Answer:
This is an Ideal Gas Process. The Ideal gas equation states that:
PxV=nxRxT
where n = number of moles of the gas
P = pressure of the gas at any state
V= Volume of the gas at any state
T=Temperature of the gas at any state
Initial State : P₁ = 1Bar=1.00x10⁵N/m², T₁=35°C = 308K, V₁=?
Final State: P₂=135Bar=1.35x10⁷N/m², T₂=195°C=468K, V₂=?
From Eqn 1, we can calculate the molar volume of the Gas at the final state Q=n/V₂ =P₂/RT₂ ....... . . . . . . . . . Eqn 1
Q=1.35x10⁷N/m²/(8.314J/mol-K x 468K)
Q=33,828mol/m³ = 33.8Kmol/m³
The molar volume of the Gas at the final state Q =33.8Kmol/m³
To estimate the Enthalpy and the Entropy change for an ideal gas process requires these equations which are derived from the law of conservation of energy (1st and 2nd law of thermodynamics):
ΔH = CpΔT
ΔS = CpInΔT - RInΔP/M
ΔT=T2-T1 and ΔP=P2-P1
Cp=Specific heat capacity of an ideal propane gas = 1.68kJ/KgK
R = Universal gas constant = 8.314J/molK
For Enthalpy change, ΔH = 1.68 kJKgK x (468 - 308)K
ΔH = 268.8KJ/kg
For Entropy change, ΔS = (CpInΔT) - (RInΔP)/M M=Propane molar mass=44.1kg/mol
ΔS=(1.68xln(468/308)) - (8.314xln(135/1)/44.1)
ΔS =0.703KJ/KgK - 0.925KJ/kgK
ΔS = - 0.222KJ/KgK
Therefore, the entropy and enthalpy changes for this process are - 0.222KJ/KgK and 268.8KJ/kg.
