The P−V diagram of a system undergoing thermodynamic transformation is shown in figure. The work done by the system and heat given to the system in going from A→B→C is 30J and 40J respectively. The change in internal energy between A and C is
A
70J
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B
10J
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C
−70J
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D
−10J
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Solution
The correct option is B10J Given,
WAB+WBC=30J QAB+QBC=40J
Applying 1st law of thermodynamics for process AB, QAB=ΔUAB+WAB(1)
Simillarly applying 1st law of thermodynamics for process BC, QBC=ΔUBC+WBC(2)
Adding (1) and (2), QAB+QBC=ΔUAB+WAB+ΔUBC+WBC 40=(ΔUAB+ΔUBC)+30
Here ΔUAB+ΔUBC=ΔUAC ⇒ΔUAC=40−30=10J