Calculate this value of 1 mole at STP with C_V

  • Thread starter grandpa2390
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In summary: That's the correct algebraic equation. I don't understand how you get that numerical value. What value if R are you using?8.3145nvm ok I found the mistake. I am really sorry for all this trouble. It was a silly mistake that I made (I forgot to multiply by V in the derivative of a)I feel so embarrassed right now but thank you for your help. :)In summary, after working through the algebraic equations, a mistake was found in the derivative of a where V was not multiplied, causing the incorrect numerical value to be obtained. The correct value is 1.0065, with an R value of 8.3145.
  • #1
grandpa2390
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Homework Statement


Given ##(\frac{∂H}{∂U})_P = (\frac{C_V+ (π_T + P)*V*a}{C_V + V*a*π_T}) ##
Calculate this value of 1 mole of ideal gas at STP that has constant heat capacity of 12.5 ##\frac{J*K}{mol}##

n=1
T = 273.15 K
P = 1 atm
##C_V = 12.5 \frac{J*K}{mol}##
a = ?
##π_T = ?##

Homework Equations


##a = \frac{1}{V}*(\frac{∂V}{∂T})_p##
##π_T = T(\frac{∂P}{∂T})_v - P##

The Attempt at a Solution


So, since this is an ideal gas. I tried to find a by solving PV=nRT for V and then taking the partial with respect to T. I got nR/P
I plugged in the numbers and for a I got .003661

For π_T I did the reverse. I solved for P and then took the partial with respect for V.
I plugged in the values and and got 0.

Plugging these numbers into the given formula above. I keep coming up with 1.00656
The answer is 1.67...
What am I doing wrong?
 
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  • #2
Try to go as far as possible algebraically before plugging in numbers. In other words, find analytical expressions for a and πT for an ideal gas and substitute the values in the equation.
 
  • #3
no cigar

I got to this after two separate attempts. If I am making an error, it is the same error over and over.

##\frac{C_v*n^2*R^2*T}{PVC_v+n^2*R^2*T-PVnR}##

and when I plug in the numbers, I get .08205
apparently everything should cancel except one R.

I don't know.
 
  • #4
What do you get for a and πT individually?
 
  • #5
DrClaude said:
What do you get for a and πT individually?

I'm sure that is where I went wrong. either that or i calculated the volume incorrectly.
I got roughly 22 for volume by solving PV=nRT for V

for a, I did v=nRT/P and differentiated. so ##\frac{nR}{PV}## .0037

for pi I got ##\frac{nRT}{V} - P## 0

assuming V is correct...
 
  • #6
grandpa2390 said:
I'm sure that is where I went wrong. either that or i calculated the volume incorrectly.
I got roughly 22 for volume by solving PV=nRT for V
Hint: you don't need the volume if you do the algebra to the end.

grandpa2390 said:
for a, I did v=nRT/P and differentiated. so ##\frac{nR}{PV}## .0037

for pi I got ##\frac{nRT}{V} - P## 0
Yes, ##\pi_T = 0## and ##a = nR / PV##. What do you get then for ##(\pi_T + P ) V a## and for ##V a \pi_T##?
 
  • #7
DrClaude said:
Hint: you don't need the volume if you do the algebra to the end.Yes, ##\pi_T = 0## and ##a = nR / PV##. What do you get then for ##(\pi_T + P ) V a## and for ##V a \pi_T##?

gah I did it 3 times already and I keep getting the same result. I'll try a fourth time...
is my answer in post 3 wrong then you say?
 
  • #8
grandpa2390 said:
is my answer in post 3 wrong then you say?
Yes. The actual equation is much simpler.

If you can show your derivation, it will be easier for me to help you figure out where it goes wrong.
 
  • #9
DrClaude said:
Yes. The actual equation is much simpler.

If you can show your derivation, it will be easier for me to help you figure out where it goes wrong.

that's what I was going to do. but typing it in LaTeX takes so long. if I were to do it in Mathematica, could I just copy and paste? is that possible?
 
  • #10
grandpa2390 said:
that's what I was going to do. but typing it in LaTeX takes so long. if I were to do it in Mathematica, could I just copy and paste? is that possible?
Anything that's readable will be fine.
 
  • #11
DrClaude said:
Anything that's readable will be fine.

then I can try uploading a scan of me handwriting it? I know it is against the rules, but I'll do it again extra slow and if it is illegible I can just scrap it and do it through mathematica.
 
  • #12
grandpa2390 said:
then I can try uploading a scan of me handwriting it? I know it is against the rules, but I'll do it again extra slow and if it is illegible I can just scrap it and do it through mathematica.
We don't like scans in OPs, but we understand it can be hard to recopy everything. Here it will be fine.
 
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  • #13
DrClaude said:
We don't like scans in OPs, but we understand it can be hard to recopy everything. Here it will be fine.

thank you :)

I'm going to resolve it from scratch. who knows. maybe this time, just because I asked a question, I'll get the right answer. That's usually what happens. I get the wrong answer until I ask someone for help.
 
  • #14
part 1 I edited the steps so you could easily point out my mistake ;)
 

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  • #15
part 2edit: this page isn't right. I accidentally multiplied by Cv in the denominator
 

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  • #16
I just had a brainstorm (or Eureka! moment)

I could replace V with nRT/P :)
 
  • #17
Correcting the accidental error in this work in the denominator got me 1.00029

replacing v with nRT/P got me the same thing. :(
 
  • #18
In step 3, simplify to ##\pi_T = 0##, as discussed above.
 
  • #19
DrClaude said:
In step 3, simplify to ##\pi_T = 0##, as discussed above.

##\frac{C_v+nR}{C_v}## ?

= 1.0065...
 
  • #20
but why?

shouldn't they be the same?

I get why it simplifies to 0. I didn't see it before but nRT/v = p and P-P = 0

but still... where did I make a mistake. it was definitely easy to in the mess I made
 
  • #21
disregard this
 
  • #22
well it is probably going to be really hard if my mistake isn't near the beginning because I messed up the denominator. here is an abbreviated version that is correct (except for it not being correct)
 

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  • #23
grandpa2390 said:
##\frac{C_v+nR}{C_v}## ?

= 1.0065...
That's the correct algebraic equation. I don't understand how you get that numerical value. What value if R are you using?
 

1. What is the value of 1 mole at STP?

The value of 1 mole at STP (standard temperature and pressure) is 22.4 liters. This value is constant for any gas at STP.

2. What does STP stand for?

STP stands for standard temperature and pressure. Standard temperature is 0 degrees Celsius or 273.15 Kelvin, and standard pressure is 1 atmosphere or 101.3 kilopascals.

3. What is the significance of calculating 1 mole at STP?

Calculating 1 mole at STP is important because it helps scientists compare the properties of different gases under the same conditions. This allows for more accurate and consistent data analysis.

4. How do you calculate the value of 1 mole at STP?

The value of 1 mole at STP can be calculated using the ideal gas law, PV = nRT. At STP, the pressure (P) is 1 atmosphere, the volume (V) is 22.4 liters, the number of moles (n) is 1, and the gas constant (R) is 0.0821 L·atm/mol·K. By rearranging the equation, we can solve for the temperature (T) in Kelvin, which should be 273.15 K.

5. How does the value of 1 mole at STP differ from other conditions?

The value of 1 mole at STP is specific to standard temperature and pressure conditions. Under different conditions, such as high pressures or temperatures, the volume of 1 mole of gas will vary. This is why it is important to specify the conditions when calculating the value of 1 mole.

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