Kinetic Theory Of Gases Question 59
Question: The average translational energy and the rms speed of molecules in a sample of oxygen gas at 300 K are $ 6.21\times {{10}^{-21}}J $ and 484 m/s respectively The corresponding values at 600 K are nearly (assuming ideal gas behavior)
Options:
A) $ 12.42\times {{10}^{-21}}J,928m/s $
B) $ 8.78\times {{10}^{-21}}J,684m/s $
C) $ 6.21\times {{10}^{-21}}J,968m/s $
D) $ 12.42\times {{10}^{-21}}J,684m/s $
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Answer:
Correct Answer: D
Solution:
[d] The formula for average kinetic energy is $ \overline{\text{K}\text{.E}\text{.}}=\frac{3}{2}kT$
$\
Therefore \frac{{( \overline{\text{K}\text{.E}\text{.}} )_600K}}{{( \overline{\text{K}\text{.E}\text{.}} )_300K}}=\frac{600}{300} $
$ \Rightarrow {{(\overline{\text{K}\text{.E}\text{.}})} _{600K}}=2\times 6.21\times {{10}^{-21}}J $
$ =,12.42,\times {{10}^{-21}},J $
Also the formula for r.m.s. velocity is $ C _{rms}=\sqrt{\frac{3KT}{m}}$
$\
Therefore \frac{(C_rms)_600K}{(C_rms)_300K}$=$\sqrt{\frac{600}{300}} $
$ \Rightarrow {{(C _{rms})} _{600K}}=\sqrt{2}\times 484=684m/s $
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