A thermodynamic process is shown in the given figure. The pressures and volumes corresponding to some points in the figure are : PA = 3 x 104 Pa, PB = 8 x 104 Pa and VA = 2 x 10-3 m3, VD = 5 x 10-3 m3.
In the process AB, 600 J of heat is added to the system. The change in internal energy of the system in process AC would be :
Answer & explanation
Correct answer: option 1
By the graph :
WAB = 0 and
WBC = 8 x 104 (5-4) x 10-3 = 240 J
WAC = WAB + WBC = 0 + 240 = 240 J
Now, \(\Delta Q_{AC} = \Delta Q_{AB} + \Delta Q_{BC} \)
\(\Delta Q_{AC}\) = 600 + 200 = 800 J
By First Law of Thermodynamics : \(\Delta Q_{AC} = \Delta U_{AC} + \Delta W_{AC}\)
800 = \(\Delta U_{AC} + 240 \)
\(\Rightarrow \Delta U_{AC} = 560 J\)