NEET Physics Thermodynamics Questions Solved


Six moles of an ideal gas performs a cycle shown in figure. If the temperature are TA = 600 K, TB = 800 K, TC = 2200 K and TD = 1200 K, the work done per cycle is -

(1) 20 kJ

(2) 30 kJ

(3) 40 kJ

(4) 60 kJ

(3) Processes A to B and C to D are parts of straight line graphs of the form y = mx

Also P=μRVT(μ = 6)

PT. So volume remains constant for the graphs AB and CD

So no work is done during processes for A to B and C to D i.e., WAB = WCD = 0 and WBC = P2(VCVB) = μR (TCTB)

= 6R (2200 – 800) = 6R × 1400 J

Also WDA = P1 (VAVD) = μR(TATB)

= 6R (600 – 1200)= – 6R × 600 J

Hence work done in complete cycle

W = WAB + WBC + WCD + WDA

= 0 + 6R × 1400 + 0 – 6R × 600

= 6R × 800 = 6 × 8.3 × 800 ≈ 40 kJ

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