e heat, work,AU and when a sample of 1.5 moles of liquid water at 1 bar is 100.0

Question

e heat, work,AU and when a sample of 1.5 moles of liquid water at 1 bar is 100.00°C degrees to a final temperature of 327.00 °C: H20(0, 100.00°C) . 327.00 °C). The molar volume of liquid water at 100.00 °C is 1.89 x 10 m2 mol the molar volume of steam at 327.00°C is 5.06 x 102 m mol. The molar enthalpy of rization of water at 1 bar is 40656 J mol". The molar heat capacity of the steam can be heated tronm assumed to be constant during this temperature interval, with a value of 33.58 J mol K- 8 pts)

Explanation / Answer

The no. of moles of the sample (n) = 1.5 mol

Pressure (P) = 1 bar

Initial temperature (T1) = 100 oC

Initial volume (V1) = 1.89*10-5 m3/mol

Final temperature (T2) = 327 oC

Final voluem (V2) = 5.06*10-2 m3/mol

DeltaHvap = 40656 J/mol

Cp = 33.58 J mol-1 K-1

Calculation of heat (q):

Cp = q/n(T2-T1)

i.e. 33.58 J mol-1 K-1 = q/1.5 mol * (327-100) K

Therefore, the heat (q) = 11.434 KJ

Cv of water = 75.29 J mol-1 K-1

Calculation of DeltaU:

DeltaU = n*Cv*DeltaT

= 1.5 mol * (33.58-8.314) J mol-1 K-1 * (327-100) K

= 1.5*25.266*227 J

Therefore, DeltaU = 8.603 KJ

Calculation of work (w):

According to the first law of thermodynamics:

q = DetalU + (-w)

i.e. 11.434 KJ = 8.603 KJ + (-w)

i.e. -w = 11.434 - 8.603 = 2.831 KJ

Therefore, the work done by the system (-w) = 2.831 KJ

The workdone on the system (w) = -2.831 KJ

Related Questions

Navigate

• Browse (All)
• Browse E
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.