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A vessel of volume $V$ contains ideal gas having mass density $\rho$ at temperature $T$ and pressure $p$. After a portion of the gas is let out, the pressure in the vessel is decreased by $\Delta p$. The mass of the released gas is
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Verified Answer
The correct answer is:
$\rho V \Delta p / p$
Given, initial value of volume, pressure and temperature of an ideal gas in vessel,
$$
V_1=V, p_1=p \text { and } T_1=T
$$
and density of gas $=\rho$
When, $\Delta V$ volume of gas is let out, hence pressure of gas is decreased to $\Delta p$.
Then,
$$
\begin{aligned}
& V_2=V-\Delta V \\
& p_2=p-\Delta p \\
& T_2=T_1
\end{aligned}
$$
By an ideal gas equation,
$$
\begin{array}{rlr}
\frac{p_1 V_1}{T_1} & =\frac{p_2 V_2}{T_2} & \\
\frac{p_1}{p_2} & =\frac{V_2}{V_1} & {\left[\because T_1=T_2\right]} \\
\frac{p}{p-\Delta p} & =\frac{V-\Delta V}{V}=1-\frac{\Delta V}{V} & \\
\frac{\Delta V}{V} & =1-\frac{p}{p-\Delta p} & \\
& =\frac{p-\Delta p-p}{p-\Delta p} & \\
\Delta V & =\frac{-\Delta p V}{p-\Delta p} &
\end{array}
$$
$\therefore$ Mass of released gas $=\rho \Delta V=\frac{\rho \Delta p V}{p-\Delta p} \approx \frac{\rho \Delta p V}{p}$
$$
V_1=V, p_1=p \text { and } T_1=T
$$
and density of gas $=\rho$
When, $\Delta V$ volume of gas is let out, hence pressure of gas is decreased to $\Delta p$.
Then,
$$
\begin{aligned}
& V_2=V-\Delta V \\
& p_2=p-\Delta p \\
& T_2=T_1
\end{aligned}
$$
By an ideal gas equation,
$$
\begin{array}{rlr}
\frac{p_1 V_1}{T_1} & =\frac{p_2 V_2}{T_2} & \\
\frac{p_1}{p_2} & =\frac{V_2}{V_1} & {\left[\because T_1=T_2\right]} \\
\frac{p}{p-\Delta p} & =\frac{V-\Delta V}{V}=1-\frac{\Delta V}{V} & \\
\frac{\Delta V}{V} & =1-\frac{p}{p-\Delta p} & \\
& =\frac{p-\Delta p-p}{p-\Delta p} & \\
\Delta V & =\frac{-\Delta p V}{p-\Delta p} &
\end{array}
$$
$\therefore$ Mass of released gas $=\rho \Delta V=\frac{\rho \Delta p V}{p-\Delta p} \approx \frac{\rho \Delta p V}{p}$
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