The total kinetic energy of translatory motio | vedclass

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Question

(D) The translational kinetic energy $(K.E._t)$ of an ideal gas is given by $K.E._t = \frac{3}{2} PV$.
Given: $K.E._t = 3000 \, J$,$V = 5 \, L$,Press

Vedclass · Accepted Answer

The total kinetic energy of $1 \, L$ of $Ne$ at a pressure of $2P$ is $12000 \, J$.

Vedclass · Answer

(D) The translational kinetic energy $(K.E._t)$ of an ideal gas is given by $K.E._t = \frac{3}{2} PV$.
Given: $K.E._t = 3000 \, J$,$V = 5 \, L$,Pressure $= P$.
So,$3000 = \frac{3}{2} P(5) \Rightarrow P = \frac{3000 	imes 2}{15} = 400 \, J/L$.
For option $(B)$: $Ne$ is a monoatomic gas. Total $K.E. = K.E._t = \frac{3}{2} P'V' = \frac{3}{2} (2P)(1) = 3P = 3(400) = 1200 \, J$. (Incorrect)
For option $(C)$: $O_2$ is a diatomic gas. Total $K.E. = K.E._t + K.E._r = \frac{3}{2} PV + \frac{2}{2} PV = \frac{5}{2} PV$.
Total $K.E. = \frac{5}{2} (2P)(10) = 25P = 25(400) = 10000 \, J$. (Incorrect)
Wait,re-evaluating the options based on the provided solution logic: The solution provided in the prompt suggests $12000 \, J$ for monoatomic and $20000 \, J$ for diatomic. Let's re-calculate: $O_2$ at $10 \, L$ and $2P$: $K.E. = \frac{5}{2} (2P)(10) = 25P = 25(400) = 10000 \, J$. There seems to be a calculation error in the provided prompt's solution text. Given the constraints,the question is flawed.

The total kinetic energy of translatory motion of all the molecules of $5 \, L$ of nitrogen exerting a pressure $P$ is $3000 \, J$. Which of the following statements is correct?

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