an ideal gas undergoes a circular cycle as shown in figure find the ratio of minimum and maximum temperatureof cycle
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The ratio of minimum and maximum temperature of cycle is
TA/ TB = (4+√2) / (4 - √2).
Explanation:
The maximum temperature will occur at point A and minimum temperature will accurate point B of the cycle, so
At point A
PA / PB = VA / VB = 2 + cos(45∘) = 22 – √2 + 1 / √2 = 4 + √2 / 2
nRTA = PAVA = (4+√2 / 2 ) PoVo
Similarly at point B.
PB / Po = VB / Vo = 2 - cos (45°)
nRTB = PBVB = (4 - √2 / 2) PoVo
TA/ TB = (4+√2) / (4 - √2)
Thus the ratio of minimum and maximum temperature of cycle is
TA/ TB = (4+√2) / (4 - √2).
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