# If 32 L of a gas at room temperature exerts a pressure of 64 kPa on its container, what pressure will the gas exert if the container's volume changes to 16 L?

Jul 9, 2016

The Pressure will change to $128 k P a$

#### Explanation:

To solve this problem we can use Boyle's Law comparing the Pressure and Volume relationships

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

${P}_{1} = 64 k P a$
${V}_{1} = 32 L$
P_2 = ?
${V}_{2} = 16 L$

(64kPa)((32L) = P_2(16L)

Use the multiplicative inverse to isolate the variable for ${P}_{2}$

$\left(64 k P a\right) \left(\frac{32 \cancel{L}}{16 \cancel{L}}\right) = {P}_{2} \frac{\cancel{16 L}}{\cancel{16 L}}$

#(64kPa)(2) = P_2

$128 k P a = {P}_{2}$

More molecules in a small volume means more collisions and therefore more pressure.