Question

A reversible heat engine carries `1 mol` of an ideal monatomic gas around the cycle `ABCA`, as shown in the diagram. The process `BC` is adiabatic. Call the processes `AB, BC` and `CA` as `1,2` and `3` and the heat `( DeltaQ)_(r)`, change in internal energy `(DeltaU)`, and work done `( DeltaW)_(r), r=1,2,3` respectively. The temperature at `A,B,C` are `T_(1)=300K,T_(2)=600K` and `T_(3)=455K`. Indicate the pressure and volume at `A,B` and` C` by `P_(r)` and `V_(r), r=1,2,3,` respectively. Assume that intially pressure `P_(1)=1.00atm.`
Suppose the gas had been taken along an isothermal process from `B` instead of along the adiabatic process `BC` shown, so as to reach the same volume `V_(3)`, and the total heat `Q_(1)` is taken during the complete cycle involving the isothermal, while `Q_(a)` is the total heat taken in the cycle `ABCA` shown in the diagram. Then
A. `Q_(i)ltQ_(a)`
B. `Q_(i)=Q_(a)`
C. `Q_(i)gtQ_(a)`
D. `Q_(i) ` will be greater or less than `Q_(a)` dependng upon the value of `T_(3)`

Answer 1

Correct Answer - A
The isothermal process at `T=600K` is shown by 4 `( ` see figure `)`. Now the thermal energy `Q_(1)` is the same through both cycles. When `Q_(2)` is zero for the adiabatic expension, `Q_(3) ` is positive and finite. Also, the isobaric compression will involve more work done on the gas for the process `PA` than for `CA`. Also temperature at `P` is `600 K`. More heat is emitted by the gas in the change `PA` than in the change `CA`. Hence the cycle `ABPA` involving the isothermal will take less total heat than the cycle `ABDA`, `i.e., Q_(i)ltQ_(a)`