_(2)=1xx5/2R(4T_(0)-2T_(0))= 5RT_(0)= 5P_(0)V_(0)`
In going from C to D, volume is constant. Pressure decreases from `2p_(0) to P_(0)`. Temp. decreases from `4T_(0) to 2T_(0)`. No heat is extracted from the source.
In going to D to A, pressure is constant
Volume decreases from `2v_(0) to V_(0)`. No heat is extracted from the source. Hence in single cycle, heat extracted from the source
`=Q_(1)+Q_(2)= 3/2P_(0)V_(0)+5P_(0)V_(0)= (13)/2p_(0)v_(0)`