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Two containers $\mathrm{A}$ and $\mathrm{B}$ contain $\mathrm{CO}_2$ gas. Pressure, volume and absolute temperature of the gas in $\mathrm{A}$ are 4 times more compared to that in $\mathrm{B}$. The mass of the gas in $\mathrm{B}$ is $x \mathrm{~g}$, then the mass of the gas in A will be
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$4 x g$
Using PV $=n R T$
$\begin{aligned} & \frac{\mathrm{PV}}{\mathrm{RT}}=\mathrm{n}=\frac{\mathrm{m}}{\mathrm{M}} \\ & \therefore \quad \mathrm{M}=\mathrm{m} \frac{\mathrm{RT}}{\mathrm{PV}}\end{aligned}$
Both the containers contain same gas $\left(\mathrm{CO}_2\right)$. So, molar mass (M) will be same.
$\begin{aligned} & \Rightarrow \frac{m_A R T_A}{P_A V_A}=\frac{m_B R T_B}{P_B V_B} \\ & \frac{m_A\left(4 T_B\right)}{\left(4 P_B\right)\left(4 V_B\right)}=\frac{m_B T_B}{P_B V_B} \\ & \frac{m_A}{4}=x g \\ & m_A=4 x g\end{aligned}$
$\begin{aligned} & \frac{\mathrm{PV}}{\mathrm{RT}}=\mathrm{n}=\frac{\mathrm{m}}{\mathrm{M}} \\ & \therefore \quad \mathrm{M}=\mathrm{m} \frac{\mathrm{RT}}{\mathrm{PV}}\end{aligned}$
Both the containers contain same gas $\left(\mathrm{CO}_2\right)$. So, molar mass (M) will be same.
$\begin{aligned} & \Rightarrow \frac{m_A R T_A}{P_A V_A}=\frac{m_B R T_B}{P_B V_B} \\ & \frac{m_A\left(4 T_B\right)}{\left(4 P_B\right)\left(4 V_B\right)}=\frac{m_B T_B}{P_B V_B} \\ & \frac{m_A}{4}=x g \\ & m_A=4 x g\end{aligned}$
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