Kinetic Theory Of Gases Question 16
Question: A light container having a diatomic gas enclosed within is moving with velocity $ v $ . Mass of the gas is $ M $ and number of mass of moles is n. The kinetic energy of gas w.r.t. ground is
Options:
A) $ \frac{1}{2}MV^{2}+\frac{3}{2}nRT $
B) $ \frac{1}{2}MV^{2} $
C) $ \frac{1}{2}MV^{2}+\frac{5}{2}nRT $
D) $ \frac{5}{2}nRT $
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Answer:
Correct Answer: C
Solution:
[c] The kinetic energy of gas W.r.t. center of mass of the system K.E. $ =\frac{5}{2}nRT $ Kinetic energy of gas w.r.t ground =kinetic energy of centre of mass w.r.t. ground +Kinetic energy of gas w.r.t. centre of mass. $ K.E.=\frac{1}{2}MV^{2}+\frac{5}{2}nRT $
BETA
