Thermodynamics Question 331
Question: Two rigid boxes containing different ideal gases are placed on a table. Box A contains one mole of nitrogen at temperature $ T _{0}, $ While Box B contains one mole of helium at temperature (7/3) $ T _{0} $ . The boxes are then put into thermal contact with each other and heat flows between them until the gases reach a common final temperature (Ignore the heat capacity of boxes). Then, the final temperature of the gases, $ T _{f}, $ in term of $ T _{0} $ is
Options:
A) $ T _{f}=\frac{7}{3}T _{0} $
B) $ T _{f}=\frac{3}{2}T _{0} $
C) $ T _{f}=\frac{5}{2}T _{0} $
D) $ T _{f}=\frac{3}{7}T _{0} $
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Answer:
Correct Answer: B
Solution:
[b] When two gases are mixed together then Heat lost by the Helium gas = Heat gained by the Nitrogen gas
$ {\mu _{B}}\times {{(C _{v})} _{He}}\times ( \frac{7}{3}T _{0}-T _{f} )={\mu _{A}}\times {{(C _{v})} _{N2}}\times (T _{f}-T _{0}) $
$ 1\times \frac{3}{2}R\times ( \frac{7}{3}T _{0}-T _{f} )=1\times \frac{5}{2}R\times (T _{0}-T _{f}) $
By solving we get $ T _{f}=\frac{3}{2}T _{0} $
BETA
