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Q. The minimum pressure required to compress $600 \,dm^{3}$ of a gas at $1$ bar to $150\, dm^{3}$ at $40^{\circ}C$ is

NEETNEET 2020Some Basic Concepts of Chemistry

Solution:

By Boyle's law
$P_{1}V_{1}=P_{2}V_{2}$
$1$ bar $\times 600dm^{3}=P_{2}\times 150\,dm^{3}$
$P_{2}=4$ bar