Question

Three bulbs are connected by tubing, and the tubing is evacuated. The volume of the tubing is 45.0 mL. The first bulb has a volume of 77.0 mL and contains 8.89 atm of argon, the second bulb has a volume of 250 mL and contains 2.82 atm of neon, and the third bulb has a volume of 21.0 mL and contains 8.42 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

Answer 1

Answer:

The final pressure of the whole system is 34.80 atm.

Explanation:

Given that,

Volume = 45.0 ml

Volume of first bulb = 77.0 mL

Pressure  = 8.89 atm

Volume of second  bulb = 250 mL

Pressure = 2.82 atm

Volume of third  bulb = 21.0 mL

Pressure = 8.42 atm

We need to calculate the final pressure of the whole system

Using formula of pressure

$P_{1}V_{1}+P_{2}V_{2}+P_{3}V_{3}+P_{t}V_{t}=P_{f}V_{f}$

Where, $P_{1}$= pressure of first bulb

$P_{2}$= pressure of second bulb

$P_{3}$= pressure of third bulb

$P_{4}$= initial pressure of tube

$V_{1}$= Volume of first bulb

$V_{2}$=Volume of second bulb

$V_{3}$= Volume of third bulb

$V_{4}$= Initial volume of tube

Put the value into the formula

$8.89\times77.0+250\times2.82+21.0\times8.42+0=P_{f}\times45$

$P_{f}=\dfrac{1566.35}{45}$

$P_{f}=34.80\ atm$

Hence, The final pressure of the whole system is 34.80 atm.