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

 

Answers (1)

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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