question_answer

Six moles of an ideal gas perfomrs 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 [BCECE 2005]

A)            20 kJ

B)            30 kJ

C)            40 kJ

D)            60 kJ

Correct Answer: C

Solution :

                   Processes A to B and C to D are parts of straight line graphs of the form y = mx Also  \[P=\frac{\mu R}{V}T\]  (m = 6) Þ P µ T. 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(VC ? VB­) = mR (TC  ? TB)               = 6R (2200 ? 800) = 6R ´ 1400 J Also WDA = P1 (VA ? VD) = mR(TA ? TB)                = 6R (600 ? 1200)= ? 6R ´ 600 J Hence work done in complete cycle W = WAB + WBC + WCD + WDA     = 0 + 6R ´ 1400 + 0 ? 6R ´ 600     = 6R ´ 900 = 6 ´ 8.3 ´ 800 » 40 kJ