Chapter 5, Problem 5.117QP

(a)

Interpretation Introduction

Interpretation:

The pressures in valves (b) and (c) has to be calculated.

Answer to Problem 5.117QP

The partial pressures in valves (b) and (c) is

${\text{P}}_{\text{B}}=\text{8}\text{.0}\text{atm}$

${\text{P}}_{\text{C}}\text{=}\text{5}\text{.3}\text{atm}$

Explanation of Solution

Bulb (b) contains the same number of particles as (a), but is half the volume.  The pressure will be double that of A.

${\text{P}}_{\text{B}}\text{=}\text{(2)(4}\text{.0}\text{atm)}\text{=}\text{8}\text{.0}\text{atm}$

The volume of bulb (c) is the same as bulb (a), but there are $\text{12}$ particles in (c) while there are 9 particles in (a).  The pressure is directly proportional to the number of moles of gas (or particles of gas) at the same temperature and volume.

${\text{P}}_{\text{C}}\text{=}\text{(}\frac{\text{12}}{\text{9}}\text{)(4}\text{.0}\text{atm)}\text{=}\text{5}\text{.3}\text{atm}$

(b)

Interpretation Introduction

Interpretation:

The total pressure and partial pressure of each gas after the valves are opened has to be calculated.

Answer to Problem 5.117QP

The total pressure in the container is the sum of the partial pressures is ${\text{P}}_{\text{total}}\text{=}\text{5}\text{.3}\text{atm}$

The partial pressure is ${\text{P}}_{\text{A}}\text{=}{\text{P}}_{\text{B}}\text{=}\text{2}\text{.65 atm}$

Explanation of Solution

The valves are opened at constant temperature, the gases expand to fill the entire container.  In bulb (a), the pressure before opening the valve is $\text{4}\text{.0}\text{atm}$ and the volume is 4.0 L.  After the valves are opened, the volume is now $\text{10 L }\left(\text{4 L + 2 L + 4 L}\right)\text{.}$ We use Boyle’s law to calculate the partial pressure of each sample of gas after the valves are opened.

Given data,

${\text{P}}_{\text{1}}$ Valves pressure is $\text{4}\text{.0}\text{atm}$

${\text{V}}_{\text{1}}\text{=}\text{4}\text{.0L}$

${\text{V}}_2\text{=}10.\text{0L}$

$\begin{array}{l}\text{(a)}{\text{P}}_{\text{1}}{\text{V}}_{\text{1}}{\text{=P}}_{\text{2}}{\text{V}}_{\text{2}}\\ \text{(4}\text{.0}\text{atm)(4}\text{.0}\cancel{\text{L}}\text{)}\text{=}{\text{P}}_{\text{2}}\text{(10}\text{.0}\cancel{\text{L}}\text{)}\\ {\text{P}}_{\text{2}}\text{=}\text{1}\text{.6}\text{atm}\end{array}$

Given data,

${\text{P}}_{\text{1}}$ Valves pressure is $\text{8}\text{.0}\text{atm}$

${\text{V}}_{\text{1}}\text{=}2.\text{0L}$

${\text{V}}_2\text{=}10.\text{0L}$

$\begin{array}{l}\text{(b)}{\text{P}}_{\text{1}}{\text{V}}_{\text{1}}{\text{=P}}_{\text{2}}{\text{V}}_{\text{2}}\\ \text{(8}\text{.0}\text{atm)(2}\text{.0}\cancel{\text{L}}\text{)}\text{=}{\text{P}}_{\text{2}}\text{(10}\text{.0}\cancel{\text{L}}\text{)}\\ {\text{P}}_{\text{2}}\text{=}\text{1}\text{.6}\text{atm}\end{array}$

Given data,

${\text{P}}_{\text{1}}$ Valves pressure is $\text{5}\text{.3 }\text{atm}$

${\text{V}}_{\text{1}}\text{=}4.\text{0L}$

${\text{V}}_2\text{=}10.\text{0L}$

$\begin{array}{l}\text{(c)}{\text{P}}_{\text{1}}{\text{V}}_{\text{1}}{\text{=P}}_{\text{2}}{\text{V}}_{\text{2}}\\ \text{(5}\text{.3}\text{atm)(4}\text{.0}\cancel{\text{L}}\text{)}\text{=}{\text{P}}_{\text{2}}\text{(10}\text{.0}\cancel{\text{L}}\text{)}\\ {\text{P}}_{\text{2}}\text{=}2.1\text{atm}\end{array}$

The total pressure in the container is the sum of the partial pressures

${\text{P}}_{\text{total}}\text{=}\text{1}\text{.6}\text{atm+}\text{1}\text{.6}\text{atm}\text{+}\text{2}\text{.1}\text{atm}\text{=}\text{5}\text{.3}\text{atm}$

There are 15 particles of gas A and 15 particles of gas B in the container. Therefore, the partial pressure of each gas will be half the total pressure of $\text{5}\text{.3}\text{atm}$

${\text{P}}_{\text{A}}\text{=}{\text{P}}_{\text{B}}\text{=}\text{2}\text{.65 atm}$