Chapter 6, Problem 6.104QP

(a)

Interpretation Introduction

Interpretation: The volume of dry ${\text{O}}_{\text{2}}$ gas sample at $25{}^{\text{ο}}\text{C}$ and $\text{1 atm}$ and the volume of dry ${\text{O}}_{\text{2}}$ gas sample at $25{}^{\text{ο}}\text{C}$ and $\text{1 atm}$ if $P_{\text{ethanol}}$ is $\text{50 mmHg}$ at ${\text{25 }}^{\text{ο}}\text{C}$ is to be calculated.

Concept introduction: The relationship between number of moles, pressure, volume and temperature for any gas is described by ideal gas law. The gas should behave as an ideal gas.

To determine: The volume of dry ${\text{O}}_{\text{2}}$ gas sample at $25{}^{\text{ο}}\text{C}$ and $\text{1 atm}$.

Answer to Problem 6.104QP

Solution

The total sample volume of dry ${\text{O}}_{\text{2}}$ gas sample is $\underset{\_}{0.465L}$.

Explanation of Solution

Explanation

Given

The total volume is $\text{0}\text{.480 L}$.

The temperature is $25{}^{\text{ο}}\text{C}$.

The total pressure is $\text{1 atm}$.

The pressure of dry ${\text{O}}_{\text{2}}$ gas sample is $\text{1 atm}$.

The conversion of Celsius to Kelvin is done as,

$\text{0}{}^{\text{ο}}\text{C}=273\text{K}$

Therefore, the conversion of $25{}^{\text{ο}}\text{C}$ to Kelvin is done as,

$\begin{array}{c}25{}^{\text{ο}}\text{C}=25+273\text{K}\\ =298\text{K}\end{array}$

The pressure of ${\text{O}}_{\text{2}}$ ( $P_{{\text{O}}_{\text{2}}}$) is calculated by the formula,

$P_{{\text{O}}_{\text{2}}}=P_{\text{total}}-P_{{\text{H}}_{\text{2}}\text{O}}$

Where,

• $P_{\text{total}}$ is the total pressure.
• $P_{{\text{H}}_{\text{2}}\text{O}}$ is the pressure of water at $25{}^{\text{ο}}\text{C}$.

The value of $P_{{\text{H}}_{\text{2}}\text{O}}=0.031\text{3 atm}$.

Substitute the value of $P_{\text{total}}$ and $P_{{\text{H}}_{\text{2}}\text{O}}$ in the above equation.

$\begin{array}{c}{P}_{{\text{O}}_{\text{2}}}=1.00\text{ atm}-0.031\text{3 atm}\\ \text{=0}\text{.969 atm}\end{array}$

The number of ${\text{O}}_{\text{2}}$ ( $n$) is calculated by the formula,

$n=\frac{PV}{RT}$

Where,

• $P$ is the pressure of ${\text{O}}_{\text{2}}$.
• $V$ is the total volume.
• $R$ is gas constant ( $0.08206\text{L}\cdot \text{atm/mol}\cdot \text{K}$).
• $T$ is temperature.

Substitute the value of $P$, $V$, $R$ and $T$ in the above equation.

$\begin{array}{c}n=\frac{0.96\text{9 atm}\times 0.\text{480 L}}{0.08206\text{L}\cdot \text{atm/mol}\cdot \text{K}\times \text{298 K}}\\ =\text{0}\text{.0190 mol}\end{array}$

The volume of dry ${\text{O}}_{\text{2}}$ ( $V_1$) is calculated by the formula,

$V_1=\frac{nRT}{P_1}$

Where,

• $P_1$ is the pressure of dry ${\text{O}}_{\text{2}}$ gas sample.
• $R$ is gas constant ( $0.08206\text{L}\cdot \text{atm/mol}\cdot \text{K}$).
• $T$ is temperature.

Substitute the value of $P_1$, $n$, $R$ and $T$ in the above equation.

$\begin{array}{c}{V}_1=\frac{0.0\text{19 mol}\times 0.08206\text{L}\cdot \text{atm/mol}\cdot \text{K}\times \text{298 K}}{\text{1 atm}}\\ =\underset{\_}{0.465L}\end{array}$

(b)

Interpretation Introduction

To determine: The volume of dry ${\text{O}}_{\text{2}}$ gas sample at $25{}^{\text{ο}}\text{C}$ and $\text{1 atm}$ if $P_{\text{ethanol}}$ is $\text{50 mmHg}$ at ${\text{25 }}^{\text{ο}}\text{C}$.

Answer to Problem 6.104QP

Solution

The volume of dry ${\text{O}}_{\text{2}}$ gas sample at $25{}^{\text{ο}}\text{C}$ and $\text{1 atm}$ if $P_{\text{ethanol}}$ is $\text{50 mmHg}$ at ${\text{25 }}^{\text{ο}}\text{C}$

is $\underset{\_}{0.497L}$.

Explanation of Solution

Explanation

The pressure of ${\text{O}}_{\text{2}}$ ( $P_{{\text{O}}_{\text{2}}}$) is calculated by the formula,

$P_{{\text{O}}_{\text{2}}}=P_{\text{total}}-P_{\text{ethanol}}$

Where,

• $P_{\text{total}}$ is the total pressure.
• $P_{\text{ethanol}}$ is the pressure of ethanol at $25{}^{\text{ο}}\text{C}$.

The value of $P_{\text{ethanol}}$ is $\text{50 mmHg}$ at ${\text{25 }}^{\text{ο}}\text{C}$.

The conversion of $\text{mmHg}$ to $\text{atm}$ is done as,

$\text{760 mmHg}=1.00\text{ atm}$

Therefore, the conversion of $\text{50 mmHg}$ to $\text{atm}$ is done as,

$\begin{array}{c}\text{50 mmHg}=\frac{\text{50 mmHg}\times 1.00\text{ atm}}{\text{760 mmHg}}\\ =0.0657\text{ atm}\end{array}$

Substitute the value of $P_{\text{total}}$ and $P_{\text{ethanol}}$ in the above equation.

$\begin{array}{c}{P}_{{\text{O}}_{\text{2}}}=1.00\text{ atm}-0.0657\text{ atm}\\ \text{=0}\text{.934 atm}\end{array}$

The volume of dry ${\text{O}}_{\text{2}}$ ( $V_2$) is calculated by the formula,

$V_2=\frac{nRT}{P_2}$

Where,

• $P_2$ is the pressure of dry ${\text{O}}_{\text{2}}$ gas sample when $P_{\text{ethanol}}$ is $\text{50 mmHg}$.
• $R$ is gas constant ( $0.08206\text{L}\cdot \text{atm/mol}\cdot \text{K}$).
• $T$ is temperature.

Substitute the value of $P_2$, $n$, $R$ and $T$ in the above equation.

$\begin{array}{c}{V}_2=\frac{0.0\text{19 mol}\times 0.08206\text{L}\cdot \text{atm/mol}\cdot \text{K}\times \text{298 K}}{\text{0}\text{.934 atm}}\\ =\underset{\_}{0.497L}\end{array}$

Conclusion

The volume of dry ${\text{O}}_{\text{2}}$ gas sample at $25{}^{\text{ο}}\text{C}$ and $\text{1 atm}$ if $P_{\text{ethanol}}$ is $\text{50 mmHg}$ at ${\text{25 }}^{\text{ο}}\text{C}$

is $\underset{\_}{0.497L}$