Heat capacity ratio yields inconsistent results

In summary: The problem statement is ambiguous and does not give a clear indication as to what is being asked. Therefore, the short approach does not work.
  • #1
uter
10
0

Homework Statement


A Gas is in a Volume V0 = 1 Liter at Pressure p0 = 3 bar.
  1. Isochoric Heating using the Heat Q1 = 182 J, the pressure raises to p1 = 6.34 bar.
  2. Gas is reset to inital state. Isobaric Heating using the Heat Q2 = 546 J, the Volume increases to V2 = 3 Liter.
Calculate Cp/CV. Calculate the degrees of freedom f.

Homework Equations



pV = nRT
CV = ΔQ/ΔT (for first step)
Cp = ΔQ/ΔT (for second step)
Cp/CV = 1+ 2/f
and optionally
U = ΔQ + ΔW = f/2 n R T

The Attempt at a Solution


The task as described above is pretty straightforward and yields the expected f = 3 solution.

However, I thought about another way of solving this exercise, but it keeps on yielding different results:
If I take the last equation (Joules Law: U = f/2 n R T) together with the ideal Gas Equation for the isochoric process, I get f = 1.09. (Again straightforward inserting and solving for f.)

Those results don't match. Neither do they, if I take the isobaric process (taking into account, that work is done). Just by taking the quotient of those two, I get the expected result. Is this a flaw in the exercise or is there any mistake within my formulas?
 
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  • #2
the last equation provided by you has an inherent sign convention which is where you are going wrong . hence check the feasability of the 2nd statement of the problem
 
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  • #3
Well in the case of an isochoric heating the sign convention wouldn't affect the result, would it? I mean ΔW = 0 anyways.
 
  • #4
Show us the details of your work for the second method, which is the method I envisioned using. This method seems very straightforward.

Chet
 
  • #5
In the case of isochoric heating we have ΔW = 0 which gives us from Joule's law and the first law of TD: ΔQ = f/2 n R ΔT.
Furthermore we know from the ideal gas equation, that Δp V = n R ΔT, which we can solve for ΔT = Δp V / (R n).
We insert the latter equation into the first one and get ΔQ = f/2 Δp V, which we can solve for f = 2 ΔQ / (Δp V).
However, inserting the above values (ΔQ = 182 J, Δp = 3.34e5 Pa, V = 1e-3 Liter) results in f = 1.09.

In the case of isobaric heating, Joule's law and the first law of TD give us: ΔQ = f/2 n R ΔT + p ΔV.
Again, using the ideal gas equation with pΔV = n R ΔT gives inserted in the previous equation for ΔT:
ΔQ = (f/2 + 1) pΔV.
Solving for f and inserting the according values results in f = -0.04 (or something around that, calculated it several times yesterday).

All this left me pretty confused…
 
  • #6
uter said:
In the case of isobaric heating, Joule's law and the first law of TD give us: ΔQ = f/2 n R ΔT + p ΔV.

this is where the sign convention matters .

and 1.09 looks like the correct answer
 
  • #7
Alright, but why does it differ from the value I get using Cp/CV ?
 
  • #8
I approached this a little differently:

Let the subscript 1 refer to the case of constant volume and the subscript 2 refer to the case of constant pressure. So, for constant volume:

##nC_v(T_1-T_0)=Q_1##

##\frac{p_0v_0}{RT_0}C_v(T_1-T_0)=Q_1##

From the ideal gas law,

##T_1=\frac{p_1}{p_0}T_0##

Combining these equations, I get:

##\frac{C_v}{R}=\frac{Q_1}{v_0(p_1-p_0)}##

By a similar procedure for the constant pressure case, I get:

##\frac{C_p}{R}=\frac{Q_2}{p_0(v_2-v_0)}##

So, from these results:

##\frac{C_p}{C_v}=\frac{Q_2}{Q_1}\frac{(\frac{p_1}{p_0}-1)}{(\frac{v_2}{v_0}-1)}##

Chet
 
  • #9
Hi Chet,

thanks for your detailed explanation. That gives us exactly the expected result of 5/3 -> f = 3. But what's wrong with my "short" approach?

Thanks!

P.S.: I somehow suspect that this exercise has been created in a "reverse" way which is ambiguous… I mean going back from the Cp/Cv equation gives me one condition for all variables, whilst going back from my short approach gives two conditions…

P.P.S.: I checked this in Maple: If I fix all state variables (pi and Vi), Q1 = 501 J and Q2 must be 1500 J. So the ratio between Q1 and Q2 was perfectly fine, but my approach fixes them to absolute values…
Is this correct?
 
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  • #10
Well, if I use the equations separately to evaluate the situation, I get:

##\frac{C_v}{R}=\frac{182}{(0.001)(3.34\times 10^5)}=0.54491##

##\frac{C_p}{R}=\frac{546}{(0.002)(3.\times 10^5)}=0.91##

Neither of these is consistent with ##C_v = 1.5 R## and ##C_p = 2.5 R##. So, there is something wrong with the problem statement. If ##Q_1## were 500 J, and ##Q_2 = 1500 J##, then everything would be consistent.

Chet
 
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Related to Heat capacity ratio yields inconsistent results

1. What is heat capacity ratio and why is it important in scientific research?

Heat capacity ratio is the ratio of the amount of heat required to raise the temperature of a substance by a certain amount to the amount of heat required to raise the temperature of the same substance by the same amount at constant pressure. It is important in scientific research because it helps to understand the thermal behavior of a substance and its ability to store and release heat.

2. Why do different substances have different heat capacity ratios?

Different substances have different heat capacity ratios because the ratio depends on the molecular structure and composition of the substance. For example, substances with more complex molecular structures tend to have higher heat capacity ratios because they have more ways for the molecules to store energy.

3. How is heat capacity ratio measured in experiments?

Heat capacity ratio can be measured in experiments by measuring the change in temperature of a substance when a known amount of heat is added or removed. This is typically done using a calorimeter, which can accurately measure the heat transfer in a controlled environment.

4. What are some common sources of error in measuring heat capacity ratio?

Some common sources of error in measuring heat capacity ratio include heat loss to the surroundings, incomplete mixing of the substance, and variations in the initial temperature of the substance. It is important to control these variables as much as possible to obtain accurate results.

5. How can inconsistent results in heat capacity ratio be explained?

Inconsistent results in heat capacity ratio can be explained by a number of factors, such as experimental error, incomplete understanding of the substance's molecular structure, or changes in the substance's state (solid, liquid, gas). It is important to carefully analyze the experimental procedures and results to identify and address any potential sources of error.

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