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Heat and Thermodynamics: Monoatomic to diatomic

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Initially, a quantity of a mono atomic gas (cv=3R/2) occupies a rigid container with volume V=0.19m^3. The temperature and pressure are Ti=287k and Pi=108Kpa. The gas reacts completely with itself to form half the original moles of a diatomic gas (cp=5R/2). The reaction is exothermic and produces an amount of heat Q=36497 J which increases the final internal energy, what is the final temperature?

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Initially, a quantity of a monatomic gas (cv=3R/2) occupies a rigid container with volume V=0.19m^3, The temperature and pressure are Ti=287k and Pi=108Kpa. The gas reacts completely with itself to form half the original moles of a diatomic gas (cp=5R/2). The reaction is exothermic and produces an amount of heat Q=36497 J which increases the final internal energy, what is the final temperature?
Solution:
The quantity of the gas in moles is given by the gas equation as

Or
Now Pi = 108 k Pa = 1.08*105 Pa
V = 0.19 m3
Ti = 287 K
And R = 8.314 J/mol. K
Substituting in above equation we get
mol
The internal energy of a gas at temperature T is given by

Where Cv is the specific heat at constant volume.
Thus the initial internal energy of the gas is given by

Or J
Now as the volume of the gas remains constant, there will be no work done by the gas on external agency and hence the heat evolved during reaction remains with the gas as its internal energy and hence new (sensible) internal energy of the gas is given by (first law of thermodynamics)
U2 = U1 + Q = 30781 + 36497 = 67278 J
Now in second state
Number of moles n2 = n1/2 = 8.60/2 = 4.30 mol
Specific heat at constant volume Cv = Cp - R = (5R/2) - R = 3R/2
If the final temperature of the gas is Tf then we get the internal energy as

Or
Substituting the values we get
K
Thus the final temperature of the gas will be 1254.6 K.

I think as the gas becomes diatomic its specific heat at constant volume will be 5R/2 because of the rotational and oscillation kinetic energy (degree of freedom 5) and hence we should have

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