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Question 71 of 90

Q.The kinetic energy per molecule of a gas at temperature T is ____.

(a) (3/2)RT(3/2)RT
(b) (3/2)KBT(3/2)K_BT
(c) (2/3)RT(2/3)RT
(d) (3/2)(RT/M)(3/2)(RT/M)
Maharashtra MsbshseMaharashtra HSC (MSBSHSE) Board 2018MCQ· 1mImportance★★★★★
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Each translational degree of freedom carries average energy 12KBT\tfrac12 K_BT; a gas molecule has 3 translational degrees of freedom, giving 32KBT\tfrac32 K_BT per molecule.

From the kinetic theory of gases, the mean square speed of gas molecules is related to pressure and volume by PV=13Nmc2‾PV = \tfrac13 Nm\overline{c^2}, and using the ideal gas law PV=NKBTPV = NK_BT (for NN molecules), we get

13mc2‾=KBT  ⇒  12mc2‾=32KBT.\tfrac13 m\overline{c^2} = K_BT \;\Rightarrow\; \tfrac12 m\overline{c^2} = \tfrac32 K_BT. …

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