An ideal gas goes from State A to state B via three different process as indicate in the `P-V` diagram.

If `Q_(2), Q_(3)` indicates the heat absorbed by the gas along the three processes and `DeltaU_(1), DeltaU_(2), DeltaU_(3)` indicates the change in internal energy along the three processes respectively, then

Options

A. `Q_(1) gt Q_(2) gt Q_(3)` and `DeltaU_(1)= deltaU_(2)=DeltaU_(3)`
B. `Q_(3) gt Q_(2) gt Q_(1)` and `DeltaU_(1)=DeltaU_(2)=DeltaU_(3)`
C. `Q_(1)=Q_(2)=Q_(1)` and `DeltaU_(1) gt DeltaU_(2) gt DeltaU_(3)`
D. `Q_(3) gt Q_(2)Q_(1)` and `DeltaU_(1) gt DeltaU_(2) gt DeltaU_(3)`

Answer

A

Explanation

In (figure). For all of three processes, initial and final states are the same. Therefore, change in internal energy is same i.e., `DeltaU_(1)= DeltaU_(2)= DeltaU+(3)`
As work done = area under `P-V` graph
`:. DeltaW_(1) gt DeltaW_(2) gt DeltaW_(3)`
As `Q= DeltaU+ DeltaW`
`:. Q_(1) gt Q_(2) gt Q_(3)`