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

[barryj]

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 ?

 

[barryj]

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.

 

[DrClaude]

That looks ok.

 

[barryj]

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?

 

[barryj]

<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

 

[DrClaude]

and the temperature at point D = 48.6.

I get something slightly different here.

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.

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.

 

[ehild]

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.

 

[barryj]

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.;

 

[barryj]

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?

 

[DrClaude]

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.

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.

 

[DrClaude]

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.

 

[barryj]

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

 

[DrClaude]

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.

 

[barryj]

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?

 

[DrClaude]

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.