Answer:
The pressure of argon inside the 155 ml flask is approximately 633 mm Hg.
Explanation:
Given:
The temperature is constant. So, the process is isothermal.
Initial pressure of argon is, [tex]P_1=785\ mm\ Hg[/tex]
Initial volume of argon is, [tex]V_1=125\ ml[/tex]
Final volume of argon is, [tex]V_2=155\ ml[/tex]
Final pressure of argon is, [tex]P_2=?[/tex]
Now, as per Boyle's law, for an isothermal process, the product of pressure and volume is always a constant or [tex]PV=constant[/tex]. Therefore, we can write it as:
[tex]P_1V_1=constant\\P_2V_2=constant\\\therefore P_1V_1=P_2V_2[/tex]
Rewriting the above in terms of [tex]P_2[/tex], we have:
[tex]P_2=\frac{P_1V_1}{V_2}[/tex]
Plug in 785 for [tex]P_1[/tex], 125 for [tex]V_1[/tex], 155 for [tex]V_2[/tex] and solve for [tex]P_2[/tex]. This gives,
[tex]P_2=\frac{785\times 125}{155}\\P_2=633.06\approx 633\ mm\ Hg[/tex]
Therefore, the pressure of argon inside the 155 ml flask is approximately 633 mm Hg.