# If 9/5 L of a gas at room temperature exerts a pressure of 15 kPa on its container, what pressure will the gas exert if the container's volume changes to 7/3 L?

Dec 26, 2016

We use the equation ${p}_{1} {V}_{1} = {p}_{2} {V}_{2}$

#### Explanation:

$15 \times \frac{9}{5} = {p}_{2} \times \frac{7}{3}$

Divide by $\frac{7}{3}$ = multiply by $\frac{3}{7}$:

$15 \times \frac{9}{5} \times \frac{3}{7} = {p}_{2} \cdot \cancel{\frac{7}{3}} \times \cancel{\frac{3}{7}}$

$\to {p}_{2} = \frac{3 \times \cancel{5} \times 9 \times 3}{\cancel{5} \times 7} = \frac{81}{7} \approx 11.6 k P a$