NEET Physics Thermodynamics Questions Solved


An ideal gas is taken through the cycle ABCA, as shown in the figure. If the net heat supplied to the gas in the cycle is 5 J, the work done by the gas in the process CA is 

(1) – 5 J

(2) – 10 J

(3) – 15 J

(4) – 20 J

(1) For cyclic process. Total work done =WAB+WBC+WCA

ΔWAB = PΔV = 10(2 – 1) = 10J and ΔWBC =0

(as V = constant)

ΔQ = ΔU + ΔW

ΔU = 0 (Process ABCA is cyclic)

⇒ ΔQ = ΔWAB + ΔWBC + ΔWCA

⇒ 5 = 10 + 0 + ΔWCA ⇒ ΔWCA = – 5 J

Difficulty Level:

  • 50%
  • 30%
  • 12%
  • 9%
Crack NEET with Online Course - Free Trial (Offer Valid Till August 25, 2019)