Chapter 6, Problem 6.80QP

(a)

Interpretation Introduction

Interpretation: Information regarding a sealed, flexible bag of potato chips that is carried from Boston to Denver is given. Various questions based on the changes taking place in volume, pressure in the bag are to be answered.

Concept introduction: According to the Boyle’s law, the pressure of a gas is inversely proportional to volume at the constant temperature and for fixed moles of gas.

According to Gay Lussac’s law, the pressure of the gas is directly proportional of the gas at constant volume and for a fixed number of moles of the gas.

To determine: The final volume of the bag upon arrival in Denver.

Answer to Problem 6.80QP

Solution

The final volume is $\underset{\_}{0.602L}$.

Explanation of Solution

Explanation

Given

The initial volume of the gas is $0.500\text{L}$.

The initial pressure of the gas is $1\text{atm}$.

The final pressure of the gas is $0.83\text{atm}$.

According to Boyle’s law, the relationship between pressure and volume is given as,

$P_1{V}_1=P_2{V}_2$

Where,

• $P_1$ is the initial pressure.
• $P_2$ is the final pressure.
• $V_1$ is the initial volume.
• $V_2$ is the final volume.

Substitute the value of $P_1$, $V_1$ and $P_2$ in the above equation,

$\begin{array}{c}{P}_1{V}_1=P_2{V}_2\\ 1.00\text{}\text{atm}\times 0.500\text{L}=0.83\text{}\text{atm}\times V_2\\ V_2=\underset{\_}{0.602L}\end{array}$

The final volume is $\underset{\_}{0.602L}$.

(b)

Interpretation Introduction

To determine: The pressure in the bag after $10\%$ expansion in the volume of the bag.

Answer to Problem 6.80QP

Solution

The pressure in the bag is $\underset{\_}{0.909atm}$.

Explanation of Solution

Explanation

Given

The initial volume of the gas is $0.500\text{L}$.

The initial pressure of the gas is $1\text{atm}$.

The final volume of the gas is calculated as,

$\begin{array}{c}{V}_2=V_1+\left(V_1\times \frac{10}{100}\right)\\ =0.500\text{L}+\left(0.500\text{L}\times \frac{10}{100}\right)\\ =0.55\text{L}\end{array}$

According to Boyle’s law, the relationship between pressure and volume is given as,

$P_1{V}_1=P_2{V}_2$

Where,

• $P_1$ is the initial pressure.
• $P_2$ is the final pressure.
• $V_1$ is the initial volume.
• $V_2$ is the final volume.

Substitute the value of $P_1$, $V_1$ and $V_2$ in the above equation,

$\begin{array}{c}{P}_1{V}_1=P_2{V}_2\\ 1.00\text{}\text{atm}\times 0.500\text{L}=P_2\times 0.55\text{L}\\ P_2=\underset{\_}{0.909atm}\end{array}$

The pressure in the bag is $\underset{\_}{0.909atm}$.

(c)

Interpretation Introduction

To determine: The pressure in the bag during the flight.

Answer to Problem 6.80QP

Solution

The pressure in the bag during the flight is $\underset{\_}{178.8torr}$.

Explanation of Solution

Explanation

The initial temperature of the gas is $25°\text{C}$

The final pressure of the gas is $\text{126}\text{torr}$.

The conversion of $°\text{C}$ into Kelvin is done as,

$°\text{C}=\left(T°\text{C}+273\right)\text{K}$

Where,

• $T°\text{C}$ is the temperature in degree Celsius.

Hence, the conversion of $25°\text{C}$ into Kelvin is,

$\begin{array}{c}25°\text{C}=\left(25+273\right)\text{K}\\ =298\text{K}\end{array}$

According to Gay Lussac’s law, the relationship between pressure and temperature is given as,

$\frac{P_1}{T_1}=\frac{P_2}{T_2}$

Where,

• $P_1$ is the initial pressure.
• $P_2$ is the final pressure.
• $T_1$ is the initial temperature.
• $T_2$ is the final temperature.

Substitute the value of $T_2$, $P_2$ and $T_1$ in the above equation,

$\begin{array}{c}\frac{P_1}{T_1}=\frac{P_2}{T_2}\\ \frac{P_1}{298\text{K}}=\frac{126\text{torr}}{210\text{K}}\\ P_1=\underset{\_}{178.8torr}\end{array}$

The pressure in the bag during the flight is $\underset{\_}{178.8torr}$.

Conclusion

1. a. The final volume is $\underset{\_}{0.602L}$.
2. b. The pressure in the bag is $\underset{\_}{0.909atm}$.
3. c. The pressure in the bag during the flight is $\underset{\_}{178.8torr}$