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(B) The total kinetic energy of a gas is given by ${E_{total}} = \frac{f}{2}NkT$,where $f$ is the degrees of freedom,$N$ is the number of molecules,an

Vedclass · Accepted Answer

${P_2} = {P_1},\;{T_2} = \frac{{{T_1}}}{2}$

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(B) The total kinetic energy of a gas is given by ${E_{total}} = \frac{f}{2}NkT$,where $f$ is the degrees of freedom,$N$ is the number of molecules,and $k$ is the Boltzmann constant.Since ${E_{total}}$ is kept constant,we have $N_1 T_1 = N_2 T_2$.Given $N_2 = 2N_1$,we get $N_1 T_1 = (2N_1) T_2$,which implies ${T_2} = \frac{{{T_1}}}{2}$.From the ideal gas equation,$PV = NkT$,we have $P \propto NT$.Therefore,$\frac{{{P_2}}}{{{P_1}}} = \frac{{{N_2}{T_2}}}{{{N_1}{T_1}}} = \left( \frac{2N_1}{N_1} ight) \left( \frac{T_1/2}{T_1} ight) = 2 	imes \frac{1}{2} = 1$.Thus,${P_2} = {P_1}$ and ${T_2} = \frac{{{T_1}}}{2}$.

$A$ box contains $N$ molecules of a perfect gas at temperature ${T_1}$ and pressure ${P_1}$. The number of molecules in the box is doubled while keeping the total kinetic energy of the gas the same as before. If the new pressure is ${P_2}$ and temperature is ${T_2}$,then:

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