# Figure below shows two paths that may be taken by a gas to go from a state A to a state C.In process AB, 400 J of heat is added to the system and in process BC, 100 J of heat is added to the system. The heat absorbed by the system in the process AC will be: Option 1) 460 J Option 2) 300 J Option 3) 380 J Option 4) 500 J

As we learnt in

Work by area -

If we plot a process on PV diagram.Then area under the curve will give work done.

- wherein

$A\rightarrow B$, isochoric process

$\therefore W_{A\rightarrow B}=0$

$\Delta Q_{A\rightarrow B}=\Delta U_{A\rightarrow B}=400 J$

$B\rightarrow C,$ isobaric process

$\Delta Q_{B\rightarrow C}=\Delta U_{B\rightarrow C}+\Delta W_{B\rightarrow C}$

$=\Delta U_{B\rightarrow C}+P\Delta V_{B\rightarrow C}$

$100=\Delta U_{B\rightarrow C}+6 \times 10^{4} \left(4 \times 10^{-3}-2 \times 10^{-3} \right )$

$=\Delta U_{B\rightarrow C}+6 \times 10^{4} \times 2 \times 10^{-3}$

$\Delta U_{B\rightarrow C}=100J-120J=-20J$

$\Delta U_{A\rightarrow C}=\Delta U_{A\rightarrow B}+\Delta U_{B\rightarrow C}=\Delta Q_{A\rightarrow C}-\Delta W_{A\rightarrow C}$

$400+(-20)=\Delta Q_{A\rightarrow C}- \left(P\Delta V_{A} +area \Delta ABC\right )$

$380=\Delta Q_{A\rightarrow C}- \left[2 \times 10^{4}(4\times 10^{-3}-2 \times 10^{-3})+\frac{1}{2}\times (4\times 10^{-3}-2\times 10^{-3})\times (6\times 10^{4}-2 \times 10^{4}) \right ]$

$=\Delta Q_{A\rightarrow C}- \left(2 \times 10^{4}-2 \times 10^{-3}+\frac{1}{2}\times 2\times 10^{-3}\times 4 \times 10^{4} \right )$

$\Delta Q_{A\rightarrow C}=380 + 40 + 40 = 460 J$

Correct option is 1.

Option 1)

460 J

Correct

Option 2)

300 J

Incorrect

Option 3)

380 J

Incorrect

Option 4)

500 J

Incorrect

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