Change of internal energy during linear transformation

  • #1
TheEyeOfInnos
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Homework Statement
An ideal gas goes through a transformation that in (pV) coordinates is a straight line. In the final state we have ##V_{2}=4V_{1}## and ##p_{2}=p_{1}/3##. We are asked to calculate the ratio between the change in internal energy during heating and the change in internal energy during cooling.
Relevant Equations
##p=a-bV##
##pV = nRT##
##pV^{\gamma}=const.##
##\Delta U = nC_{V}\Delta T##
A linear transformation is described by:
$$p=a-bV$$
From that we can find $a$ and $b$:
$$b=\frac{p_{2}-p_{1}}{V_{1}-V_{2}} = \frac{2}{9}\frac{p_{1}}{V_{1}}$$
$$a=p_{1}+bV_{1}=\frac{11}{9}p_{1}$$

I considered an adiabatic process that intersects the linear transformation to find the point up until the system received heat:
$$pV^{\gamma}=\text{const.}$$
$$\left(a-bV\right)V^{\gamma}=\text{const.}$$
Differentiating wrt to volume:
$$V_{P} = \frac{\gamma}{\gamma+1}\frac{a}{b}=\frac{\gamma}{\gamma+1}\frac{11}{2}V_{1}$$
now we can obtain:
$$p_{P} = a-bV_{P}=\frac{22}{18}\frac{\gamma}{\gamma+1}p_{1}$$

In order to find the temperature ##T_{P}## I used ##p_{1}V_{1}=nRT_{1}## and ##p_{P}V_{P}=nRT_{P}##
$$T_{P} = T_{1}\cdot p_{P}\cdot V_{P} = \frac{121}{18}\frac{\gamma}{\gamma+1}T_{1}$$
In order to find the temperature ##T_{2}## I used ##p_{1}V_{1}=nRT_{1}## and ##p_{2}V_{2}=nRT_{2}##
$$T_{2}=T_{1}\cdot p_{2}\cdot V_{2}=\frac{4}{3}T_{1}$$

Now:
$$\frac{\Delta U_{heating}}{\Delta U_{cooling}} = \frac{T_{P}-T_{1}}{T_{2}-T_{P}}$$
But I cannot get rig of ##\gamma## and the problem does not specify which type of gas we are dealing with. The solution apparently is -1.96.
 
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  • #2
TheEyeOfInnos said:
Homework Statement: An ideal gas goes through a transformation that in (pV) coordinates is a straight line. In the final state we have ##V_{2}=4V_{1}## and ##p_{2}=p_{1}/3##. We are asked to calculate the ratio between the change in internal energy during heating and the change in internal energy during cooling.
Relevant Equations: ##p=a-bV##
##pV = nRT##
##pV^{\gamma}=const.##
##\Delta U = nC_{V}\Delta T##

A linear transformation is described by:
$$p=a-bV$$
From that we can find $a$ and $b$:
$$b=\frac{p_{2}-p_{1}}{V_{1}-V_{2}} = \frac{2}{9}\frac{p_{1}}{V_{1}}$$
$$a=p_{1}+bV_{1}=\frac{11}{9}p_{1}$$
So $$p=\frac{11}{9}p_1-\frac{2}{9}\frac{p_1}{V_1}V$$and $$pV=\frac{11}{9}p_1V-\frac{2}{9}\frac{p_1}{V_1}V^2$$So, for an ideal gas, $$nRT=\frac{11}{9}p_1V-\frac{2}{9}\frac{p_1}{V_1}V^2$$
If there is both heating and cooling during the process, T must pass through a maximum or minimum during the change. What is the value of V at this extremum? What is the value of T at this extremum? What are the initial and final values of T?
 
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  • #3
TheEyeOfInnos said:
I considered an adiabatic process that intersects the linear transformation to find the point up until the system received heat:
$$pV^{\gamma}=\text{const.}$$
$$\left(a-bV\right)V^{\gamma}=\text{const.}$$
Differentiating wrt to volume:
$$V_{P} = \frac{\gamma}{\gamma+1}\frac{a}{b}=\frac{\gamma}{\gamma+1}\frac{11}{2}V_{1}$$
now we can obtain:
$$p_{P} = a-bV_{P}=\frac{22}{18}\frac{\gamma}{\gamma+1}p_{1}$$
Your method looks correct for your interpretation of "heating" as positive heat being added to the gas and "cooling" as heat leaving the gas.

However, the answer -1.96 corresponds to @Chestermiller's interpretation of "heating" as temperature increasing and "cooling" as temperature decreasing. In this case, you don't need to know ##\gamma##.
 
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  • #4
TSny said:
Your method looks correct for your interpretation of "heating" as positive heat being added to the gas and "cooling" as heat leaving the gas.

