Carnot Engine Between Reservoirs with Varying Temperatures

[doombanana]

Homework Statement

Two containers with fixed volume are each filled with n moles of a monatomic ideal gas with constant heat capacity. Initially, the volumes of gas are at temperatures $T_1,i$and $T_2,i$. What is the maximum amount of work that can be obtained by operating an engine between the two containers?

Homework Equations

Let $T_1 > T_2$

(1) $\eta = \frac{|dW|}{|dQ|} = 1 - \frac{T_2}{T_1}$

(2) $|dQ| = -c_vdT_1$ because |dT₁| is negative so |dQ| = -dQ

(3) $c_v = \frac{3}{2}nR$

The Attempt at a Solution

Substituting (2) into (1) and solving for |dw|gives
$|dW| = -c_v \Big(1 - \frac{T_2}{T_1}\Big)dT_1$

I need to integrate this somehow, but both T₁ and T₂ are changing so I'm not sure how to do this. I'm assuming there's a relationship between T₁ and T₂ that I can use that will allow me to integrate? Any help pointing me in the right direction would be greatly appreciated.

 

[vela]

Try coming up with a relationship between T₁ and T₂. You should be able to calculate how much heat of the dQ taken from the hot reservoir is expelled into the cold reservoir, which you can then use to get the change in T₂.

 

[TSny]

Hello doombanana. Welcome to PF!

Try setting up an equation for the the cooler gas similar to your equation (2). Also, I think you'll need to use the 1st law of thermodynamics to relate dW, dQ₁, and dQ₂.

 

[doombanana]

Try coming up with a relationship between T₁ and T_{2[/sup]. You should be able to calculate how much heat of the dQ taken from the hot reservoir is expelled into the cold reservoir, which you can then use to get the change in T2.}

I guess finding this relationship is where I'm having trouble.

Using Tsny's suggestion, from the first law I found that
$1- \frac{|dQ_2|}{|dQ_1|}= 1-\frac{c_vdT_2}{-c_vdT_1}=1-\frac{T_2}{T_1}$
solving the last two parts of the equation for T₂ gives
$T_2=\frac{T_1c_vdT_2}{-c_vdT_1}$

substituting that into my eq for |dW| gives
$|dW|= -c_v( dT_1 - dT_2)$

integrating gives
$|W| = c_v(T_{1,i}-T_{2,i})$I know this isn't right because |W| ends up being larger than |Q_1|, but I'm not sure where I went wrong, or if this is even remotely close to the correct way to go about this. (I have a feeling it's not.)

EDIT: when integrating I assumed that the final temperature in both reservoirs was equal to (T_1,i + T_2,i)/2 which now I'm thinking isn't the case. If so, I'm even more lost than before.:uhh:

Hello doombanana. Welcome to PF!

Try setting up an equation for the the cooler gas similar to your equation (2). Also, I think you'll need to use the 1st law of thermodynamics to relate dW, dQ₁, and dQ₂.

Thank you! I don't know how I've made it this far without creating an account here.

 

[TSny]

I guess finding this relationship is where I'm having trouble.

Using Tsny's suggestion, from the first law I found that
$1- \frac{|dQ_2|}{|dQ_1|}= 1-\frac{c_vdT_2}{-c_vdT_1}=1-\frac{T_2}{T_1}$
solving the last two parts of the equation for T₂ gives
$T_2=\frac{T_1c_vdT_2}{-c_vdT_1}$

OK!

substituting that into my eq for |dW| gives
$|dW|= -c_v( dT_1 - dT_2)$

Check for a sign error inside the parentheses.

integrating gives
$|W| = c_v(T_{1,i}-T_{2,i})$

You'll see that with the sign corrected in the previous expression, the integrated expression for W will include the final equilibrium temperature.

EDIT: when integrating I assumed that the final temperature in both reservoirs was equal to (T_1,i + T_2,i)/2 which now, I think, isn't the case. If so, I'm even more lost than before.:uhh:

See if you can find the final temperature by integrating your equation
$T_2=\frac{T_1c_vdT_2}{-c_vdT_1}$

 

[TSny]

Just a note. You can apply the 1st law to the entire process as

W = |Q₁|-|Q₂|

Each Q can be expressed in terms of C_v and the overall temperature changes. This avoids having to first set up an expression for dW and integrating. But the result is the same.

You're still going to need to determine the final equilibrium temperature in terms of the initial temperatures.

 

[doombanana]

Check for a sign error inside the parentheses.

You'll see that with the sign corrected in the previous expression, the integrated expression for W will include the final equilibrium temperature.

Sign errors are the bane of my existence, but at least I see where it came from.

See if you can find the final temperature by integrating your equation
$T_2=\frac{T_1c_vdT_2}{-c_vdT_1}$

Got it!

Just a note. You can apply the 1st law to the entire process as

W = |Q₁|-|Q₂|

Each Q can be expressed in terms of C_v and the overall temperature changes. This avoids having to first set up an expression for dW and integrating. But the result is the same.

You're still going to need to determine the final equilibrium temperature in terms of the initial temperatures.

Oooh, that is a bit of a shortcut and a more intuitive of setting up the problem.

Thank you very much for your help! You've saved me hours of beating my head against the wall.