[Physics] Clarify homework results regarding heat, work and the ideal gas law


So I have a bunch of homework problems relating to thermodynamics and work done by an ideal gas. I've been having trouble with the isobaric questions, and want to make sure I'm getting the concept down. I went through the following question and got the answers "right", but their answers are significantly different from mine and I'm not sure why (I had to answer part C using Part D because my answer to part C was too high).

The questions are on a website with a pass/fail system and very little feedback if the answer is wrong. If my answer is "close enough" it will accept it then give me their correct answer. In general, I use too many significant digits and let it round, rather than rounding myself, as too many digits isn't graded against me.

Exercise 19.17
A cylinder contains $0.250 mol$ of carbon dioxide $(CO_2)$ gas at a temperature of $27.0°C$. The cylinder is provided with a frictionless piston, which maintains a constant pressure of $1.00atm$ on the gas. The gas is heated until its temperature increases to $127.0°C$. Assume that the $CO_2$ may be treated as an ideal gas.

At this point, I know it's an isobaric process because it's at constant pressure. I can convert 27 and 127 °C to 300.15 and 400.15 K, and convert 1.00 atm to 101325 Pa (not that either conversion is actually required in this case).

The equations I have for isobaric processes are:
$pV=nRT$ [Ideal Gas Equation]
$\Delta U=Q-W$ [First Law Equation]
$W=p(V_2-V_1)$ [Area under PV curve for constant PV is a square]
$Q=nC_p\Delta T$ [From the book/wikipedia]

Part A. How much work is done by the gas in this process?

So I don't have the volume to get work directly, but I can calculate it from the ideal gas equation:
$pV=nRT \leftrightarrow V_i={nRT_i \over p}$
$W=p(V_2-V_1)=p({nRT_2 \over p}-{nRT_1 \over p})=nR(T_2-T_1)$

Plugging in known values, I get:
$W=0.25 mol \cdot 8.3145 {J \over mol\cdot K}\cdot (400.15 K-300.15 K)$
$W=207.9 J$ which the webpage rounds to 208 J.

Part B. On what is this work done? (Piston or Cylinder)

It's the piston because the piston is being moved and the cylinder isn't. Work is force across distance, so moving something does work on it. Yay for easy parts.

Part C. What is the change in internal energy of the gas?

Originally, I was confused by how to do this because I'd copied the work equation into my $\Delta U$ equation which didn't make sense. So I went ahead and calculated heat supplied, which is part D, then subtracted work done.

Looking back through, it looks like $U=nC_VT$ so $\Delta U=nC_V(T_2-T_1)$.

Now, the book says $C_V=k{1 \over 2}R$ where k is the number of degrees of freedom. A single atom has 3 translational degrees of freedom. Then my understanding is that each additional atom adds 2 rotational degrees of freedom. So $C_V={3 \over 2}R$ for one-atom "molecules", $C_V={5 \over 2}R$ for two-atom molecules (such as $O_2$ or $N_2$), and $C_V={7 \over 2}R$ for three-atom molecules (such as the $CO_2$ in this question).

So I can substitute known values to get:
$\Delta U=0.25 mol\cdot {7 \over 2}8.3145 {J \over mol\cdot K}(400.15K-300.15K)$
$\Delta U=727.5J$ which the site says is "close" but not good enough.

Part D. How much heat was supplied to the gas?

Here, I have an equation for solving Q directly.
$Q=nC_p\Delta T=n(C_V+R)\Delta T$

Because I'm looking for $C_p=C_V+R$, I can add ${2 \over 2} R$ to the $C_V={7 \over 2}R$ I got above, for a total of $C_p={9 \over 2}R$. Plugging in known values, I get
$Q=0.25 mol\cdot {9 \over 2} 8.3145 {J \over mol\cdot K} (400.15 K – 300.15K)$
$Q=935.4J$ which the website accepts, but "rounds" to 924 J which seems like I'm making a mistake rather than rounding a little weird.

From here, I can go back to Part C.
$\Delta U=Q-W=935.4J-207.9J=727.5J$ which is what I calculated directly above and isn't accepted. By using the values they got for parts A and D, I can calculate $924J – 208J = 716J$ which the site accepts but rounds to $712J$.

Part E: How much work would have been done if the pressure had been 0.50 atm?

Because the original equation for work cancelled out the pressure, I should get the same answer regardless of pressure, $208 J$, which is accepted as correct.

So my equation for work seems to be spot on, but my equations for heat and internal energy are off by quite a bit. What am I doing wrong?

Best Answer

There is something strange about your work.

Cutting through the rather verbose solution, you basically say

$$W = nR\Delta T\tag1 $$ $$\Delta U = \frac{7}{2} nR \Delta T \tag2$$ $$Q = \frac92 nR \Delta T \tag3$$

Mathematically, it is obvious that (1) + (2) = (3) - yet you don't get that when you evaluate each of these directly.

When I substitute your numbers, I get

W = 207.9 J
U = 727.5 J
Q = 935.4 J

which is what you got. The results the site gives as "correct" are off in a strange way - the value you got in part (A) is not even the difference between the values they gave for (D)-(C) (which they ought to be).

I see several possible things going on here: 1) they are using a different value for the heat capacity (actual heat capacity of CO2 is a function of temperature, see for example this table), and 2) there is some strange rounding in the way they generated their results. It wouldn't be the first time.

You appear to have a good grasp of the subject and are being needlessly confused by the web quiz. Oh, to have intelligent, live teachers instead of dumb, lifeless computers...