Chemistry - Thermodynamics - II Question with Solution | TestHub
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 8√2 and 4, respectively. The maximum temperature in the cycle is 800 K.
The temperature of gas at state 'C' is
Options:
Answer:
Solution:
Tmax occurs at D and A (isothermal expansion DA at the top temperature): TA = TD = 800 K. Tmin = 800/4 = 200 K = TB (coldest point, end of isobaric compression). Isobaric AB: VA/VB = TA/TB = 4. Also VA is maximum and VC = VD is minimum: VA/VC = 8√2, so VB/VC = 8√2/4 = 2√2. Adiabatic BC (γ = 5/3): TC = TB(VB/VC)^(γ−1) = 200 × (2√2)^(2/3) = 200 × (2^(3/2))^(2/3) = 200 × 2 = 400 K. Matches (c).
