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If $\mathrm{N}_2$ gas is compressed at 2 atmosphere from $9.0 \mathrm{~L}$ to $3.0 \mathrm{~L}$ at $300 \mathrm{~K}$, find the final pressure at same temperature.
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The correct answer is:
$6.0 \mathrm{~atm}$
According to Boyle's law, $\mathrm{P}_1 \mathrm{~V}_1=\mathrm{P}_2 \mathrm{~V}_2$ (at constant $\mathrm{n}$ and $\mathrm{T}$ )
$\therefore \quad P_2=\frac{P_1 V_1}{V_2}=\frac{2 \times 9.0}{3.0}=6.0 \text { bar }$
$\therefore \quad P_2=\frac{P_1 V_1}{V_2}=\frac{2 \times 9.0}{3.0}=6.0 \text { bar }$
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