For most diatomic gases at room temperatures gamma = pm1.40 pm 0.01. Find the specific heat of nitrogen in these conditions.

For most diatomic gases at room temperatures gamma = pm1.40 pm 0.01. F

1.	For most diatomic gases at room temperatures gamma = pm1.40 pm 0.01. Find the specific heat of nitrogen in these conditions.
Answer» Solution :For diatomic gases in the specified temperature RANGE we have `R/(0.41) le C_(mv) le R/(0.39)` i.e., `2.02 xx 10^4 le C_(mv) le 2.13 xx 10^4` J/(kmole.K) `2.85 xx 10^4 le C_(MP) le 2.96 xx 10^4` J/(kmole .K)` .

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For most diatomic gases at room temperatures gamma = pm1.40 pm 0.01. Find the specific heat of nitrogen in these conditions.

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