# If 18 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 24 L?

Mar 30, 2016

${P}_{2} = 64 \cdot \frac{18}{24} = \frac{3}{4} \cdot 64 = 48 k P a$

#### Explanation:

Given: Container with initial, ${V}_{1} = 18 L \mathmr{and} {P}_{1} = 64 k P a$
Final ${V}_{2} = 24 L$
Required: New Pressure
Solution Strategy: Straight application of Boyle's Law
P_1V_1=P_2V_2; P_2=P_1V_1/V_2 substituting the known values
${P}_{2} = 64 \cdot \frac{18}{24} = \frac{3}{4} \cdot 64 = 48 k P a$