Calculate ΔU. n is constant. V, T and p change

[jamesrb]

Homework Statement

Calculate ΔU when 1.00 mol of H₂ goes from 1.00 atm, 10.0 L, and 295 K to 0.793 atm, 15.0 L, and 350 K.

Homework Equations

ΔU = q + w
q=mcΔT

The Attempt at a Solution

From the moles of H₂ we can get the mass of H₂ and use q=mcΔT.
1.00 mol H₂=(2.0158 g H₂/1 mol H₂)=2.0158 g H₂
Thus:
q=mcΔT
q=(2.0158 g)(14.314J/mol*K)(55K)
q=1586 J

The answer section in my book says ΔU = 1590 J. My answer is close enough but what about w? The problem doesn't explicitly say the gas is subject to a constant external pressure so I can't use w=p_extΔV.

 

[theOrange]

Does your book say that ΔU = 1590 J, or q = 1590 J?

ΔU ≠q

=================================

Work is calculated with this integral:

W = \int_{V_1}^{V_2} p\,dv

Which means you need to find a formula for pressure, because it is not constant.

 

[Borek]

I would assume answer in the book is identical to your answer - it is just rounded to follow the number of significant digits given in the problem. All numbers are given to 3 sf, your answer gives 4.

 

[Chestermiller]

This is follow-up to what TheOrange said. In this problem, q is not equal to ΔU. The equation you wrote for q should really be the equation for ΔU. In the ideal gas region, ΔU=mCvΔT always, irrespective of whether the volume is constant, since Cv is defined by the equation:

mC_v=\frac{∂U}{∂T}

Chet

 

[theOrange]

Look at an ideal gas, the formula below may be helpful as well:

P*V^n = constant

 

[Chestermiller]

Look at an ideal gas, the formula below may be helpful as well:

P*V^n = constant

He was only asked to determine ΔU, not the work or the heat.

Chet

 

[theOrange]

He was only asked to determine ΔU, not the work or the heat.

Chet

\Delta U = Q - W

Which btw he already posted as a relevant equation...

 

[Chestermiller]

\Delta U = Q - W

Which btw he already posted as a relevant equation...

He may have posted this as a relevant equation, but it is obviously not needed for this problem. For any ideal gas,
ΔU=mC_vΔT
and, in this problem,
m=1 mole
C_v=28.836 J/(mole K)
ΔT=55 K

The clear purpose of this exercise was to test the understanding of the student to determine whether he was aware that, for an ideal gas, the change in internal energy could be calculated from $ΔU=mC_vΔT$, irrespective of the changes in pressure and volume.

In any case, there certainly isn't enough information provided to determine either Q or W, particularly since no indication is given of whether the process is reversible or irreversible.

In thermo, it is very important to recognize that the internal energy is a physical property of the material.

Chet

 

[theOrange]

He may have posted this as a relevant equation, but it is obviously not needed for this problem. For any ideal gas,
ΔU=mC_vΔT
and, in this problem,
m=1 mole
C_v=28.836 J/(mole K)
ΔT=55 K

The clear purpose of this exercise was to test the understanding of the student to determine whether he was aware that, for an ideal gas, the change in internal energy could be calculated from $ΔU=mC_vΔT$, irrespective of the changes in pressure and volume.

In any case, there certainly isn't enough information provided to determine either Q or W, particularly since no indication is given of whether the process is reversible or irreversible.

In thermo, it is very important to recognize that the internal energy is a physical property of the material.

Chet

He asked himself, "My answer is close enough but what about w? The problem doesn't explicitly say the gas is subject to a constant external pressure so I can't use w=pextΔV."

So I was just trying to help by saying that he needs to find a formula for P (since it isn't constant), and that W is an integral.

 

[Chestermiller]

He asked himself, "My answer is close enough but what about w? The problem doesn't explicitly say the gas is subject to a constant external pressure so I can't use w=pextΔV."

So I was just trying to help by saying that he needs to find a formula for P (since it isn't constant), and that W is an integral.

It was already pointed out by Borek in post #3 with regard to the comparison between the book and the OP's calculation, that the OP was dealing with a significant figures issue.

The work W is always equal to $\int{P_{ext}dV}$, irrespective of whether the process is reversible or irreversible. However, we can be certain that P_ext=P (with P determined from the equation of state for the material) only if the process is reversible.

Chet

 

[theOrange]

It was already pointed out by Borek in post #3 with regard to the comparison between the book and the OP's calculation, that the OP was dealing with a significant figures issue.

The work W is always equal to $\int{P_{ext}dV}$, irrespective of whether the process is reversible or irreversible. However, we can be certain that P_ext=P (with P determined from the equation of state for the material) only if the process is reversible.

Chet

Chet, seriously what are you talking about. What Borek said had nothing to do with what I said. The formula for work using \Delta V does not work in this case because P is not constant. Therefore I was letting him know he needs to use the integral form and find an equation for P. Which can be found using the ideal gas formula. I also said that it MIGHT be helpful.

I really don't see what the problem is. To avoid spamming this topic even more maybe you should just message me, if this is SO important to you.

 

[Chestermiller]

Chet, seriously what are you talking about. What Borek said had nothing to do with what I said. The formula for work using \Delta V does not work in this case because P is not constant. Therefore I was letting him know he needs to use the integral form and find an equation for P. Which can be found using the ideal gas formula. I also said that it MIGHT be helpful.

I really don't see what the problem is. To avoid spamming this topic even more maybe you should just message me, if this is SO important to you.

It is important to me, because I didn't want the OP to get confused. But I like your idea about discussing this in private messages so that we can reach a consensus. Then we can report back jointly to the thread. Is this agreeable to you?

Chet