# If 9/4 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?

Oct 12, 2017

$14.5 k P a$

#### Explanation:

Boyle's law states that for an ideal gas $P \setminus \propto \frac{1}{V}$, or $P = \frac{k}{V}$ where $k$ is a constant.

In this question we are given the pressure and volume, but not the constant, though it is easy to work out.

$P V = k$

$15 k P a = 1.5 \cdot {10}^{4} P a$

$\frac{9}{4} L = 2.25 L = 2.25 \cdot {10}^{- 3} {m}^{3}$

Therefore, $\left(1.5 \cdot {10}^{4}\right) \left(2.25 \cdot {10}^{- 3}\right) = 33.75$

Our new volume of $\frac{7}{3} L$ gives us the equation of $P = \frac{33.75}{\frac{7}{3000}} = \frac{101250}{7} \approx 14 , 464 P a = 14.464 k P a \approx 14.5 k P a$