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Question

(B) The total kinetic energy of an ideal gas is given by $E = \frac{3}{2} NkT$.Given that the total kinetic energy remains constant,we have $E_1 = E_2

Vedclass · Accepted Answer

$P_2 = P_1, T_2 = T_1/2$

Vedclass · Answer

(B) The total kinetic energy of an ideal gas is given by $E = \frac{3}{2} NkT$.Given that the total kinetic energy remains constant,we have $E_1 = E_2$.Substituting the expressions: $\frac{3}{2} N_1 k T_1 = \frac{3}{2} N_2 k T_2$.Since $N_2 = 2N_1$,we get $N_1 T_1 = (2N_1) T_2$,which simplifies to $T_2 = \frac{T_1}{2}$.From the ideal gas law,$PV = NkT$. Since the volume $V$ of the box is constant,we have $P = \frac{NkT}{V}$.Therefore,$P_2 = \frac{N_2 k T_2}{V} = \frac{(2N_1) k (T_1/2)}{V} = \frac{N_1 k T_1}{V} = P_1$.Thus,the new pressure is $P_2 = P_1$ and the new temperature is $T_2 = \frac{T_1}{2}$.

$A$ box contains $N$ molecules at temperature $T_1$ and pressure $P_1$. If the number of molecules in the box is doubled but their total kinetic energy remains unchanged,what are the new pressure $P_2$ and temperature $T_2$?

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