Thermodynamics Two kilograms of air and 4 kg of carbon dioxide (CO 2 ) are confi

Question

Thermodynamics

Two kilograms of air and 4 kg of carbon dioxide (CO2) are confined to opposite sides of a rigid, well-insulated container by a partition, as shown in Figure 8. The partition is free to move and allows conduction from one gas to the other without energy storage in the partition itself. The air is initially at 500 kPa and 350 K, while the CO2 is initially at
200 kPa and 450 K. The air and CO2 each behave as ideal gases with constant specific heat ratio k = 1.395 (calculate any necessary constant specific heat values using k). Determine:

a. The final equilibrium temperature, in K,

b. The final equilibrium pressure, in kPa,

Explanation / Answer

The answers are close to the answers given by you. Here is the procedure....

The dU for mixing of ideal gases is 0.

Here, dUco2 + dUair = 0

Now, dU= nCvdT

Each of them have k= 1.395.

k= Cp/Cv

or, Cp - Cv /Cv = k - 1 / 1

or, R / Cv = k - 1

Each of them has same Cv = R/k-1 = 8.314/0.395 = 0.021 kJ/mol

So, dUco2 + dUair = 0

or, 4 * 0.021 * ( T2 - 450)    +   2 * 0.021 * (T2- 350) =0

or, T2 = 416.6 K

Vfinal = Vair + VCO2

= (mRT/P)CO2 + (mRT/P)air

= ( 4 kg * 0.2968 kJ/kg K * 450 K)/ 2 * 10^5 Pa    +   ( 2 * 0.2870   kJ/kg K * 350 K)/ 5 * 10^5 Pa

= 3.08 m3

P f = mRT/Vf = (mRT)CO2 + (mRT)air / Vf ..... [use T2 calculated before]

= [{( 4 kg * 0.2968 kJ/kg K * 416.6 K)   +   ( 2 * 0.2870   kJ/kg K * 416.6 K)} / 3.07 m3 ] * [1000J/kJ][N.m/J][1bar/ 10^5 N/m]

= 2.40 bar

= 240000 Pa

=240 kPa

VCO2

= (mRT/P)CO2 ...... {use final temp]

= 2.1 m3

Vair

= (mRT/P)air .... (use final temp]

= 1 m3

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