Question 126 of 168
The temperature of a gas is and the average translational kinetic energy of its molecules is . The temperature at which the average translational kinetic energy of the molecules of the same gas becomes is :
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Answer: (B)
$\begin{aligned}& \mathrm{K} . \mathrm{E}=\frac{\mathrm{nf}_1 \mathrm{RT}}{2} \\& \mathrm{~T}_{\mathrm{i}}=-78^{\circ} \mathrm{C} \rightarrow 273+\left[-78^{\circ} \mathrm{C}\right]=195 \mathrm{~K}
\end{aligned}$
K.E
To double the K.E energy temp also become double
$\begin{aligned}& \mathrm{T}_{\mathrm{f}}=390 \mathrm{~K} \\
& \mathrm{~T}_{\mathrm{f}}=117^{\circ} \mathrm{C}\end{aligned}$
\end{aligned}$
K.E
To double the K.E energy temp also become double
$\begin{aligned}& \mathrm{T}_{\mathrm{f}}=390 \mathrm{~K} \\
& \mathrm{~T}_{\mathrm{f}}=117^{\circ} \mathrm{C}\end{aligned}$