Chapter 6, Problem 6.48QP

(a)

Interpretation Introduction

Interpretation: The behavior of $485\text{mL}$ of an ideal gas in response to pressure is studied in a vessel of the gas. The final volume if the pressure of the sample is increased from $715\text{mmHg}$  to $3.55\text{atm}$, decreased from $1.15\text{atm}$ to $520\text{mmHg}$, and increased by $26\%$ is to be calculated.

Concept introduction: An Ideal gas is the theoretical gas whose atoms possess negligible space and no interactions, and which thus obeys the gas laws exactly. The gas laws deal with the behavior of gases with respect to pressure, temperature and volume.

According to Boyle’s law,

$P\alpha \frac{1}{V}$

To determine: The final volume of the gas if pressure is increased from $715\text{mmHg}$ to $3.55\text{atm}$ at constant temperature.

Answer to Problem 6.48QP

Solution

The final volume of the gas is $\underset{\_}{136mL}$ when pressure is increased from $715\text{mmHg}$ to $3.55\text{atm}$ at constant temperature.

Explanation of Solution

Explanation

Given

The initial pressure is $715\text{mmHg}$.

The final pressure is $3.55\text{atm}$.

The initial volume is $485\text{mL}$.

Conversion of $1\text{mmHg}$ into $\text{atm}$ is done as,

$\text{1}\text{mmHg}=\frac{1}{760}\text{atm}$

Therefore, conversion of $715\text{mmHg}$ into $\text{atm}$ is,

$\begin{array}{c}715\text{mmHg}=\frac{1\times 715}{760}\text{atm}\\ =\text{0}\text{.941}\text{atm}\end{array}$

The gas laws deal with the behavior of gases with respect to pressure, temperature and volume.

According to Boyle’s law,

$P\alpha \frac{1}{V}$

It shows that as the volume increases, pressure of gas decreases at constant temperature.

Therefore, the equation is written as,

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

Where,

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

Substitute the given value in above equation,

$\begin{array}{c}{V}_2=\frac{P_1\times V_1}{P_2}\\ =\frac{0.941\text{atm}\times 485\text{mL}}{3.35\text{atm}}\\ =\underset{\_}{136mL}\end{array}$

Therefore, final volume of the gas is $\underset{\_}{136mL}$.

(b)

Interpretation Introduction

To determine: The final volume of the gas if pressure is decreased from $1.15\text{atm}$ to $529\text{mmHg}$ at constant temperature.

Answer to Problem 6.48QP

Solution

The final volume of the gas is $\underset{\_}{815mL}$ when pressure is decreased from $1.15\text{atm}$ to $529\text{mmHg}$ at constant temperature.

Explanation of Solution

Explanation

Given

The initial pressure is $1.15\text{atm}$.

The final pressure is $529\text{mmHg}$.

The initial volume is $485\text{mL}$.

Conversion of $1\text{mmHg}$ into $\text{atm}$ is done as,

$\text{1}\text{mmHg}=\frac{1}{760}\text{atm}$

Therefore, conversion of $529\text{mmHg}$ into $\text{atm}$ is,

$\begin{array}{c}529\text{mmHg}=\frac{1\times 529}{760}\text{atm}\\ =\text{0}\text{.684}\text{atm}\end{array}$

The gas laws deal with the behavior of gases with respect to pressure, temperature and volume.

According to Boyle’s law,

$P\alpha \frac{1}{V}$

It shows that as the volume increases, pressure of gas decreases at constant temperature.

Therefore, the equation is written as,

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

Where,

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

Substitute the given value in above equation,

$\begin{array}{c}{V}_2=\frac{P_1\times V_1}{P_2}\\ =\frac{1.15\text{atm}\times 485\text{mL}}{0.684\text{atm}}\\ =\underset{\_}{815mL}\end{array}$

Therefore, final volume of the gas is $\underset{\_}{815mL}$.

(c)

Interpretation Introduction

To determine: The final volume of the gas if pressure is increased by $26\%$ at constant temperature.

Answer to Problem 6.48QP

Solution

The final volume of the gas is $\underset{\_}{385mL}$ when pressure is increased by $26\%$ at constant temperature.

Explanation of Solution

Explanation

Given

The initial pressure is $P_1$.

The initial volume is $485\text{mL}$.

The final pressure is increased by $26\%$.

The gas laws deal with the behavior of gases with respect to pressure, temperature and volume.

According to Boyle’s law,

$P\alpha \frac{1}{V}$

It shows that as the volume increases, pressure of gas decreases at constant temperature.

Therefore, the equation is written as,

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

Where,

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

It is given that the pressure is increased by $26\%$ at constant temperature which means,

$\begin{array}{c}{P}_2=P_1+\left(P_1\times \frac{26}{100}\right)\\ =P_1+\frac{13P_1}{50}\\ =\frac{63P_1}{50}\end{array}$

Substitute the given value in above equation,

$\begin{array}{c}{V}_2=\frac{P_1\times V_1}{P_2}\\ =\frac{P_1\times 485}{63P_1/50}\\ =\underset{\_}{385mL}\end{array}$

Therefore, final volume of the gas is $\underset{\_}{385mL}$.

Conclusion

1. a. The final volume of the gas is $\underset{\_}{136mL}$ when pressure is increased from $715\text{mmHg}$ to $3.55\text{atm}$ at constant temperature.
2. b. The final volume of the gas is $\underset{\_}{815mL}$ when pressure is decreased from $1.15\text{atm}$ to $529\text{mmHg}$ at constant temperature.
3. c. The final volume of the gas is $\underset{\_}{385mL}$ when pressure is increased by $26\%$ at constant temperature