One mole of an ideal gas is compressed isothermally but irreversibly at 130oc from 2.5 bar to 6.5bar in a piston/cylinder device. the work required is 30% greater than the work of reversible, isothermal compression. the heat transferred from the gas during compression flows to a heat reservoir at 25oc. claculate the entropy changes of the gas, the heat reservoir, and δstotal.

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