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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
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$2: 1$
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
$p^{\prime}=\frac{1}{3} \frac{(2 m) N}{V} \cdot\left(\frac{\mathbf{v}}{2}\right)^2=\frac{1}{2}\left\{\frac{1}{3} \frac{m N}{V}(\mathbf{v})^2\right\}$
$\begin{aligned} & =\frac{1}{2} p^{\prime} \\ & \Rightarrow \quad \frac{p}{p^{\prime}}=\frac{2}{1} \Rightarrow p: p^{\prime}=2: 1\end{aligned}$
$p=\frac{1}{3} \frac{m N}{V}\left(\mathbf{v}^2\right)$
When mass of molecule is doubled and speed is halved, pressure becomes
$p^{\prime}=\frac{1}{3} \frac{(2 m) N}{V} \cdot\left(\frac{\mathbf{v}}{2}\right)^2=\frac{1}{2}\left\{\frac{1}{3} \frac{m N}{V}(\mathbf{v})^2\right\}$
$\begin{aligned} & =\frac{1}{2} p^{\prime} \\ & \Rightarrow \quad \frac{p}{p^{\prime}}=\frac{2}{1} \Rightarrow p: p^{\prime}=2: 1\end{aligned}$
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