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A gas at a pressure of $2 \mathrm{~atm}$ is heated from $25^{\circ} \mathrm{C}$ to $323^{\circ} \mathrm{C}$ and simultaneously compressed of $\frac{2}{3}$ rd of its original value. Then the final pressure is
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$6 \mathrm{~atm}$
Given, $p_1=2 \mathrm{~atm}$
$\begin{aligned} p_2 & =? \\ V_1 & =1 \mathrm{~L} \\ V_2 & =\frac{3}{2} \\ T_1 & =298 \mathrm{~K} \\ T_2 & =596 \mathrm{~K}\end{aligned}$
Using combined gas law, we get
$\begin{aligned} \frac{p_1 V_1}{T_1} & =\frac{p_2 V_2}{T_2} \\ \Rightarrow \quad p_2 & =\frac{2 \times 1}{298} \times \frac{3 \times 596}{2}=6 \mathrm{~atm}\end{aligned}$
$\begin{aligned} p_2 & =? \\ V_1 & =1 \mathrm{~L} \\ V_2 & =\frac{3}{2} \\ T_1 & =298 \mathrm{~K} \\ T_2 & =596 \mathrm{~K}\end{aligned}$
Using combined gas law, we get
$\begin{aligned} \frac{p_1 V_1}{T_1} & =\frac{p_2 V_2}{T_2} \\ \Rightarrow \quad p_2 & =\frac{2 \times 1}{298} \times \frac{3 \times 596}{2}=6 \mathrm{~atm}\end{aligned}$
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