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

Mar 2, 2017

Answer:

The pressure is $= 22.5 k P a$

Explanation:

We apply Boyle's Law

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

${P}_{1} = 5 k P a$

${V}_{1} = 3 L$

${V}_{2} = \frac{2}{3} L$

${P}_{2} = {V}_{1} / {V}_{2} \cdot {P}_{1}$

$= \frac{3}{\frac{2}{3}} \cdot 5 = \frac{45}{2} = 22.5 k P a$