# If 9/4 L of a gas at room temperature exerts a pressure of 42 kPa on its container, what pressure will the gas exert if the container's volume changes to 12/7 L?

Nov 19, 2016

The pressure is $= \frac{441}{8} k P a$

#### Explanation:

Here, we use Boyle's Law.

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

${P}_{1} = 42 k P a$

${V}_{1} = \frac{9}{4} l$

${V}_{2} = \frac{12}{7} l$

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

${P}_{2} = 42 \cdot \frac{\frac{9}{4}}{\frac{12}{7}} = \frac{42 \cdot 9 \cdot 7}{12 \cdot 4} = \frac{441}{8} k P a$