# Mixing two ideal gases with different V, T at constant pressure

• chocopanda
In summary: T_2) = -8/2 * 3*R*(150...5)$$In summary, the student is struggling with the ideal gas law and does not understand how to apply it. They are also confused about the homework statement. chocopanda Homework Statement I'm translating my exercise to English so please bear with me. We are mixing air (an ideal gas) with Volume V1 = 3,00m^3 and Temperature T1 = 150,0C with additional air V2 = 8,00m^3 and T_2 = 5,0C. What is the final temperature and total volume if the pressure p remains constant throughout? Relevant Equations We learned that for a gas with constant pressure, the following applies: V/V0 = T/T0 To be honest, thermodynamics is really not my strong suit and I get confused when and how to apply formulas. My thought process is as follows: - there are two ideal gases (ideal gas law applies) - the pressure remains constant (isobaric process), so p1= p2 = p - I imagine there being two adjacent boxes separated by a wall: 1 box is filled with V1, T1 and the other one is filled with V2, T2. To start the mixing process, we get rid of the "connecting wall" and the two gases begin to mix until an equilibrium sets in. Since the two gases have different temperatures, the final temperature at the end is lower than T1. since V2 is much, much colder than V1 and has almost thrice the volume. I don't think we can simply sum up both volumes and then calculate the temperature, since a volume shrinks when its temperature drops. I think I would try calculating the final temperature first, then with it the volume, but I get so confused since every formula kinda needs both V and T. For example for the final volume, I could apply something like V = (V_1 + V_2) * T where T is the final calculated temperature, but my confidence on how to apply laws kinda shattered last week since I bombed the mandatory practice exercises (I got around 20% of the points :( I was so sure I got them right...) I will be forever grateful for any help. If it helps, we are in our 4th week of theromdynamics lectures, so we haven't had stuff like enthalpy, specific heat capacity, heat Q,... Did you learn about the ideal gas law ? BvU said: Did you learn about the ideal gas law ? Hey, thanks for replying! Yes, we have learned about the ideal gas law. I am having trouble applying it. For this case, I would perhaps choose the form p*V = n*R*T to get the amount of gas in moles for both gases and maybe sum them up? As in n1 + n2 = n (sum of all the gas in moles, since the sum wouldn't change in terms of differences in temperature or volume), but I don't know exactly how that helps me? How about (notionally) not removing the wall, but letting heat flow through, keeping each side at pressure p? haruspex said: How about (notionally) not removing the wall, but letting heat flow through, keeping each side at pressure p? Thanks for replying :) I thought about that, but the exercise requires that both gases should be mixed and calculate the final volume at the end, I'm still struggling :( chocopanda said: Thanks for replying :) I thought about that, but the exercise requires that both gases should be mixed and calculate the final volume at the end, I'm still struggling :( My point is that there is no difference whether you let the temperatures equalise and then remove the wall or in the other order. haruspex said: My point is that there is no difference whether you let the temperatures equalise and then remove the wall or in the other order. I'm very sorry, I really do not know where to go from here... is this possible with the ideal gas law? I know that both gases have the same pressure. I could solve for the pressure p for both gases: p1=n1*R*T1/V1, p2=n2*R*T2/V2 The mixing process happens under a constant pressure p which should be equal for both sides: p1=p2=p Can I insert the equations for both p1 and p2? I don't really see how that would help me though :( chocopanda said: Homework Statement: I'm translating my exercise to English so please bear with me. We are mixing air (an ideal gas) with Volume V1 = 3,00m^3 and Temperature T1 = 150,0C with additional air V2 = 8,00m^3 and T_2 = 5,0C. What is the final temperature and total volume if the pressure p remains constant throughout? Relevant Equations: We learned that for a gas with constant pressure, the following applies: V/V0 = T/T0 I imagine there being two adjacent boxes Perhaps imagine two adjacent balloons with external pressure P, instead of a ( rigid ) box. chocopanda said: I really do not know where to go from here... I have given you my suggestion: treat it as though the temperature equalisation (at constant pressures) happens first, the mixing later. The mixing will not change anything, I believe. Let P be the pressure. In terms of P, what are ##n_1## and ##n_2##? Thanks for the hints, I'll try again. So the internal energy of a system is: ##U = f/2 * n*R*T## and since air is not an monoatomic gas we can insert f=5? The change of the internal energy ##\delta U_1 = -\delta U_2## from the hotter gas to the colder and since a change in ##U## means ##\delta T## as well I get this (##T_1 \leq T_f \leq T_2)## :$$5/2 * n*R*(T_f - T_1) = 5/2 * n*R*(T_2 - T_f)$$and solving for ##T_f## would mean that the final temperature would be (T1+T2)/2 ? But how can that be if we have different volumes :( Last edited: Chestermiller said: Let P be the pressure. In terms of P, what are ##n_1## and ##n_2##? Since the ideal gas law ##p*V=n*R*T## applies. ##n## is the amount of particles in a gas in moles. Each gas has a certain amount of moles ##n_1## and ##n_2##, solving for ##n##: ##n_i= \frac {p*V_i} {R*T_i}## chocopanda said:$$5/2 * n*R*(T_f - T_1) = 5/2 * n*R*(T_1 - T_f)$$What happened to T2, n1, n2? Chestermiller haruspex said: What happened to T2, n1, n2? I completely overlooked that we have different ##n := n_1, n_2## I'm so sorry. I feel like I'm getting very close to ##T_f## :) Editing the equation:$$5/2 * n_1*R*(T_f - T_1) = 5/2 * n_2*R*(T_2 - T_f)$$Simplifying and solving for ##T_f##:$$n_1*(T_f - T_1) = n_2*(T_2 - T_f)T_f = \frac {n_1 * T_1 + n_2 * T_2*} {n_1+n_2}$$For ##n: n_i= \frac {p*V_i} {R*T_i}## and inserting that for n1, n2:$$T_f = \frac {V_1*T_1 + V_2*T_2} {\frac {V_1 + V_2} {T_1+ T_2}} T_f = \frac {(V_1+V_2)* (T_1+T_2} {V_1+V_2}$$Is that the correct approach? Edit: Oh, is a fraction in a fraction not supported? Did I break Latex? I somehow can't fix it Edit2: Fixed it. Damn those  chocopanda said: I completely overlooked that we have different ##n := n_1, n_2## I'm so sorry. I feel like I'm getting very close to ##T_f## :) Editing the equation:$$5/2 * n_1*R*(T_f - T_1) = 5/2 * n_2*R*(T_2 - T_f)$$Simplifying and solving for ##T_f##:$$n_1*(T_f - T_1) = n_2*(T_2 - T_f)T_f = \frac {n_1 * T_1 + n_2 * T_2*} {n_1+n_2}$$For ##n: n_i= \frac {p*V_i} {R*T_i}## and inserting that for n1, n2:$$T_f = \frac {V_1*T_1 + V_2*T_2} {\frac {V_1 + V_2} {T_1+ T_2}} T_f = \frac {(V_1+V_2)* (T_1+T_2} {V_1+V_2}$$Is that the correct approach? Edit: Oh, is a fraction in a fraction not supported? Did I break Latex? I somehow can't fix it Edit2: Fixed it. Damn those  You did the algebra wrong to get your 2nd equation for Tf. Also, for these two air masses which exchange heat with one another at constant pressure, it is the total enthalpy that is constant, not the total internal energy. PhDeezNutz Chestermiller said: You did the algebra wrong to get your 2nd equation for Tf. Also, for these two air masses which exchange heat with one another at constant pressure, it is the total enthalpy that is constant, not the total internal energy. Thank you, so the equation would be$$T_f = \frac {V_1*T_1+V_2*T_1} {V_1+V_2} $$Hope there aren't any mistakes in it And for the final volume ##V_f##, since the air cools down to around ##T_f=45,5C##, I can't just sum up ##V_1+V_2## since the overall volume decreases as well chocopanda said: Thank you, so the equation would be$$T_f = \frac {V_1*T_1+V_2*T_1} {V_1+V_2} $$Hope there aren't any mistakes in it And for the final volume ##V_f##, since the air cools down to around ##T_f=45,5C##, I can't just sum up ##V_1+V_2## since the overall volume decreases as well Try again..$$T_f=\frac{V_1+V_2}{\frac{V_1}{T_1}+\frac{V_2}{T_2}}