However, the answer -1.96 corresponds to @Chestermiller's interpretation of "heating" as temperature increasing and "cooling" as temperature decreasing. In this case, you don't need to know ##\gamma##.
The linear P~V behavior is inconsistent with an adianatic change.
 
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  • #5
Chestermiller said:
The linear P~V behavior is inconsistent with an adianatic change.
I think the OP's approach is to find the point of the linear process that is tangent to an adiabat. So, at this point, an infinitesimal displacement along the linear process does not involve any heat transfer to or from the gas. So this is the point during the linear process where you switch between heat being added to the gas and heat being removed from the gas.

EDIT: However, the OP referred to the adiabat as intersecting the linear process rather than being tangent to the linear process. But I think his calculation amounts to finding the point where the linear process is tangent to an adiabat.
 
  • #6
Thanks @Chestermiller! I am able to get the correct answer using this method. However, I am not exactly sure why the reasoning of finding the point up until the system receives heat does not work.
@TSny you're right, I meant the adiabatic process being tangent to the linear process.
Another way to get to that volume is by using the first law of thermodynamics:

$$dU=\delta Q - \delta W$$
$$\delta Q = dU + \delta W$$
$$\delta Q = nC_{V}dT + pdV$$
$$\delta Q = \frac{nR}{\gamma-1}dT + pdV$$
where ##T=\frac{pV}{nR}=\frac{aV-bV^2}{nR}\Rightarrow dT = \frac{a-2bV}{nR}##
$$\delta Q = \left[\frac{a-2bV}{\gamma-1} + \left(a-bV\right)\right]dV$$
$$\delta Q = \left[a\left(\frac{1}{\gamma-1}+1\right) - b\left(\frac{2}{\gamma-1}+1\right)V\right]dV$$
$$\delta Q = \left( a\frac{\gamma}{\gamma-1} - b\frac{\gamma+1}{\gamma-1} V\right)dV$$
So in order for the system to receive heat on any portion on the transformation it should satisfy ##\delta Q \geq 0##. Since ##dV>0##, we can get the volume up until the system receives heat:
$$V_{P}=\frac{a}{b}\frac{\gamma}{\gamma+1}$$

As long as ##V<V_{P}## we see that ##\delta Q>0##. As ##V## tends to ##V_{P}## the ##\delta Q## tends to zero and reaches a maximum, then ##\delta Q<0##.

By comparison with the volume I got using @Chestermiller suggestion (which is ##V_{P}=\frac{a}{2b}##) I see that in the problem ##\gamma## should be ##1##. So is this approach wrong? I did not bring any adiabatic process into disscusion this time. The adiabatic exponent only appears because of Robert-Mayer relation.
 
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  • #7
Please plot a graph showing the comparison between the two choices for P vs V. I would like to see how the two representations compare. In my judgment, the adiabatic P-V diagram will be very different from the linear P-V diagram. In addition, it seems very clear to me that the P-V diagram in the problem statement should describe the state of the gas at each and every point along the process.
 
  • #8
In the diagram below, the blue line represents the linear process ##P = a - bV## and the red lines are various adiabatic curves ##PV^\gamma = C ## for ##\gamma = 5/3##.

1709224866557.png


The green dot is the point of maximum temperature for the linear process and the red dot is the point that is tangent to an adiabat. Heat is being added to the gas for points on the linear process that are to the left of the red dot. Heat is being removed for points to the right of the red dot.
 
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  • #9
Here's a graph showing isotherms instead of adiabats.

1709225744226.png
 
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  • #10
TheEyeOfInnos said:
So is this approach wrong?
I think your approach and @Chestermiller's approach are both correct, but they represent different interpretations of the word "heating" in the problem statement. For you, heating is interpreted as heat being added to the gas. For Chestermiller, heating is interpreted as increasing the temperature of the gas. The answer of -1.96 for the energy-change ratio corresponds to Chestermiller's interpretation.

As you go from the green dot to the red dot on the blue line in the previous graph, heat is being added to the gas while the temperature is dropping.
 
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  • #11
But the same adiabat doesn’t pass through the initial and final states, let alone not being linear wrt P vs V
 
  • #12
Chestermiller said:
But the same adiabat doesn’t pass through the initial and final states, let alone not being linear wrt P vs V
That's right.

But for an infinitesimal step of the linear process centered at the red dot of the graph, there is no difference between moving along the linear process or moving along the adiabat that is tangent to the blue process at the red dot. So, as you pass through the red dot along the blue process, heat is neither being added to the gas nor removed from the gas. So, the red dot separates the "heating" portion of the linear process from the "cooling" portion of the linear process if "heating" is interpreted as "adding heat".
 
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