Chapter U3.11, Problem 5E

(a)

Interpretation Introduction

Interpretation: The volume of air inside the bottle needs to be calculate, if the volume of glass bottle is 180 mL at 1.0 atm and 25°C temperature.

Concept Introduction:

Due to random movement of gas particles, they colloid with other gas particles and also colloid with wall of container. The collision between gas particles of air and wall of container exert pressure on the wall of container.

The Boyle’s law states that at constant temperature and amount of gas molecules, the volume is inversely proportional to the pressure of gas.

  $\text{Pressure }\propto \frac{\text{1}}{\text{volume }}$

Answer to Problem 5E

Since the pressure is inversely proportional to the volume therefore the volume of air inside the bottle should increase.

Explanation of Solution

• Initial volume = 180 mL
• Initial temperature =25°C
• Initial pressure = 1.0 atm
• Final pressure = 0.75 atm
• Final temperature = 5°C

The air pressure outside the glass bottle decreases at mountain. Since the pressure is inversely proportional to the volume therefore the volume of air inside the bottle should increase.

(b)

Interpretation Introduction

Interpretation: The temperature of air inside the bottle at the top of mountain to cool to 5°C needs to be determined, if the volume of glass bottle is 180 mL at 1.0 atm and 25°C temperature.

Concept Introduction:

Due to random movement of gas particles, they colloid with other gas particles and also colloid with wall of container. The collision between gas particles of air and wall of container exert pressure on the wall of container.

The Boyle’s law states that at constant temperature and amount of gas molecules, the volume is inversely proportional to the pressure of gas.

  $\text{Pressure }\propto \frac{\text{1}}{\text{volume }}$

Answer to Problem 5E

Yes, the temperature of inside air will decrease as volume is increases and pressure is decreases.

Explanation of Solution

• Initial volume = 180 mL
• Initial temperature =25°C
• Initial pressure = 1.0 atm
• Final pressure = 0.75 atm
• Final temperature = 5°C

Yes, the temperature of inside air will decrease as volume is increases and pressure is decreases. As pressure decreases the gas is expanded and expansion of gas requires energy that will decrease the temperature of inside air.

(c)

Interpretation Introduction

Interpretation: The pressure of air inside the bottle needs to be determined, if the volume of glass bottle is 180 mL at 1.0 atm and 25°C temperature.

Concept Introduction:

Due to random movement of gas particles, they colloid with other gas particles and also colloid with wall of container. The collision between gas particles of air and wall of container exert pressure on the wall of container.

The Boyle’s law states that at constant temperature and amount of gas molecules, the volume is inversely proportional to the pressure of gas.

  $\text{Pressure }\propto \frac{\text{1}}{\text{volume }}$

Answer to Problem 5E

The outside pressure decreases at mountain so inside pressure must increases to balance it.

Explanation of Solution

The outside pressure decreases for bottle therefore the inside pressure will increase the balance the outside pressure. It will increase the volume of inside air in the glass bottle.

(d)

Interpretation Introduction

Interpretation: The new pressure of gas needs to be determined, if the volume of glass bottle is 180 mL at 1.0 atm and 25°C temperature.

Concept Introduction:

Due to random movement of gas particles, they colloid with other gas particles and also colloid with wall of container. The collision between gas particles of air and wall of container exert pressure on the wall of container.

The Boyle’s law states that at constant temperature and amount of gas molecules, the volume is inversely proportional to the pressure of gas.

  $\text{Pressure α }\frac{\text{1}}{\text{volume }}$

Answer to Problem 5E

New pressure = 0.93 atm

Explanation of Solution

• Initial volume = 180 mL = 0.180 L
• Initial temperature =25°C = 273 +25 = 298 K
• Initial pressure = 1.0 atm
• Final pressure = 0.75 atm
• Final temperature = 5°C = 273 +5 = 278 K

At constant volume, the new pressure must be:

  $\begin{array}{l}\frac{{\text{P}}_{\text{1}}}{{\text{T}}_{\text{1}}}\text{ =}\frac{{\text{ P}}_{\text{2}}}{{\text{T}}_{\text{2}}}\\ \frac{\text{1}\text{.0 atm }}{\text{298 K}}\text{ =}\frac{{\text{ P}}_{\text{2}}}{278K}\\ {\text{P}}_{\text{2}}\text{=}\frac{\text{1}\text{.0 atm }\times 278K}{\text{298 K}}\\ {\text{P}}_{\text{2}}\text{= 0}\text{.93 atm}\end{array}$