Question
Physics Question on Thermodynamics
A sample of an ideal gas is taken through the cyclic process ABCA as shown in figure. It absorbs, 40 J of heat during the part AB, no heat during BC and rejects 60 J of heat during CA. A work of 50 J is done on the gas during the part BC. The internal energy of the gas at A is 1560 J. The work done by the gas during the part CA is:
Fig.
A
20 J
B
30 J
C
–30 J
D
–60 J
Answer
30 J
Explanation
Solution
The correct answer is (B) : 30 J
ΔUAB = 40 J as process is isochoric.
ΔUBC + WBC = 0
ΔUBC = +50 (WBC = –50 J)
UC = UA + ΔUAB + ΔUBC = 1650
For CA process,
QCA = – 60 J
ΔUCA + WCA = –60
–90 + WCA = –60
⇒ WCA = +30 J
The graph given is inconsistent with the statement BC may be adiabatic and CA cannot be like isobaric as shown, as increasing volume while rejecting heat at same time.