7) The heat capacity of a gaseous mixture of hydrocarbons is required for some p

Question

7) The heat capacity of a gaseous mixture of hydrocarbons is required for some process calculations. For this purpose, a sample of the gas is put in a glass flask and is initially at 25.0°C and 910.0 mmHg absolute pressure. A stopcock is then opened momentarily to allow the pressure inside the flask to equalize with that of the atmosphere. Then the stopcock is shut. The flask is next warmed, and when the gas is again at 25.0°C, its pressure is determined to be 780.0 mmHg. Atmospheric pressure is 760.0 mmHg. What is the cp of the gas mixture in J/mol K? It may be assumed that the gas mixture is ideal and that the cp is constant.

Explanation / Answer

Gas sample follow idea gas law then :- PV = RT

Intial conditions:

T = 25 degC = 298 K, P = 910 mmHg = 121323 Pa,

121323*V = 8.314*298

V = 0.204 m3/mol

after opening stopcock momentarily pressure becomes 760 mmHg inside the flask and volume remains constant,

volume is constant:

P/T = constant, P1/T1 = P2/T2

121323*T2 = 101325*298

T2 = 248.88 K, P2 = 101325

On heating Final conditions:

T3 = 25 degC, P3 = 780 mmHg = 10665.8 Pa,

Hydrocarbon mixture is polyatomic then specific heat ratio k = 4/3

gas misture is ideal so Cp will be:

Cp = 4R

Cp = 4*8.314 = 33.256 J/mol.K

Cp = 33.256 J/mol.K

Related Questions

Navigate

• Browse (All)
• Browse 7
• Subjects

---

---

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