2. consider a process where propane gas, which is initially at 1 bar and 35 ºc, is compressed to a final state of 135 bar and 195 ºc. estimate the molar volume of the gas in its final state. estimate also the entropy and enthalpy changes for this process. you can assume that propane behaves as an ideal gas in the initial state.

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.