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- Homework Statement
- An isolated container has two separate halves (V = V1 = V2 = 5,00l). One half contains n1 = 1,00 mol carbon dioxide (CO2) at a temperature of T1 = 0C. The other half contains n1 = 1,00 mol Xenon (Xe) at a temperature of T2 = 100C. Now the partition between the two halves are opened, the two gases mix and an equilibrium sets in. Calculate the final temperature T3 and pressure p2.

Data for CO2: m_A,CO2 = 131,3g, f_CO2 = 7. Data for Xe: m_A,Xe = 131,3g, f_Xe = 3.

- Relevant Equations
- dU = δQ + δW

pV = nkT

U = f/2*nRT

dQ = C_V*n*dT

Hey everyone, I have an attempt at fully solving this problem (my final pressure is ##p_f = 5373,64 hPa##, final temp. is ##T_f = 303,15K = 30C##), but this exercise confuses me very much.

First, I have not used the masses in my calculations and I'm pretty sure my prof. accidentally copypasted the mass for ##Xe## into ##CO_2##, since the molar mass doesn't add up, so ##m_A,CO2 = 44,01g##. I'm still unsure how they are relevant though.

Secondly, it is not stated in the exercise that the gases are ideal gases, but since this is a physics thermodynamic class and literally every other problem we had to solve we had to assume the gases are ideal. I could be very wrong in assuming this, but otherwise I would have no idea how to solve this problem.

Now for the system, I am imagining (as stated in the exercise) an isolated container with two gases separated by a partition. Both halves fill a volume of ##V=5l## so after removing the partition the final volume should simply add up to ##V_f = 10l##.

In a mixture of gases, the final pressure is the sum of the partial pressures of the gases (Dalton's law). Assuming we are dealing with ideal gases, we can simply use

$$p_i = \frac {n_i k T_i}{V}$$

$$p_1*V_1 = p_2*V_2$$

solving for ##p_2## and using ##V_2 = V_f##. My final partial pressures are ##p_2,CO2 = 2271,10 hPa## and ##p_2,Xe = 3102,54 hPa## so ##p_f = p_2,CO2 + p_2,Xe = 5373,64 hPa##. Is this even plausible?

Then for the final temperature, using the first law of thermodynamics ##dU = δQ + δW## and there not being any work done on the system ##dU = δQ## we could simply calculate the final temperature by using

$$ΔU = \frac {f} {2} * nR(T_f - T_i)$$

with ##f## already being provided by the homework statement and ##ΔU_1 = -ΔU_2##, so what I end up getting for ##T_f## is:

$$T_f = \frac {\frac {7} {2} T_1 + \frac {3} {2} T_2} {5}$$

$$T_f = 303,15K$$

Does that even make sense? How come the temperature of the mix (even though the mols and volume the gases occupy are the same) after equilibrium isn't close to 50C? Did I solve the problem wrong? Is it because of the degree of freedom? (Molecule vs atom)

I feel like I'm either being baited by the mass or I completely missed the mark on this one, I'm thankful for any input.

First, I have not used the masses in my calculations and I'm pretty sure my prof. accidentally copypasted the mass for ##Xe## into ##CO_2##, since the molar mass doesn't add up, so ##m_A,CO2 = 44,01g##. I'm still unsure how they are relevant though.

Secondly, it is not stated in the exercise that the gases are ideal gases, but since this is a physics thermodynamic class and literally every other problem we had to solve we had to assume the gases are ideal. I could be very wrong in assuming this, but otherwise I would have no idea how to solve this problem.

Now for the system, I am imagining (as stated in the exercise) an isolated container with two gases separated by a partition. Both halves fill a volume of ##V=5l## so after removing the partition the final volume should simply add up to ##V_f = 10l##.

In a mixture of gases, the final pressure is the sum of the partial pressures of the gases (Dalton's law). Assuming we are dealing with ideal gases, we can simply use

$$p_i = \frac {n_i k T_i}{V}$$

$$p_1*V_1 = p_2*V_2$$

solving for ##p_2## and using ##V_2 = V_f##. My final partial pressures are ##p_2,CO2 = 2271,10 hPa## and ##p_2,Xe = 3102,54 hPa## so ##p_f = p_2,CO2 + p_2,Xe = 5373,64 hPa##. Is this even plausible?

Then for the final temperature, using the first law of thermodynamics ##dU = δQ + δW## and there not being any work done on the system ##dU = δQ## we could simply calculate the final temperature by using

$$ΔU = \frac {f} {2} * nR(T_f - T_i)$$

with ##f## already being provided by the homework statement and ##ΔU_1 = -ΔU_2##, so what I end up getting for ##T_f## is:

$$T_f = \frac {\frac {7} {2} T_1 + \frac {3} {2} T_2} {5}$$

$$T_f = 303,15K$$

Does that even make sense? How come the temperature of the mix (even though the mols and volume the gases occupy are the same) after equilibrium isn't close to 50C? Did I solve the problem wrong? Is it because of the degree of freedom? (Molecule vs atom)

I feel like I'm either being baited by the mass or I completely missed the mark on this one, I'm thankful for any input.

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