gmax137

## What happens to the final temperature when two ideal gases with different initial temperatures are mixed at constant pressure?

The final temperature of the mixture will be a weighted average of the initial temperatures of the two gases. The weights are determined by the specific heat capacities and the masses (or moles) of the gases. Assuming no heat is lost to the surroundings, the final temperature can be calculated using the principle of conservation of energy.

## How does the volume change when two ideal gases with different initial volumes are mixed at constant pressure?

When two ideal gases are mixed at constant pressure, the final volume of the mixture will be the sum of the initial volumes of the two gases. This is because the pressure is constant and the gases are ideal, so they follow the ideal gas law, which implies additive volumes under constant pressure conditions.

## Will the mixing of two ideal gases at constant pressure result in any change in the total number of moles?

No, the total number of moles will remain the same when two ideal gases are mixed. The number of moles is a conserved quantity, so mixing does not alter the total mole count of the system.

## How does the entropy change when two ideal gases are mixed at constant pressure?

The entropy of the system increases when two ideal gases are mixed. This is because mixing increases the randomness or disorder of the system. The change in entropy can be calculated using the Gibbs free energy equation for mixing ideal gases, which takes into account the mole fractions of the gases.

## What assumptions are made when mixing two ideal gases at constant pressure?

The main assumptions are that the gases behave ideally, meaning they follow the ideal gas law (PV=nRT), and that there are no intermolecular forces between the gas molecules. Additionally, it is assumed that the mixing process is adiabatic (no heat exchange with the surroundings) and that the gases are perfectly mixed, leading to a uniform final state.

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