Answer:
The pressure of a sample of argon gas was increased from 2.12atm to 6.96atm at constant temperature. If the final volume of the argon sample was 16.9L . The initial volume of the argon sample was 55.5L
Explanation:
Considering the initial volume, initial pressure to be V1 and P1 respectively
Assuming the final volume and final temperature of the gas to be V2 and P2 respectively
Given,
Pressure of the gas was increased from 2.12 atm to 6.96 atm
Therefore, P1 = 2.12 atm
P2 = 6.96 atm
Final volume, V2 = 16.9 L
Applying Boyle’s law,
[tex]P1\times V1 = P2\times V2[/tex]
Substituting the value
[tex]2.12 \times (V1) = 6.96\times 16.9[/tex]
[tex]V1 = \frac{(6.96 \times 16.9)}{2.12}[/tex]
V1 = 55.5 L
Therefore, the initial volume was 55.5L
