Question

A gas occupies 0.60 m3 at a 5.0 atm. If the temperature of the gas remains the same and the pressure decreases to 2.5 atm, what would be the new volume occupied by the gas?

Known:
Unknown:
Formula/Work Shown:


Answer:

Answer 1

The new volume of the gas is $1.20 m^3$

Explanation:

For a gas kept at constant temperature, we can apply Boyle's law, which states that:

"For a gas kept at constant temperature, the product of the gas pressure and of its volume remains constant"

Mathematically:

$pV=const.$

where

p is the gas pressure

V is its volume

For this problem, we can rewrite the equation as

$p_1 V_1 = p_2 V_2$

where:

$p_1 = 5.0 atm$ is the initial pressure of the gas

$V_1 = 0.60 m^3$ is the initial volume of the gas

$p_2 = 2.5 atm$ is the final pressure of the gas

$V_2$ is the final volume of the gas

And solving for $V_2$, we find:

$V_2=\frac{p_1 V_1}{p_2}=\frac{(5.0)(0.60)}{2.5}=1.20 m^3$

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Answer 2

The new volume of the gas will be $1.2 \;\rm m^{3}$.

Given data:

The volume occupied by the gas is, $V = 0.60\;\rm m^{3}$.

The initial pressure is, P = 5.0 atm.

The final pressure is, P' = 2.5 atm.

Unknown value : - Final Volume of the gas

Formula/work shown:

Here the concept of Boyle's law will be applied. As per the Boyle's law, the pressure varies inversely with the volume occupied by the gas. Then the expression is,

${P}{V}=P'V'$

Here, V' is the new volume of gas.

Solving as,

$\dfrac{P}{P'}=\dfrac{V'}{V}\\\\\dfrac{5.0}{2.5}=\dfrac{V'}{0.60}\\\\V'=0.60 \times \dfrac{5.0}{2.5}\\\\V' =1.2 \;\rm m^{3}$

Thus, we can conclude that the new volume of the gas will be $1.2 \;\rm m^{3}$.

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