Three bulbs are connected by tubing, and the tubing is evacuated. The volume of the tubing is 47.0 mL. The first bulb has a volume of 51.0 mL and contains 7.25 atm of argon, the second bulb has a volume of 250 mL and contains 1.18 atm of neon, and the third bulb has a volume of 45.0 mL and contains 5.96 atm of hydrogen. If the stopcocks (valves) that isolate all three bulbs are opened, what is the final pressure of the whole system in atm?

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Mergus Mergus · Answer 1 · 2019-11-06T06:01:59+01:00

Answer:

2.30 atm

Explanation:

Using Boyle's law :-

[tex]P\times V=Constant[/tex]

This can be written as:-

[tex] {P_1}\times {V_1}+{P_2}\times {V_2}+{P_3}\times {V_3}={P}\times {V}[/tex]

Given ,

For Bulb 1

Pressure = 7.25 atm

Volume = 51.0 mL

For Bulb 2

Pressure = 1.18 atm

Volume = 250 mL

For Bulb 3

Volume = 45.0 mL

Pressure = 5.96 atm

Given that the volume of the tubing = 47.0 mL

Volume of the whole system = Sum of the volume of the three bulbs + Volume of the tubing

Also, Total volume, V = 51.0 + 250 + 45.0 +47.0 mL = 393 mL

Using above equation as:

[tex] {7.25}\times {47.0}+{1.18}\times {250}+{5.96}\times {45.0}={P}\times {393}[/tex]

We get, Total pressure, P = 2.30 atm