Respuesta :
Answer:
Entropy change of gas = -7.944 J/mol*K
Entropy change of the reservoir = 13.96 J/mol*K
Total entropy change = 6.02 J/mol*K
Explanation:
Step 1: Data given
1 mol of ideal gas is compressed isothermally ( = constant temperature)
Temperature : 130 °C
Pressure = 2.6 bar to 6.5 bar
The work required is 30% greater than the work of reversible
The heat transferred from the gas during compression flows to a heat reservoir at 25°C
Step 2: Calculate entropy change for gas
ΔSgas = Cp*ln(T2/T1) - R*ln (P2/P1)
 ⇒ Since the temperature is constant, T2 = T1 so ln(T2/T1) = 0
ΔSgas = - R*ln (P2/P1)
⇒ with R = gas constant =  8.314
⇒ with P1 = initial pressure = 2.5 barr
⇒ with P2 = final pressure = 6.5 bar
ΔSgas = -8.314 *ln(6.5/2.5)
ΔSgas = -7.944 J/mol*K
Step 3: Calculate reversible work done
Reversible work done = R*T*ln (P2/P1)
Reversible work done = 8.314 * 403.15 * ln (6.5/2.5)
Reversible work done = = 3202. 67 J/mol
Step 4: Actual work done
Actual work done = 1.3 * 3202.67 = 4163.47 J/mol
For isothermal compression:
Q = -W
Q = -4163.47 J/mol
Step 5: Calculate entropy change for reservoir
Entropy change for reservoir ΔSres = -Q/Tres = 4163.47 /298.15
ΔSres = 13.96 J/mol*K
Step 6: Calculate total entropy change
ΔStotal = ΔSgas + ΔSres = -7.944 J/mol*K +13.96 J/mol*K
ΔStotal =  6.02 J/mol*K
The entropy change of the gas is 6.02 J/mol.
How to calculate entropy?
Firstly, the entropy change for gas will be:
= 1 Ă— 8.314 Ă— In(2.5/6.5)
= -7.944 J/K
The reversible work done will be:
= 8.314 Ă— 403.15 Ă— In(6.5/2.5)
= 3202.67 J/mol
The actual work done will be:
= 1.3 Ă— 3202.67
= 4163.47 J/mol
The entropy change for reservoir will be:
= -7.944 + 13.96
= 6.02 J/mol
In conclusion, the entropy change is 6.02 J/mol
Learn more about entropy on:
https://brainly.com/question/13971861
