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First Law of Thermodynamics and temperature

  • Thread starter Gee Wiz
  • Start date
  • #1
137
0

Homework Statement



A toy balloon is initially at a temperature of T = 20°C. Its initial volume is 0.0042 m3 (It is a 10 cm radius sphere). Assume that the pressure in the balloon always equals atmospheric pressure, 101.3 kPa.
The new temp was found to be 52.23°C
By how many Joules did the internal energy of the gas in the balloon increase? Air is almost entirely diatomic gas.
ΔU =?

Homework Equations


T*N*α*K
pv=rnt
pv=kNt

The Attempt at a Solution


5/2*(52.23-20)*1.381e-23*N

n=pv/rt
N=n*6.022e23

I keep getting 1.17e-4, and it says i am off by a power of 10. I am not sure where I am messing up
 

Answers and Replies

  • #2
747
36
What is your equation for change in internal energy. I am trying to work this problem but I get a much different answer so I am probably using the wrong equation.

What did you get for n, the number of moles?
 
Last edited:
  • #3
ehild
Homework Helper
15,477
1,854

Homework Statement



A toy balloon is initially at a temperature of T = 20°C. Its initial volume is 0.0042 m3 (It is a 10 cm radius sphere). Assume that the pressure in the balloon always equals atmospheric pressure, 101.3 kPa.
The new temp was found to be 52.23°C
By how many Joules did the internal energy of the gas in the balloon increase? Air is almost entirely diatomic gas.
ΔU =?

Homework Equations


T*N*α*K
pv=rnt
pv=kNt

The Attempt at a Solution


5/2*(52.23-20)*1.381e-23*N

n=pv/rt
N=n*6.022e23

I keep getting 1.17e-4, and it says i am off by a power of 10. I am not sure where I am messing up
What did you get for N?

ehild
 
  • #4
137
0
it is the degrees of freedom (5/2 for diatomic)*(Temperature)*(K which is a constant 1.381e-23)*(# of molecules)
 
  • #5
747
36
So, what did you get for N or n?
 
  • #6
137
0
n= 1.746e-7
N=1.05e17
 
  • #7
747
36
I think your value for n is incorrect. Did you get it from the equation

PV = nRT

?
 
  • #8
137
0
Yes. I did 1*.0042e-3/(.08206*293)..converting the kPa's to atms and Celsius to k
 
  • #9
747
36
Using PV = nRT I have always used P in Pascals, V in meters^3 n in moles R = 8.32 and T = 293.

If you use P in atm, then is the constant 0.082 correct?
 
  • #10
137
0
yes i think so
 
  • #11
747
36
My previous post is incorrect. I should have said the volume is in liters, not meter^3
 
  • #12
137
0
right, and I think that m^3=mL=.001L
 
  • #13
747
36
So when I change my calculation from m^3 to liters I get an answer closer to yours but not exact.

so, to find n, we use pressure in Pascals, volume in liters, R = 8.3, and K = 293.
so what do you get for n?
 
  • #14
137
0
1.74e-4
 
  • #15
747
36
I agree. Now, change in internal energy is (5/2)Rn(delta T)?
 
  • #16
137
0
I believe so
 
  • #17
747
36
I get (5/2)(8.31)(1.74E-4)(32.2) = I can't give the answer
 
  • #18
137
0
I get the same message with that answer that I do with my previous one. "It appears you have made a power of ten error"
 
  • #19
CAF123
Gold Member
2,902
88
In PV = nRT, [P] = N/m2, [n] = mol, [T] = K and with R = 8.31, [R] = J K-1mol-1 => [V] = J m2/N = m3, no?
 
  • #20
747
36
Are you using Quest? sometimes their answers are wrong. I get .116 if I use the equation above or if I use your original equation. I think you made a math error when you got 1.17 E-4.
 
  • #21
137
0
I am not 100% sure about your units for P and V (it may just be notation i am not familiar with)
I usually see [P]=atm or pascals, [V]=L (m^3=mL)
 
  • #22
137
0
It's smart physics
 
  • #23
747
36
I think we agree on n = -1.74E-4, the units of p and v are done with.

This means that N = (1.74E-4)(6.022E23) =1.04E20

Just recalculate your first answer. I think you made a math error.

1.17E-4 is not correct didn't you get .117?
 
  • #24
137
0
Here is everything I am doing

=(5/2)*(1.381e-23)*(32.23)*( (1*.0042)/(.08206*293) )* (6.022e23)
 
  • #25
747
36
I don't recognize some of your constants and conversions but.....

Take your equation from post #1, 5/2*(52.23-20)*1.381e-23*N and plug in the value for N

N = (1.74E-4)(6.02E23) = 1.04E20 and what do you get? .116 or .117 I think this is the correct answer.

Check your math and we will agree even though smart Physics says we are off. I don't believe it.
 

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