The `P - V` diagram of 2 gm of helium gas for a certain process `A rarr B` is shown in the figure. What is the heat given to the gas during the process `A rarr B`?

Options

A. `4 P_(0) V_(0)`
B. `6 P_(0) V_(0)`
C. `4.5 P_(0) V_(0)`
D. `2 P_(0) V_(0)`

Answer

B

Explanation

Change in internal energy from `A rarr B` is
`Delta U = (f)/(2) muRDelta = (f)/(2) mu(P_(f)V_(f)-P_(i)V_(i))`
`=(3)/(2)(2P_(0)2V_(0) P_(0)xxV_(0))=(9)/(2)P_(0)V_(0)`
Work done in process `A rarr B` is equal to the Area covered by the graph with volume axis i.e.,
`W_(A to B)=(1)/(2)(P_(0)+2P_(0))xx(2V_(0)-V_(0))=(3)/(2)P_(0)V_(0)`
Hence, `DeltaQ=DeltaU+DeltaW=(9)/(2)P_(0)V_(0)+(3)/(2)P_(0)V_(0)=6P_(0)V_(0)`