As shown in the figure to the right, we have an insulated container which is div
ID: 1477305 • Letter: A
Question
As shown in the figure to the right, we have an insulated container which is divided into two equal halves. Into the left side we place n, n moles of a monatomic gas at temperature To and into the right side we place 2n moles of a diatomic gas at temperature 21g- The partition between the two halves will not move, but it is thin enough to allow thermal energy to flow between the two sides until thermal equilibrium between the two is reached. In all following parts, express your answers in terms of n, R, , and , or a subset of these quantities only mono dia a. (3 pts) What is the common final temperature reached by the two gasses? b. (3 pts) How much thermal energy flowed out of the diatomic gas into the monatomic gas during the process of reaching thermal equilibrium? (3 pts) Next we start over again, keeping the same box and immovable partition, but this time replacing the diatomic c. gas in the right half with a monatomic gas, again 2n m oles and initial temperature 2To What is the common final temperature reached by these two gasses? mono monoExplanation / Answer
a)
Common final temperature, T = [nTo + 2n x 2To] / (n+2n) = 5/3 To
b)
Change in energy = nCvdT
Cv = 5/2 R for diatomic gas, R being the universal gas constant.
dT = 2To - 5/3 To = 1/3 To
n = 2n
Change in energy = 2n x 5/2 R x 1/3 To = 5/3 nRT
c)
Final temperature will be same as in the previous case
T = [nTo + 2n x 2To] / (n+2n) = 5/3 To
d)
Change in energy = nCvdT
For monoatomic gas, Cv = 3/2 R
Here, number of mole = 2
dT = 2To - 5/3 To = 1/3 To
Change in energy = 2n x 3/2 R x 1/3 To = nRTo.
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