# A gas at a pressure of "203 kPa" occupies "40.0 L". If the pressure is decreased to "35.0 kPa", what is the new volume? Assume the temperature is held constant.

Jan 4, 2017

The final volume of helium gas will be $\text{232 L}$.

#### Explanation:

This is an example of Boyle's law, which states that the volume of a given amount of gas, held at a constant temperature, will vary inversely with its pressure. This means that when the pressure is increased, the volume will decrease and vice versa

The equation to use is "P_1V_1=P_2V_2, where $P$ represents pressure, and $V$ represents volume.

Known
${P}_{1} = \text{203 kPa}$
${V}_{1} = \text{40.0 L}$
${P}_{2} = \text{35.0 kPa}$

Unknown
${V}_{2}$

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

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

V_2=(203cancel"kPa"xx40.0"L")/(35.0cancel"kPa")="232 L"

The final volume of helium gas will be $\text{232 L}$.