Question

a gas that exerts a pressure of 6.20 atm in a container with an unknown volume will exert a pressure of 9.150 atm when transferred to a container with a volume of 322 ml. what is the initial volume of the gas?
v = ? ml
assume that the number of moles and the temperature remain constant.

Explanation:

Step1: Use Boyle's Law formula

$P_1V_1 = P_2V_2$
Given: $P_1=6.20$ atm, $P_2=9.150$ atm, $V_2=322$ mL, find $V_1$.

Step2: Rearrange for $V_1$

$V_1 = \frac{P_2V_2}{P_1}$

Step3: Calculate $V_1$

$V_1 = \frac{9.150 \times 322}{6.20} = \frac{2946.3}{6.20} ≈ 475.21$ mL

Answer:

475 mL (rounded to reasonable significant figures)