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
- 20 kJ
- 30 kJ
- 40 kJ
- 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 
(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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