Final temperature of the gas mixture,neglecting heat losses

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The discussion centers on calculating the final temperature of a gas mixture consisting of 2 moles of diatomic gas at temperature T and 2 moles of monoatomic gas at temperature 5T, while neglecting heat losses. The equation used for the solution is based on the principle of conservation of energy, stating that the initial heat of the system equals the final heat. The user is confused about the masses (m1 and m2) in the equation, as they have noted both gases contain 2 moles. The specific heat capacities for diatomic and monoatomic gases are also relevant to the calculation. The final temperature can be determined by solving the energy balance equation with the correct values for m1, m2, and their respective heat capacities.
Intekhab Alam
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A box containing 2 moles of rigid diatomic gas molecules at temperature T connected to another box containing 2 moles of ideal monoatomic gas at temperature 5T. What is the final temperature of the gas mixture,neglecting heat losses?
2. The attempt at a solution:
Q(initial)=Q(final)
m1C(v1)T(1)+m2C(v2)T(2)=(m1+m2)C(v)T(final)
but i don't know m1 & m2.
 
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I read 2 moles for m1 and also for m2 ?!
 
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