a gas that exerts a pressure of 6.20 atm in a...
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?\nv = ? ml\nassume that the number of moles and the temperature remain constant.
Answer
# Explanation:
## Step1: Write down Boyle's Law formula
Boyle's Law is \(P_1V_1 = P_2V_2\), where \(P_1\) is the initial pressure, \(V_1\) is the initial volume, \(P_2\) is the final pressure, and \(V_2\) is the final volume.
## Step2: Substitute the given values into the formula
We are given \(P_1 = 6.20\space atm\), \(P_2=9.150\space atm\), and \(V_2 = 322\space mL\). Plugging these into \(P_1V_1 = P_2V_2\) gives \(6.20\times V_1=9.150\times322\).
## Step3: Solve for \(V_1\)
First, calculate \(9.150\times322 = 2946.3\). Then, \(V_1=\frac{2946.3}{6.20}\).
\[V_1 = 475.209677\approx475\space mL\]
# Answer:
\(475\space mL\)