How Does the Ideal Gas Law Apply in Multi-Part Physics Problems?

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Homework Help Overview

The discussion revolves around the application of the Ideal Gas Law in a multi-part physics problem involving calculations of temperature, internal energy, work, and heat transfer for a gas under various conditions. Participants explore the implications of using different values for the gas constant R in their calculations.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants attempt to calculate temperatures and internal energy using the Ideal Gas Law, questioning the appropriate value of R to use in different contexts. Some express confusion over discrepancies in their internal energy calculations and the implications of using different units.

Discussion Status

There is ongoing exploration of the calculations related to internal energy and work done, with some participants providing guidance on the correct use of R. Multiple interpretations of the results are being discussed, particularly regarding the units of energy and the calculations of work.

Contextual Notes

Participants note the importance of unit consistency, especially when transitioning between different forms of the gas constant R, and the potential for confusion when calculating energy in different units (e.g., joules vs. atm liters).

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Homework Statement


See attached diagram

Homework Equations


See below

The Attempt at a Solution


[/B]
Part 1) Temperature at point A.
Since P is in Atmospheres and V is in Liters R = 0.08205
Using ideal gas law, PV = nRT
T = PV/R = (1)(2)/0.080205 = 24.375 K

Part 2) As above T = (3)(2)/0.08205 = 73.125 K

Part 3) This is where I am not sure...
Delta E = (3/2)nR (deltaT) = (1.5)(0.08205)(73.125-24.375) = 6 Joules ?
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Continuing...

Part 4) How much work is done from point A to point B?
Answer 0 work is done sine there is no change in volume.

Part 5) How much heat is transferred from point A to point B
Answer, it must be 6 Joules, the same as the change in internal energy.
 
That looks ok.
 
As above, I calculated the temperature at point C = 146,
and the temperature at point D = 48.6.

I am assuming that I can use either PV/T = PV/T or PV = nRT to calculate the temperatures, yes?

So here is my big concern.
The change in U from point B to C is .. E = (3/2)R (deltaT) = (1.5)(0.08205)(73) = 9 J
The work is..,W = P(deltaV) = (3)(2) = 6 J,
so the heat added must be 9 + 6 = 15 J Is this correct?
 
<Moderator's note: Threads merged>

1. Homework Statement
We have 1 mole of gas at P = 1 atm and V = 2 Liters
Since I am using atm and literes, R = 0.0821
So, since PV = nRT I calculate that T = PV/R = (1)(2)(0.0821) = 24.4 K
The internal energy = (3/2)nRT
In this case R = 8.315 J/(mol X K)
so E = (3/2)(8.315)(24.4) = 298 J

Lets add heat so the pressure goes to 3 atm.
Then the temperature is 73.2 K
and the internal energy = (3/2)(8.315)(73.2) = 912 J

This says that the change in internal energy is 912 - 298 = 614 J

However, this is my confusion..
I read that the change in internal energy = (3/2)nR(delta T)
Since I am using atm and liters R should be 0.0821 and this gives
(delta E) = (3/2)(0.0821)(73.2 - 23) = 6 J

I get different answers for the change in internal energy.

What is going on here? What value of R should I be using where?

Homework Equations

The Attempt at a Solution

 
Last edited by a moderator:
barryj said:
and the temperature at point D = 48.6.
I get something slightly different here.

barryj said:
I am assuming that I can use either PV/T = PV/T or PV = nRT to calculate the temperatures, yes?
In this case, the first equality comes from the second. The equation of state for the ideal gas is PV=nRT, and since nR is constant, so will the ratio PV/T.

barryj said:
So here is my big concern.
The change in U from point B to C is .. E = (3/2)R (deltaT) = (1.5)(0.08205)(73) = 9 J
The work is..,W = P(deltaV) = (3)(2) = 6 J,
so the heat added must be 9 + 6 = 15 J Is this correct?
The procedure is correct, but I just realized that your values for U are wrong. The gas constant you used is okay when dealing with pressure in atm and value in l, but for calculating U you need to use R = 8.314 J K-1 mol-1.
 
barryj said:

Homework Statement


We have 1 mole of gas at P = 1 atm and V = 2 Liters
Since I am using atm and literes, R = 0.0821
So, since PV = nRT I calculate that T = PV/R = (1)(2)(0.0821) = 24.4 K
The internal energy = (3/2)nRT
In this case R = 8.315 J/(mol X K)
so E = (3/2)(8.315)(24.4) = 298 J

Lets add heat so the pressure goes to 3 atm.
Then the temperature is 73.2 K
and the internal energy = (3/2)(8.315)(73.2) = 912 J

This says that the change in internal energy is 912 - 298 = 614 J

However, this is my confusion..
I read that the change in internal energy = (3/2)nR(delta T)
Since I am using atm and liters R should be 0.0821 and this gives
(delta E) = (3/2)(0.0821)(73.2 - 23) = 6 J

I get different answers for the change in internal energy.

