Chapter 6, Problem 6.49QP

(a)

Interpretation Introduction

Interpretation: The final pressure of the given ammonia gas corresponding to the given changes in the volume 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 pressure of ammonia gas if volume is gradually decreased from $78.0\text{mL}$ to $39.0\text{mL}$ at constant temperature.

Answer to Problem 6.49QP

Solution

The final pressure of ammonia gas is $\underset{\_}{2.00atm}$ when volume is gradually decreased from $78.0\text{mL}$ to $39.0\text{mL}$ at constant temperature.

Explanation of Solution

Explanation

Given

The initial pressure is $1.00\text{atm}$.

The initial volume is $78.0\text{mL}$.

The final volume is $39.0\text{mL}$.

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,

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

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,

$\begin{array}{c}{P}_2=\frac{P_1\times V_1}{V_2}\\ =\frac{1.00\text{atm}\times 78.0\text{mL}}{39.0\text{mL}}\\ =\underset{\_}{2.00atm}\end{array}$

Therefore, final pressure of ammonia gas is $\underset{\_}{2.00atm}$.

(b)

Interpretation Introduction

To determine: The final pressure of ammonia gas if volume is increased from $43.5\text{mL}$ to $65.5\text{mL}$ at constant temperature.

Answer to Problem 6.49QP

Solution

The final pressure of ammonia gas is $\underset{\_}{0.664atm}$ when volume is increased from $43.5\text{mL}$ to $65.5\text{mL}$ at constant temperature.

Explanation of Solution

Explanation

Given

The initial pressure is $1.00\text{atm}$.

The initial volume is $43.5\text{mL}$.

The final volume is $65.5\text{mL}$.

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,

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

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,

$\begin{array}{c}{P}_2=\frac{P_1\times V_1}{V_2}\\ =\frac{1.00\text{atm}\times 43.5\text{mL}}{65.5\text{mL}}\\ =\underset{\_}{0.644atm}\end{array}$

Therefore, final pressure of ammonia gas is $\underset{\_}{0.644atm}$.

(c)

Interpretation Introduction

To determine: The final pressure of ammonia gas if volume is increased by $40\%$ at constant temperature.

Answer to Problem 6.49QP

Solution

The final pressure of ammonia gas is $\underset{\_}{1.67atm}$ when volume is increased by $40\%$ at constant temperature.

Explanation of Solution

Explanation

Given

The initial pressure is $1.00\text{atm}$.

The initial volume is $V_1$.

The final volume is $40\%$.

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,

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

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 volume is increased by $40\%$ at constant temperature which means,

$\begin{array}{c}{V}_2=V_1-\left(V_1\times \frac{40}{100}\right)\\ =V_1-\frac{2V_1}{5}\\ =\frac{3V_1}{5}\end{array}$

Substitute the given value in above equation,

$\begin{array}{c}{P}_2=\frac{P_1\times V_1}{V_2}\\ =\frac{1.00\text{atm}\times V_1}{3V_1/5}\\ =\underset{\_}{1.67atm}\end{array}$

Therefore, final pressure of ammonia gas is $\underset{\_}{1.67atm}$.

Conclusion

1. a. The final pressure of ammonia gas is $\underset{\_}{2.00atm}$ when volume is gradually decreased from $78.0\text{mL}$ to $39.0\text{mL}$ at constant temperature.
2. b. The final pressure of ammonia gas is $\underset{\_}{0.664atm}$ when volume is increased from $43.5\text{mL}$ to $65.5\text{mL}$ at constant temperature.
3. c. The final pressure of ammonia gas is $\underset{\_}{1.67atm}$ when volume is increased by $40\%$ at constant temperature