Thermodynamics - Two gases in a container

[Charlie Anderson]

Homework Statement

Two kilograms of air at 5 bar, 350 K, and 5 kg of carbon monoxide, initially at 4 bar, 450 K, are confined to opposite sides of a rigid, well-insulated container by a partition. The partition is free to move and allows conduction from one gas to the other without energy storage in the partition itself. The air and carbon monoxide each behave as ideal gases with constant specific heat ratio, k = 1.395. Determine the equilibrium temperature.

Homework Equations

Q - W = delta U

Cp - Cv = R

Cp/Cv = k

R_air = 8.314 / Mwt_air

R_co = 8.314 / Mwt_co

Q = mCv(T2-T1)

The Attempt at a Solution

Carbon Monoxide
Cp - Cv = 0.297
Cp / Cv = k

Cv(k) - Cv = 0.297

1.395Cv - Cv = 0.297
0.395Cv = 0.297
Cv = 0.75

Air
Cp - Cv = 0.287
Cp / Cv = k

1.395Cv - Cv = 0.287
Cv = 0.73

This is where i feel i am going wrong.

Q = mCv(T2-T1)
Q = 0 as insulated closed system

Carbon Monoxide Air
0 = 4 x 0.75 (T2 - 450) 0 = 2 x 0.73 (T2 - 350)
0 = 3T2 - 1350 0 = 1.46T2 - 511

I made the two equations equal each other (not sure if that is correct)

1.46T2 - 511 = 3T2 - 1350
839 = 1.54T2
T2 = 544.8 K

I know this is the wrong answer but I am not sure why.

[/B]

 

[Chestermiller]

There are 5 kg of CO, not 4. Also, since the change in internal energy is zero, $${5}(0.75)(T_2-450)+{2}(0.73)(T_2-350)=0$$

 

[Charlie Anderson]

There are 5 kg of CO, not 4. Also, since the change in internal energy is zero, $${5}(0.75)(T_2-450)+{2}(0.73)(T_2-350)=0$$

Ah I see! Thank you for your help.

 

[Pol]

How would you also work out the volume and pressure at equilibrium ?

 

[Chestermiller]

How would you also work out the volume and pressure at equilibrium ?

Does the total volume change? What is the initial volume of each gas? Let V1 represent the final volume of air and V2 represent the final volume of CO. In terms of V1 and V2, what is the final pressure of each gas? At equilibrium, how do these pressures compare?

 

[Pol]

The total volume does change , but I don't know what equation to use to work out the pressure ?

 

[Chestermiller]

The total volume does change , but I don't know what equation to use to work out the pressure ?

Actually, the total volume doesn't change. The only thing that happens is that the partition moves, so one chamber gets larger and the other gets smaller. To get the final pressure, I need you to answer the other questions that I asked.

Chet

 

[Pol]

I'm not really sure , I don't know the initial volume of each gas

 

[Chestermiller]

I'm not really sure , I don't know the initial volume of each gas

Use the ideal gas law. You have the mass (i.e., the number of moles), the temperature, and the pressure of each gas.

 

[Pol]

R the gas constant is 8.214 which I use aswell?

 

[Chestermiller]

R the gas constant is 8.214 which I use aswell?

If you use that (and the value is 8.314), you need to convert bars to Pa. You are aware of having to use consistent units, correct?

 

[fatih]

why did we use cv?, spesific volume of the gases were changing as piston moves.

 

[Chestermiller]

why did we use cv?, spesific volume of the gases were changing as piston moves.

What is your definition of Cv? What is the effect of specific volume on the internal energy per unit mass of an ideal gas?

 

[fatih]

amount of heat needed to raise the temperature of one kilogram of mass by 1 kelvin if the substance is at "Constant Volume". But volume isn't constant

 

[Chestermiller]

amount of heat needed to raise the temperature of one kilogram of mass by 1 kelvin if the substance is at "Constant Volume". But volume isn't constant

In thermodynamics, that's an incorrect definition. In thermodynamics, it is the change in internal energy at constant volume of one kg of material resulting from a change in temperature of 1 kelvin. For an ideal gas, what is the effect of volume on internal energy?