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  • A thermodynamic process is shown in Fig. The pressures and volumes corresponding to some points in the figure are,<img alt="\mathrm{P_A=3 \times 10^4 \mathrm{~Pa}, V_A=2 \times 10^{-3} \math

A thermodynamic process is shown in Fig. The pressures and volumes corresponding to some points in
the figure are,\mathrm{P_A=3 \times 10^4 \mathrm{~Pa}, V_A=2 \times 10^{-3} \mathrm{~m}^3, P_B=8 \times 10^4 \mathrm{~Pa}, V_D=5 \times 10^{-3} \mathrm{~m}^3}. In process AB, 600 J of heat and in process BC, 200 J of heat is added to the system. The change in the internal energy in process AC would be

Option: 1

560 J


Option: 2

800 J


Option: 3

600 J


Option: 4

640 J


Answers (1)

best_answer

Process AB is isochoric, i.e. the volume remains constant. Thus ?V = 0. Hence work
done P?V = 0. Process BC is isobaric, i.e. the pressure remains constant and external
work has to be done. The work done = \mathrm{P_B \times\left(V_D-V_A\right)=8 \times 10^4 \times\left(5 \times 10^{-3}-\right.\left.2 \times 10^{-3}\right)=240 \mathrm{~J}}Therefore, change in internal energy is

\mathrm{d U=d Q-d W=800-240=560 \mathrm{~J}}

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