Three glass bulbs, joined by closed stopcocks, have the following volumes and initial pressures of the specified gases.
Bulb A: 200. mL of Kr(g) at 190. torr
Bulb B: 400. mL of H2S(g) at 1.00 atm
Bulb C: 1.00 L of N2(g) at 75.994 kPa

After both stopcocks are opened and the gases allowed to diffuse throughout, what will be the ultimate total pressure?

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Mergus Mergus · Answer 1 · 2019-11-05T04:14:46+01:00

Answer:

P = 0.75 atm

Explanation:

Using Boyle's law

[tex] {P_A}\times {V_A}+{P_B}\times {V_B}+{P_C}\times {V_C}={P}\times {V}[/tex]

Given ,

For Bulb A

Pressure = 190 torr

The conversion of P(torr) to P(atm) is shown below:

[tex]P(torr)=\frac {1}{760}\times P(atm)[/tex]

So,

Pressure = 190 / 760 atm = 0.25 atm

Volume = 200 mL

For Bulb B

Pressure = 1 atm

Volume = 400 mL

For Bulb C

Volume = 1.00 L = 1000 mL

Pressure = 75.994 kPa

The expression for the conversion of pressure in KiloPascal to pressure in atm is shown below:

P (kPa) = [tex]\frac {1}{101.325}[/tex] P (atm)

75.994 kPa = [tex]\frac {75.994}{101.325}[/tex] atm

Pressure = 0.75 atm

Also, Total volume, V = 200 + 400 + 1000 mL = 1600 mL

Using above equation as:

[tex] {0.25}\times {200}+{1.00}\times {400}+{0.75}\times {1000}={P}\times {1600}[/tex]

We get, Total pressure, P = 0.75 atm