1 – What is called the BC transformation? 2 – Calculate the number of moles n of this gas? 3 – Calculate: a) The PB pressure and TB temperature of the gas at point B. b) The PC pressure of the gas at point C.
We give:
AB: adiabatic trigger such as VB = 2 VA AC: isothermal compression
PA =1bar=105 Pa , VA = 10^-2 m3
Gas temperature at point A: TA= 300 Kelvin
Perfect gas constant: R = 8,31 J.mol-1.K-1
Molar thermal capacity at constant pressure:Cp = 29 J.mol-1.K-1
Calorific capacity ratio of this gas: Cp/Cv = 1,40
1 – What is called the BC transformation?
2 – Calculate the number of moles n of this gas?
3 – Calculate:
a) The PB pressure and TB temperature of the gas at point B.
b) The PC pressure of the gas at point C.
4 – Calculate the amount of Q(BC) heat and W(BC) work received by the system for the transformation that changes the system from state B to state C.
5 – a – Why is the UCA internal energy variation of the gas zero during the CA transformation? b – Apply the first principle to the ABCA cycle and deduct the WAB work received by the gas during the
adiabatic relaxation AB.
c – Interpret the WAB sign.
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