Question:
The minimum pressure required to compress $600 \,dm^{3}$ of a gas at $1$ bar to $150\, dm^{3}$ at $40^{\circ}C$ is
The minimum pressure required to compress $600 \,dm^{3}$ of a gas at $1$ bar to $150\, dm^{3}$ at $40^{\circ}C$ is
Updated On: Apr 14, 2025
- $4.0$ bar
- $0.2$ bar
- $1.0$ bar
- $2.5$ bar
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The Correct Option is A
Solution and Explanation
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
$P_{1}V_{1}=P_{2}V_{2}$
$1$ bar $\times 600dm^{3}=P_{2}\times 150\,dm^{3}$
$P_{2}=4$ bar
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Concepts Used:
Gas Laws
The gas laws were developed at the end of the 18th century, when scientists began to realize that relationships between pressure, volume and temperature of a sample of gas could be obtained which would hold to approximation for all gases.
The five gas laws are:
- Boyle’s Law, which provides a relationship between the pressure and the volume of a gas.
- Charles’s Law, which provides a relationship between the volume occupied by a gas and the absolute temperature.
- Gay-Lussac’s Law, which provides a relationship between the pressure exerted by a gas on the walls of its container and the absolute temperature associated with the gas.
- Avogadro’s Law, which provides a relationship between the volume occupied by a gas and the amount of gaseous substance.
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