Question

Three glass bulbs, joined by closed stopcocks, have the following volumes and initial pressures of the specified gases.

Answer 1

Answer:

P = 0.75 atm

Explanation:

Using Boyle's law  

${P_A}\times {V_A}+{P_B}\times {V_B}+{P_C}\times {V_C}={P}\times {V}$

Given ,  

For Bulb A

Pressure = 190 torr

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

$P(torr)=\frac {1}{760}\times P(atm)$

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) = $\frac {1}{101.325}$ P (atm)

75.994 kPa = $\frac {75.994}{101.325}$ atm

Pressure = 0.75 atm

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

Using above equation as:

${0.25}\times {200}+{1.00}\times {400}+{0.75}\times {1000}={P}\times {1600}$

We get, Total pressure, P = 0.75 atm