• Support PF! Buy your school textbooks, materials and every day products Here!

Heat capacity ratio yields inconsistent results

  • #1
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?
 

Answers and Replies

  • #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
 
Last edited:
  • #3
10
0
Well in the case of an isochoric heating the sign convention wouldn't affect the result, would it? I mean ΔW = 0 anyways.
 
  • #4
19,919
4,095
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
10
0
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
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
10
0
Alright, but why does it differ from the value I get using Cp/CV ?
 
  • #8
19,919
4,095
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
10
0
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?
 
Last edited:
  • #10
19,919
4,095
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
 

Related Threads on Heat capacity ratio yields inconsistent results

  • Last Post
Replies
2
Views
1K
Replies
11
Views
467
Replies
2
Views
1K
Replies
0
Views
3K
  • Last Post
Replies
1
Views
1K
Replies
2
Views
8K
Replies
1
Views
14K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
5
Views
514
Replies
4
Views
15K
Top