In an ideal gas, if the masses of all molecules are doubled and their speeds are halved, then the ratio of…
In an ideal gas, if the masses of all molecules are doubled and their speeds are halved, then the ratio of initial and final pressures of the gas is
$2: 1$
$1: 2$
$4: 1$
$1: 4$
Solution
Pressure of an ideal gas is given by
$p=\frac{1}{3} \frac{m N}{V}\left(\mathbf{v}^2\right)$
When mass of molecule is doubled and speed is halved, pressure becomes
$p^{\prime}=\frac{1}{3} \frac{(2 m) N}{V} \cdot\left(\frac{\mathbf{v}}{2}\right)^2=\frac{1}{2}\left\{\frac{1}{3} \frac{m N}{V}(\mathbf{v})^2\right\}$
$\begin{aligned} & =\frac{1}{2} p^{\prime} \\ & \Rightarrow \quad \frac{p}{p^{\prime}}=\frac{2}{1} \Rightarrow p: p^{\prime}=2: 1\end{aligned}$