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 30.0 mL and contains 4.49 atm of argon, the second bulb has a volume of 250 mL and contains 2.64 atm of neon, and the third bulb has a volume of 21.0 mL and contains 9.36 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?

The final pressure of the whole system is 2.849 atm

Explanation:

According to Boyle's law, for eacht gas applies the following equation:

PiVi = PfVf or Pf = PiVi/Vf

with Pi = the initial pressure

with Vi = the initial volume

with Pf = the final pressure

with Vf = the final volume

The final volume for each gas is (47.0 + 30.0 + 250 + 21.0) mL, which is in total = 348 mL

2. For the first bulb, with argon:

Pf = (4.49 atm) (30.0 mL) / (348 mL)

= 0.387 atm

For the second bulb, with neon

Pf = (2.64 atm) (250 mL) / (348 mL)

= 1.897 atm

For the third bulb, with hydrogen:

Pf = (9.36 atm) (21 mL) / (348 mL)

= 0.565 atm

Step 3:Calculate the final pressure

P total = P (Ar) + P (Ne) + P (H2)

= (0.387 + 1.897 + 0.565) atm

= 2.849 atm

The final pressure of the whole system is 2.849 atm