# If 7/5 L of a gas at room temperature exerts a pressure of 3 kPa on its container, what pressure will the gas exert if the container's volume changes to 2/3 L?

Nov 28, 2016

The final pressure will be $\frac{63}{10} \text{kPa}$$=$$\text{6.3 kPa}$

#### Explanation:

This is a question involving Boyle's law, which states that the volume of a gas varies inversely with its pressure, as long as the amount and temperature are held constant.

${P}_{1} = \text{3 kPa}$
${V}_{1} = \text{7/5 L}$
${V}_{2} = \text{2/3 L}$
${P}_{2} = \text{???}$

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

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

${P}_{2} = \left(3 \text{kPa"·7/5cancel"L")/(2/3 cancel"L}\right)$

Simplify.

${P}_{2} = \frac{\frac{21}{5} \text{kPa}}{\frac{2}{3}}$

Multiply by the reciprocal of $\frac{2}{3}$.

P_2=(21/5"kPa")·(3/2)

Simplify.

${P}_{2} = \frac{63}{10} \text{kPa}$$=$$\text{6.3 kPa}$