# If 7/4 L of a gas at room temperature exerts a pressure of 16 kPa on its container, what pressure will the gas exert if the container's volume changes to 13/12 L?

##### 1 Answer
Jan 26, 2017

The pressure is $= 25.85 k P a$

#### Explanation:

We apply Boyle's Law

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

${P}_{1} = \frac{7}{4} L$

${V}_{1} = 16 k P a$

${V}_{2} = \frac{13}{12} L$

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

$= \frac{\frac{7}{4} \cdot 16}{\frac{13}{12}} = \frac{28}{13} \cdot 12 = 25.85 k P a$