# A gas occupies "1.46 L" at a pressure of "1.00 bar". What is the volume when the gas is dropped into the ocean to a depth such that the pressure is increased to "60.0 bar"?

Jul 20, 2017

The final volume $\left({V}_{2}\right)$ is $\text{0.0243 L}$.

Refer to the explanation for the process.

#### Explanation:

This question involves Boyle's law, which states that the volume of a gas held at constant temperature varies inversely with the pressure. The means that if the volume increases, the pressure decreases, and vice versa. The equation for Boyle's law is:

${P}_{1} {V}_{1} = {P}_{2} {V}_{2}$

Organize the data.

${P}_{1} = \text{1.00 bar}$

${V}_{1} = \text{1.46 L}$

${P}_{2} = \text{60.0 bar}$

${V}_{2} = \text{?}$

Solution

Rearrange the equation to isolate ${V}_{2}$. Substitute the data into the equation and solve.

${V}_{2} = \frac{{P}_{1} {V}_{1}}{P} _ 2$

V_2=(1.00color(red)cancel(color(black)("bar"))xx1.46"L")/(60.0color(red)cancel(color(black)("bar")))="0.0243 L" (rounded to three significant figures)