Chapter 5, Problem 5.58QP

(a)

Interpretation Introduction

Interpretation:

The total partial pressures of the given mixture at STP has to be calculated.

Concept Introduction:

• In a mixture of gases, every gas has an incomplete pressure which is the theoretical stress of that gas if it alone engaged the entire volume of the original combination at the same temperature.
• The sum pressure of an ideal gas mixture is the amount of the partial pressures of the gases in the mixture.

$\frac{{\text{P}}_{\text{1}}{\text{V}}_{\text{1}}}{{\text{n}}_{\text{1}}{\text{T}}_{\text{1}}}\text{=}\frac{{\text{P}}_{\text{2}}{\text{V}}_{\text{2}}}{{\text{n}}_{\text{2}}{\text{T}}_{\text{2}}}$

${\text{V}}_{\text{2}}\text{=}\frac{{\text{P}}_{\text{1}}{\text{V}}_{\text{1}}{\text{T}}_{\text{2}}}{{\text{P}}_{\text{2}}{\text{T}}_{\text{1}}}$

• STP in chemistry is the short form for Standard Temperature and Pressure.  STP most usually is used when performing arts calculations on gases, such as gas density.  The normal temperature is $\text{273 K }\left(\text{0° Celsius or 32° Fahrenheit}\right)$ and the standard pressure is $\text{1 atm}$ pressure.  This is the freezing point of pure water at sea level atmospheric pressure.

Answer to Problem 5.58QP

The total pressure of the mixture is $\text{0}\text{.89}\text{atm}$

Explanation of Solution

Dalton’s law states that the total pressure of the mixture is the sum of the partial pressures.

${\text{P}}_{\text{total}}\text{=}\text{0}\text{.32}\text{atm}\text{+}\text{0}\text{.15}\text{atm}\text{+}\text{0}\text{.42}\text{atm}\text{=}\text{0}\text{.89}\text{atm}$

(b)

Interpretation Introduction

Interpretation:

The volume in litres at STP has to be calculated.

Concept Introduction:

• In a mixture of gases, every gas has an incomplete pressure which is the theoretical stress of that gas if it alone engaged the entire volume of the original combination at the same temperature.
• The sum pressure of an ideal gas mixture is the amount of the partial pressures of the gases in the mixture.
  • $\frac{{\text{P}}_{\text{1}}{\text{V}}_{\text{1}}}{{\text{n}}_{\text{1}}{\text{T}}_{\text{1}}}\text{=}\frac{{\text{P}}_{\text{2}}{\text{V}}_{\text{2}}}{{\text{n}}_{\text{2}}{\text{T}}_{\text{2}}}$
  • ${\text{V}}_{\text{2}}\text{=}\frac{{\text{P}}_{\text{1}}{\text{V}}_{\text{1}}{\text{T}}_{\text{2}}}{{\text{P}}_{\text{2}}{\text{T}}_{\text{1}}}$
• STP in chemistry is the short form for Standard Temperature and Pressure.  STP most usually is used when performing arts calculations on gases, such as gas density.  The normal temperature is $\text{273 K }\left(\text{0° Celsius or 32° Fahrenheit}\right)$ and the standard pressure is $\text{1 atm}$ pressure.  This is the freezing point of pure water at sea level atmospheric pressure.

Answer to Problem 5.58QP

The volume of given gas is $\text{1}\text{.4}\text{L}\text{at}\text{STP}$

Explanation of Solution

$\begin{array}{l}\text{InitialConditions}\text{Final}\text{pressure}\\ {\text{P}}_{\text{1}}\text{=}\text{0}\text{.15}\text{atm}\text{+}\text{0}\text{.42}\text{atm}\text{=}\text{0}\text{.57}\text{atm}{\text{P}}_{\text{2}}\text{1}\text{.0}\text{atm}\\ {\text{T}}_{\text{1}}\text{=}\text{15°C}\text{+}\text{273}\text{=}\text{288}\text{K}{\text{T}}_{\text{2}}\text{=}\text{273}\text{K}\end{array}$

The volume

$\frac{{\text{P}}_{\text{1}}{\text{V}}_{\text{1}}}{{\text{n}}_{\text{1}}{\text{T}}_{\text{1}}}\text{=}\frac{{\text{P}}_{\text{2}}{\text{V}}_{\text{2}}}{{\text{n}}_{\text{2}}{\text{T}}_{\text{2}}}$

$\begin{array}{l}\text{Because}{\text{n}}_{\text{1}}{\text{=n}}_{\text{2}}\text{,}\text{we}\text{can}\text{write:}\\ {\text{V}}_{\text{2}}\text{=}\frac{{\text{P}}_{\text{1}}{\text{V}}_{\text{1}}{\text{T}}_{\text{2}}}{{\text{P}}_{\text{2}}{\text{T}}_{\text{1}}}\\ {\text{V}}_{\text{2}}\text{=}\frac{\text{(0}\text{.57}\text{atm}\text{)}\left(\text{2}\text{.5}\text{L}\right)\left(\text{273K}\right)}{\text{(1}\text{.0atm)(288K)}}\text{=1}\text{.4}\text{L}\text{at}\text{STP}\\ \end{array}$

The volume of given gas is calculated by plugging in the values of the given pressure of gas and temperature of gas.  The volume of given gas was found to be $\text{1}\text{.4}\text{L}\text{at}\text{STP}$.