Chemistry - Thermodynamics - II Question with Solution | TestHub
ChemistryThermodynamics - IICarnot Cycle/Cyclic processEasy2 minQB
ChemistryEasysingle choice
Passage / Comprehension
One mole of a monoatomic ideal gas is used as a working substance in an engine working in the following reversible cycle: process AB = isobaric compression; process BC = adiabatic compression; process CD = isochoric heating; process DA = isothermal expansion. The ratio of maximum to minimum volume and temperature during the cycle is 6√2 and 3, respectively. The maximum temperature in the cycle is 900 K.
The temperature of gas at state 'C' is
Options:
Answer:
B
Solution:
TA = TD = Tmax = 900 K (isothermal expansion DA at the top temperature); TB = Tmin = 900/3 = 300 K. Isobaric AB: VA/VB = TA/TB = 3. VA is maximum, VC = VD is minimum: VA/VC = 6√2, so VB/VC = 6√2/3 = 2√2. Adiabatic BC (γ = 5/3): TC = 300 × (2√2)^(2/3) = 300 × 2 = 600 K. Matches (b).
Stream:JEESubject:ChemistryTopic:Thermodynamics - IISubtopic:Carnot Cycle/Cyclic process
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