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A container of $10 \mathrm{~L}$ is filled with an ideal gas at a temperature of $27^{\circ} \mathrm{C}$ at a pressure $12 \mathrm{~atm}$. The volume of the container is reduced to $6 \mathrm{~L}$ and the temperature of the gas in increased by $30^{\circ} \mathrm{C}$, then the final pressure of the gas is
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Verified Answer
The correct answer is:
$20 \mathrm{~atm}$
Given, $V_1=10 \mathrm{~L}=10 \times 10^{-3} \mathrm{~m}^3$
$$
\begin{aligned}
& \mathrm{T}_1=27^{\circ} \mathrm{C}=27+273=300 \mathrm{~K} \\
& p_1=12 \mathrm{~atm} \\
& V_2=6 \mathrm{~L}=6 \times 10^{-3} \mathrm{~m}^3 \\
& T_2=30^{\circ} \mathrm{C}=273+30=303 \mathrm{~K} \\
& p_2=?
\end{aligned}
$$
By ideal gas equation,
$$
\begin{aligned}
\frac{p_1 V_1}{T_1} & =\frac{p_2 V_2}{T_2} \\
\Rightarrow \quad p_2 & =\frac{p_1 V_1 T_2}{T_1 V_2} \\
& =\frac{12 \times 10 \times 10^{-3} \times 303}{300 \times 6 \times 10^{-3}}=20.2 \mathrm{~atm} \\
& \simeq 20 \mathrm{~atm}
\end{aligned}
$$
$$
\begin{aligned}
& \mathrm{T}_1=27^{\circ} \mathrm{C}=27+273=300 \mathrm{~K} \\
& p_1=12 \mathrm{~atm} \\
& V_2=6 \mathrm{~L}=6 \times 10^{-3} \mathrm{~m}^3 \\
& T_2=30^{\circ} \mathrm{C}=273+30=303 \mathrm{~K} \\
& p_2=?
\end{aligned}
$$
By ideal gas equation,
$$
\begin{aligned}
\frac{p_1 V_1}{T_1} & =\frac{p_2 V_2}{T_2} \\
\Rightarrow \quad p_2 & =\frac{p_1 V_1 T_2}{T_1 V_2} \\
& =\frac{12 \times 10 \times 10^{-3} \times 303}{300 \times 6 \times 10^{-3}}=20.2 \mathrm{~atm} \\
& \simeq 20 \mathrm{~atm}
\end{aligned}
$$
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