Final temperature of the gas mixture,neglecting heat losses

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SUMMARY

The discussion focuses on calculating the final temperature of a gas mixture consisting of 2 moles of rigid diatomic gas at temperature T and 2 moles of ideal monoatomic gas at temperature 5T, while neglecting heat losses. The equation used is Q(initial) = Q(final), leading to the formula m1C(v1)T(1) + m2C(v2)T(2) = (m1 + m2)C(v)T(final). The user seeks clarification on the values of m1 and m2, confirming that both are 2 moles in this scenario.

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  • Understanding of thermodynamics principles, specifically heat transfer.
  • Knowledge of specific heat capacities for diatomic and monoatomic gases.
  • Familiarity with the concept of moles in gas calculations.
  • Basic algebra for solving equations involving multiple variables.
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  • Review the specific heat capacities of diatomic gases versus monoatomic gases.
  • Study the principles of energy conservation in thermodynamic systems.
  • Learn how to apply the ideal gas law in mixed gas scenarios.
  • Explore advanced thermodynamics topics such as heat loss calculations and their implications.
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Students and professionals in physics, particularly those studying thermodynamics, as well as engineers working with gas mixtures and heat transfer systems.

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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