Consider a container like that shown in Figure 20.12, with n₁ moles of a monatomic gas on one side and n₂ moles of a diatomic gas on the other. The monatomic gas has initial temperature T₁ᵢ. The diatomic gas has initial temperature T₂ᵢ.

b. Show that the equilibrium temperature is

Question

Consider a container like that shown in Figure 20.12, with n₁ moles of a monatomic gas on one side and n₂ moles of a diatomic gas on the other. The monatomic gas has initial temperature T₁ᵢ. The diatomic gas has initial temperature T₂ᵢ.

b. Show that the equilibrium temperature is

Accepted Answer

Hi, everyone. In this practice problem, we are being asked to determine an equation for final temperature or T F. We'll have a rigid insulated container divided into two compartments by a thin membrane. The left compartment contains an A R moles of organ gas which is mono atomic. While the right compartment can then contains an N I moles of nitrogen gas, which is diatomic two temperature sensors, one placed in the left, one in the right measure the A R N T N I respectively. And the membrane actually prevent all any of these gasses to mix. We're being asked to find the final temperatures T F of the system in terms of N A R and N I T A R N T N I, when the thermal equilibrium is reached, the options given are multiple different equations for the final temperature or T F. So in order for us to uh solve this problem, we have to uh notice that organ is a mono atomic gas with three degrees of freedom. And nitrogen is a diatomic gas with five degrees of freedom. Therefore, the right compartment or nitrogen will begin with an initial thermal energy which is going to be represented by E N I initial, which will be equals to five divided by two or 5.5 of an, and I multiplied by R multiplied by T. And I, on the other hand, for the uh left compartment or organ will begin with an initial Permal energy of E A R E initial which is two divided by two multiplied by N A R multiplied by R multiplied by D A R just like. So the total thermal energy of the system is going to be the sum of the two initial energies of the two compartment. Uh um considering no in their energy loss, which is um result of the insulating container. So the E total is going to be equal to five divided by two of N and I multiplied by R multiplied by T and I plus three divided by two of N A R multiplied by R multiplied by T A R just like. So we can actually pull the R and a half out. So the E total will then just be equal to R divided by two multiplied by open parentheses. Five N N ID N I plus three N A R multiplied by T A R just like. So this will be equation one which I will refer back in the next steps. But uh the next thing that we want to note this is that the membrane is going to be very thin, that the nitrogen and the A Argan atoms can collide at the boundary without moving from one side to the other. So the two gasses will actually interact until the equilibrium temperature of T F is reached. And at the F, the, to the total thermal energy of the system based on our equation here or equation one will then be equal to E total equals to R multiplied by T F multiplied by five N and I plus three N A R and RT F will be divided by two. And this will be the second equation that we have in our system. So the first equation is going to be at initial which I'm going to indicate with this I right here. And the second is going to be at final condition which I'm going to indicate with an F. So we want to actually equalize equation one and equation two saying that the equilibrium reach or equation one and equation two is actually equal to one another giving the final temperature of T F which will then come out to an equation of R divided by two, multiplied by five N N I multiplied by T N I plus three N A R multiplied by T A R. Close parenthesis equals to R multiplied by T F divided by two multiplied by five N N I plus three N A R close parenthesis. I'm going to cross out the R and the twos and then um rearranging our equation. To get an equation for T F, we can actually then obtain the equation or the final equation for T F which will then be equals to five N N I multiplied by T N I plus three N A R multiplied by T A R divided by five N N I plus three N A R. So that will be the final equation for the T F or the final temperature of our system after the equilibrium is reached. And this equation will be corresponding to option D in our answer choices. So option D with P F equals to five N N IP N I plus three N A RT A R of that divided by five N N I plus D N three N A R will be the answer to the final temperature equation of the system. So that'll be it for this video. If you guys still have any sort of confusion, please feel free to check out our other lesson videos on similar topics available on our website. But other than that, that will be it for this video. Thank you.

Consider a container like that shown in Figure 20.12, with n₁ moles of a monatomic gas on one side and n₂ moles of a diatomic gas on the other. The monatomic gas has initial temperature T₁ᵢ. The diatomic gas has initial temperature T₂ᵢ. b. Show that the equilibrium temperature is

Verified Solution

Key Concepts

Thermodynamic Equilibrium

Heat Capacity

First Law of Thermodynamics