Physics-
General
Easy

Question

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

  1. 20 kJ    
  2. 30 kJ    
  3. 40 kJ    
  4. 60 kJ    

The correct answer is: 40 kJ


    Processes A to B and C to D are parts of straight line graphs of the form y = mx
    Also P equals fraction numerator mu R over denominator V end fraction 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

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