Mixing two gases in an isolated system and calculating final p and T

[chocopanda]

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.

 

[Chestermiller]

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.

Correct

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.

Correct again.

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
$$pV = nkT$$
$$p_1*V_1 = p_2*V_2$$
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.

You calculated the final temperature correctly (nice job), but not the final partial pressures. The final partial pressures should satisfy:

$$p_fV_f=(1)RT_f$$They should be equal.

 

[chocopanda]

Correct

Correct again.

You calculated the final temperature correctly (nice job), but not the final partial pressures. The final partial pressures should satisfy:

$$p_fV_f=(1)RT_f$$They should be equal.

Thank you so much for the quick answer! I figured out where I went wrong (I think):

Instead of calculating the final (partial) pressure via the final temperature, I mistakingly used the gases inital temp. and volumes. So to calculate the partial pressures, which should be equal like you mentioned, the equation should be:

$$p_i = \frac {n_i R T_f} {V_f}$$
$$p_f = \sum_{i}^\infty \frac{n_i R T_f} {V_f} = \sum_{i}^\infty p_i$$

For their partial pressures I'm getting $p_i = 2520,53 hPa$ so $p_f = 5041,06 hPa$

Hope I'm right! What still isn't that clear to me is why the temperature is rather low, I get where that comes from mathematically but physically? Does that just mean that diatomic molecules like $CO_2$ that have a higher heat capacity ( $C_V = \frac {7}{2} R$ here) than atoms with lesser degrees of freedom just require more energy to heat up? I guess that makes sense considering the definition of heat capacity
And I'm still not sure why their masses were provided in the homework statement

 

[Chestermiller]

Thank you so much for the quick answer! I figured out where I went wrong (I think):

Instead of calculating the final (partial) pressure via the final temperature, I mistakingly used the gases inital temp. and volumes. So to calculate the partial pressures, which should be equal like you mentioned, the equation should be:

$$p_i = \frac {n_i R T_f} {V_f}$$
$$p_f = \sum_{i}^\infty \frac{n_i R T_f} {V_f} = \sum_{i}^\infty p_i$$

For their partial pressures I'm getting $p_i = 2520,53 hPa$ so $p_f = 5041,06 hPa$

I have no idea what you did here, especially those infinite sums. I get for the final partial pressures $$p_f=\frac{(8.314)(303.15)}{0.01}=252\ kPa$$

 

[chocopanda]

I have no idea what you did here, especially those infinite sums. I get for the final partial pressures $$p_f=\frac{(8.314)(303.15)}{0.01}=252\ kPa$$

Hi yes, that's what I got as well as per my previous post for the partial pressures, and I have to add them up for the overall final pressure