What is going on here? What value of R should I be using where?

Homework Equations

The Attempt at a Solution

If you use R=0.0821 L atm K-1mol-1 you get the internal energy in L atm units, not in joules.
 
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and the temperature at point D = 48.6.
I get something slightly different here.

PV = nRT, T = PV/R = (1)(4)/0.0821 = 48.7

Where is the error ?

I am still confused over the calculation of internal energy and the change in internal energy.

I am using P in atm and V in Liters so R should be 0.0821, yes
But i read that the internal energy = (3/2)nRT and in this casre R = 8.315

Homework helper said that if I use R = 0.0821 the energy is in atm liter units. I do not understand.;
 
Let me restate the question.

State A: P = 1 atm, V = 2 liters, 1 mole, r = 0.0821
Using PV = nRT T = PV/R = (1)(2)/.0821 = 24.36 K
E = (3/2)nRT = (1.5)(8.315)(24.36) = 303.83 J

State B: P = 3 atm, V = 2 liters, 1 mole
T = PV/R = (3)(1)/0.0821 = 73.08 K
E = (3/2)(nRT = (1.5)(8.315)(73.08) = 911.49

The increase in internal energy (deltaE) = 911.49 - 303.83 = 607.66 J

Are these calculations correct?
 
  • #10
barryj said:
and the temperature at point D = 48.6.
I get something slightly different here.

PV = nRT, T = PV/R = (1)(4)/0.0821 = 48.7

Where is the error ?
To me, 48.6 and 48.7 are not the same. This is what I wanted to point out.

barryj said:
I am still confused over the calculation of internal energy and the change in internal energy.

I am using P in atm and V in Liters so R should be 0.0821, yes
But i read that the internal energy = (3/2)nRT and in this casre R = 8.315

Homework helper said that if I use R = 0.0821 the energy is in atm liter units. I do not understand.;
R = 8.315 J K-1 mol-1 = 0.08205 atm l K-1 mol-1
If you use the second one when multiplying by a ΔT in K, you get back an energy is units of atm l, not J.
 
  • #11
barryj said:
Let me restate the question.

State A: P = 1 atm, V = 2 liters, 1 mole, r = 0.0821
Using PV = nRT T = PV/R = (1)(2)/.0821 = 24.36 K
E = (3/2)nRT = (1.5)(8.315)(24.36) = 303.83 J

State B: P = 3 atm, V = 2 liters, 1 mole
T = PV/R = (3)(1)/0.0821 = 73.08 K
E = (3/2)(nRT = (1.5)(8.315)(73.08) = 911.49

The increase in internal energy (deltaE) = 911.49 - 303.83 = 607.66 J

Are these calculations correct?
Yes, these are the correct energies.
 
  • #12
Referring to my initial diagram,

T = PV/R where R in this case = 0.0821
U = (3/2)RT where R = 8.315

At B: Pb = 3, Vb = 2, Tb = (3)(2)/0.0821 = 73.08 K
Ub = (3/2)(8.315)(73.08) = 911.49 J

At C: Pc = 3, Vc = 2, Tc = (3)(2)/0.0821 = 146.16 K
Uc = (3/2)(8.315)(146.16) = 1822.98

Work from B to C , W = P(deltaV) = (3)(2) = 6 J

Therefore required heat to be added , Qbc = 917.49 J

Uc = Ub + Qbc - Wbc 1822.98 = 911.49 + 917.49 - 6

Can someone confirm my numbers here? Thanks
 
  • #13
barryj said:
Work from B to C , W = P(deltaV) = (3)(2) = 6 J
That's not correct. If you used units throughout, you would see clearly why.
 
  • #14
I think I get it.
3 atm = 3E5 pascals, 2 litrs = 2E-3 meters^3 so w = P*V = 3E5 X 2E-3 = 600 J Correct?
 
  • #15
barryj said:
I think I get it.
3 atm = 3E5 pascals, 2 litrs = 2E-3 meters^3 so w = P*V = 3E5 X 2E-3 = 600 J Correct?
It depends what precision you need. 1 atm = 101325 Pa, so only approximately 105.
 

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