Calculate the rms speed of nitrogen at 27°C. Given N=6×10^23 molecules/ moles and K= 1.38×10^(-16) ergs/K.
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Answer:
156.9 m/ s.speed of nitrogen at 27 degree celcius .
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Explanation:
Given:
Vrms=832ms−1
R=8320J/kmolK
Molar mass, M=28
R.M.S velocity, Vrms=M3RT
Calculation: From formula,
832=283×8320×T
T=3×8320832×832×28=776.5K
The R.M.S. speed of nitrogen molecules is 832ms−1 at temperature 776.5K
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