Calculate q, w, dU and dH for step b and c A system consists of 2.320 g of carbo

Question

Calculate q, w, dU and dH for step b and c

A system consists of 2.320 g of carbon monoxide gas initially at temperature 400.0 K and pressure 0.6250 bar. Assume the heat capacity of carbon monoxide is constant at the value found for 298.15 K in the data tables of the Appendix (see also Figure 2.8). The system undergoes the following cyclic process: Step b) Constant volume extraction of heat from the system (to a temperature and pressure consistent with step c) Step c) Reversible adiabatic compression to the initial conditions. calculate: q, w, delta U, and delta H.

Explanation / Answer

step b. For dV = 0

w = 0

we know, dU = q - w

or, dU = q

q = nCvdT

Cv = 0.72 J/g.K

n = 2.320 g/28 g/mol = 0.083 mols

dT = 298.15-400 = -101.85 K

feed values,

q = 0.083 x 0.72 x 101.85 = -6.09 J

dU = 6.09 J

dH = nCpdT

Cp = 1.02 J/g.K

feed values,

dH = 0.083 x 1.02 x -101.85 = -8.62 J

step c. In an adiabatic process,

q = 0

w = -dU

dU = nCvdT

     = 0.083 x 0.72 x -101.85 = -6.09 J

w = 6.09 J

dH = -6.09 J

     

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

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