An ideal gas undergoes a circular cycle as shown in the figure. Find the ratio of maximum temperature of cycle to minimum temperature of cycle:

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1+22−12

2+22−22

3+23−22

4+24−22

answer is D.

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Detailed Solution

The maximum temperature will occur at point A and minimum temperature will occur at point B of the cycle.From symmetry we can say that coordinates of center of circle is 2P0,2V0 and radius is P0 or V0.∴PA=OC+OD=2P0+P0cos450 and PB=OC-OE=2P0-P0cos450So, at point A PAP0=VAV0=2+cos450==2+12=22+12=4+22 ⇒PA=4+22P0 andVA=4+22VnRTA=PAVA=4+22P04+22V0=PA=4+222P0V0---(1)Similarly at point BPBP0=VBV0=2−cos450=2−12=22-12=4−22nRTB=PBVB=4+222P0V0-----(2) (1)/(2)we get⇒TATB=4+24−22

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