Container with two gases and equation of state

[gerardpc]

Homework Statement

We have a cylindrical container filled with a gas: the container has an upper compartment with 2 moles of this gas, and another compartment below with 1 mole of the same gas, separated by a diathermic boundary, and also has an adiabatic mobile plunger over the upper compartment.
The lower compartment has a kind of grinder (I don't know how to translate it, sorry) which can do work on it. A picture of the container:



The initial pressure in both compartments and in the outside is $p_0$, and the temperature is $T_0$. The heat capacity at constant pressure is $C_p = 3R$.
Now imagine the "grinder" does 500J of work at the lower compartment. What will be the final $(p,V,T)$ state of the two gases?

Homework Equations

The equation of state of the gas is
$p(V-b) = nRT$

Also, the internal energy (as deduced in a previous part of the problem) is

$U = 2nRT$

The Attempt at a Solution

It is clear that the upper compartment remains at constant pressure and that the temperature at both parts of the tank will be the same along the expansion of the upper gas.
The work done by the grinder is in form of heat, I think. But i don't know how to proceed. Any clues will be appreciated.

Thanks!

 

[Chestermiller]

Apply the first law to the two chambers individually. Let Q be the heat transferred from the lower chamber to the upper chamber. The stirrer (grinder) does 500 J of work on the gas in the lower chamber, and the process in the lower chamber takes place at constant volume. The process in the upper chamber takes place at constant pressure, so here Q = ΔH = ΔU + pΔV for that chamber. The final temperatures are the same, but the final pressures are not.

 

[gerardpc]

Ok. What about this:

Lower chamber:

$\Delta U_1 = Q_{stirrer} - Q_{up} + p\Delta V$

So using the expression for the internal energy and the constant volume:

$2R\Delta T = 500 - Q_{up}$

Upper chamber:

$\Delta U_2 = Q_{up} - p\Delta V$

So:

$4R\Delta T = Q_{up} - p_0\Delta V$

Adding up these two expressions we get:

$6R\Delta T = 500 - p_0\Delta V$

And now, from the equation of state:

$\dfrac{nRT}{p} + b = V \Rightarrow \Delta V = \dfrac{nR}{p}\Delta T$, so in the upper chamber $\Delta V = \dfrac{2R}{p_0}\Delta T$

Replacing it in the last equation we get:

$6R\Delta T = 500 - p_0\Delta V = 500 - p_0 \dfrac{2R}{p_0}\Delta T = 500 - 2R \Delta T$

And finally $8R\Delta T = 500$, from where we can derive the final state of the two chambers.

Is it correct?

 

[Chestermiller]

Yes. I think this is correct. In the equation for V, I think there should have been an nb rather than a b, but this canceled out when you wrote the equation for ΔV. You seem to have a very good feel for how to do this type of modelling. Nice work.

Chet

 

[Nickie]

I would approach this problem by first considering the initial state of the system and then using the equation of state to determine the final state after the work is done by the grinder.

In the initial state, both compartments have the same pressure and temperature, which means they also have the same volume since they contain the same number of moles of gas. Using the ideal gas law, we can determine the initial volume of each compartment:

p_0V_0 = nRT_0

V_0 = nRT_0/p_0 = (2+1)RT_0/p_0 = 3RT_0/p_0

Now, when the grinder does 500J of work on the lower compartment, it will increase the volume of the lower compartment and decrease the volume of the upper compartment. However, since the boundary between the two compartments is diathermic, there will be heat transfer between the two compartments to maintain the same temperature. This means that the final state of the upper compartment will also be at a pressure of p_0 and a temperature of T_0.

To determine the final volume of the lower compartment, we can use the work-energy principle, which states that the work done on a system is equal to the change in the system's energy. In this case, the work done by the grinder is equal to the change in the internal energy of the lower compartment:

W = U_f - U_i

500J = (2nRT_f - 2nRT_0) - (nRT_f - nRT_0)

500J = nRT_f - nRT_0

T_f = (500J + nRT_0)/nR

Now, we can use the equation of state to determine the final volume of the lower compartment:

p_0V_f = nRT_f

V_f = nRT_f/p_0 = 3RT_f/p_0 = 3RT_0/p_0 + 500J/p_0

Therefore, the final state of the two gases will be:

Upper compartment: (p_0, V_0, T_0)

Lower compartment: (p_0, V_0 - 500J/p_0, (500J + nRT_0)/nR)

I hope this helps!