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In a vessel, the gas is at pressure $P$, if the mass of all the molecules is halved of their speed is double, then the resultant pressure will be :
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Verified Answer
The correct answer is:
$2 P$
The equation of pressure is
$$
P=\frac{1}{3} \frac{m \mathrm{~N}}{v^2} v_{r m s}^2
$$
where $m=$ mass of each molecules
$$
\begin{aligned}
& N=\text { number of molecules } \\
& V=\text { volume of the gas }
\end{aligned}
$$
According to the question:
mass of all molecules is halved an velocity is double then
$$
\begin{aligned}
\mathrm{P}^{\prime}=\frac{1}{3}\left(\frac{m}{2}\right) \times \frac{\mathrm{N}}{\mathrm{V}} \times( & \left(2 \mathrm{~V}_{\mathrm{ras}}\right)^2 \\
& =\frac{2}{3} \frac{m \mathrm{~N}}{\mathrm{~V}} \mathrm{~V}_{\mathrm{ras}}^2 \\
& =2 \mathrm{P}
\end{aligned}
$$
compare with eq (i)
$$
\begin{aligned}
& \mathrm{P}^{\prime}=2 \mathrm{P} \\
& \mathrm{P}^{\prime}=2 \mathrm{P}
\end{aligned}
$$
$$
P=\frac{1}{3} \frac{m \mathrm{~N}}{v^2} v_{r m s}^2
$$
where $m=$ mass of each molecules
$$
\begin{aligned}
& N=\text { number of molecules } \\
& V=\text { volume of the gas }
\end{aligned}
$$
According to the question:
mass of all molecules is halved an velocity is double then
$$
\begin{aligned}
\mathrm{P}^{\prime}=\frac{1}{3}\left(\frac{m}{2}\right) \times \frac{\mathrm{N}}{\mathrm{V}} \times( & \left(2 \mathrm{~V}_{\mathrm{ras}}\right)^2 \\
& =\frac{2}{3} \frac{m \mathrm{~N}}{\mathrm{~V}} \mathrm{~V}_{\mathrm{ras}}^2 \\
& =2 \mathrm{P}
\end{aligned}
$$
compare with eq (i)
$$
\begin{aligned}
& \mathrm{P}^{\prime}=2 \mathrm{P} \\
& \mathrm{P}^{\prime}=2 \mathrm{P}
\end{aligned}
$$
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