A sample of an ideal gas is taken through the cyclic process abca as shown in the figure. The change in the internal energy of the gas along the path ca is -180 J. The gas absorbs 250 J of heat along the path ab and 60 J along the path bc. The work done by the gas along the path abc is:
A sample of an ideal gas is taken through the cyclic process abca as shown in the figure. The change in the internal energy of the gas along the path ca is -180 J. The gas absorbs 250 J of heat along the path ab and 60 J along the path bc. The work done by the gas along the path abc is: Correct Answer 130 J
Concept:
From question, we need to find work done along abc which is:
Wac = Wab + Wbc
Now, the internal energy, work done, and heat absorbed or released is given by the relationship:
⇒ ΔQ = ΔU + ΔW
Where,
ΔQ = Change in heat transferred in the system
ΔU = Change in internal energy of the system
ΔW = Change in work done of the system
Calculation:
Now,
⇒ ΔW = ΔQ – ΔU
Now, the work done for path ab is:
⇒ ΔWab = ΔQab – ΔUab
Now, the work done for path bc is:
⇒ ΔWbc = ΔQbc – ΔUbc
Now, the work done for path ac is:
⇒ Wac = (ΔQab – ΔUab) + (ΔQbc – ΔUbc)
⇒ Wac = ΔQab – ΔUab + ΔQbc – ΔUbc
⇒ Wac = ΔQab + ΔQbc – (ΔUab + ΔUbc)
∵ ΔUab + ΔUbc = ∆Uac
From question, ΔUca = -180 J
⇒ ΔUac = -(ΔUca)
⇒ ΔUac = -(-180)
∴ ΔUac = 180 J
⇒ Wac = ΔQab + ΔQbc – ∆Uac
Where,
ΔQab = 250 J (Given)
ΔQbc = 60 J (Given)
Now, substituting the values,
⇒ Wac = 250 + 60 – 180
∴ Wac = 130 J