JEE MAIN - Physics (2020 - 6th September Morning Slot - No. 4)
Molecules of an ideal gas are known to have three translational degrees of freedom and two
rotational degrees of freedom.The gas is maintained at a temperature of T. The total internal
energy, U of a mole of this gas, and the value of
$$\gamma \left( { = {{{C_p}} \over {{C_v}}}} \right)$$ are given, respectively by:
$$\gamma \left( { = {{{C_p}} \over {{C_v}}}} \right)$$ are given, respectively by:
U = $${5 \over 2}RT$$ and $$\gamma = {7 \over 5}$$
U = 5RT and $$\gamma = {6 \over 5}$$
U = 5RT and $$\gamma = {7 \over 5}$$
U = $${5 \over 2}RT$$ and $$\gamma = {6 \over 5}$$
Explanation
Total degree of freedom (f) = 3 + 2 = 5
U = $${{nfRT} \over 2}$$ = $${{5RT} \over 2}$$
$$\gamma $$ = $${{{C_p}} \over {{C_v}}}$$ = $$1 + {2 \over f}$$ = $$1 + {2 \over 5}$$ = $${7 \over 5}$$
U = $${{nfRT} \over 2}$$ = $${{5RT} \over 2}$$
$$\gamma $$ = $${{{C_p}} \over {{C_v}}}$$ = $$1 + {2 \over f}$$ = $$1 + {2 \over 5}$$ = $${7 \over 5}$$
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