The pressure acting on 500 cubic meters of gas is reduced from 4 atm to 2 atm. If the temperature remains constant, what is the new volume?

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The pressure acting on 500 cubic meters of gas is reduced from 4 atm to 2 atm. If the temperature remains constant, what is the new volume?

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Answer 1 · 2016-03-02T04:04:44

$V_{new} = {1000 m}^{3}$ $"V"_"new"="1000 m"^3$

Explanation:

We will use Boyle's law, which states that pressure and volume are inversely proportional at constant temperature. We can express this mathematically as:

$P_{1} V_{1} = P_{2} V_{2}$ $"P"_1"V"_1="P"_2"V"_2$

Here, we see that

$P_{1} = 4 atm$ $"P"_1="4 atm"$
$P_{2} = 2 atm$ $"P"_2="2 atm"$

$V_{1} = {500 m}^{3}$ $"V"_1="500 m"^3$
$V_{2} = unknown$ $"V"_2="unknown"$

We want to manipulate Boyle's law to solve for $V_{2}$ $"V"_2$ :

$V_{2} = \frac{P_{1} V_{1}}{P_{2}}$ $"V"_2=("P"_1"V"_1)/("P"_2)$

Plugging in what we know:

$V_{2} = \frac{4 atm \cdot {500 m}^{3}}{2 atm}$ $"V"_2=("4 atm" * "500 m"^3)/("2 atm")$

$V_{2} = 2 \cdot {500 m}^{3}$ $"V"_2=2 * "500 m"^3$

$V_{2} = {1000 m}^{3}$ $"V"_2="1000 m"^3$

We can see that the volume increased from ${500 m}^{3}$ $"500 m"^3$ to ${1000 m}^{3}$ $"1000 m"^3$ , so it doubled. This is paired with the pressure being halved, as should be the case in an inversely proportional relationship.

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The pressure acting on 500 cubic meters of gas is reduced from 4 atm to 2 atm. If the temperature remains constant, what is the new volume?

1 Answer

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