Chapter U3.61, Problem 5E

(a)

Interpretation Introduction

Interpretation: Whether the volume of the balloon increases or decreases need to be explained.

Concept Introduction: The combined gas law is combination of three gas laws that is Boyle’s law, Charles’ law and Gay-Lussac’s law. According to this law the ratio of temperature to the product of pressure and volume is a constant.

Explanation of Solution

The volume and the pressure of the helium gas at 25 ^oC is 1.5 L and 1.0 atm respectively. The new pressure and temperature of the helium gas is 0.95 atm and 20 ^oC respectively.

According to combined gas law, the ratio of temperature to the product of pressure and volume is a constant.

As temperature decreases and pressure also decreases, the volume of the gas does not change much. Since, the change in temperature is slightly more than pressure thus, new volume of the gas will be slightly more than the initial volume.

(b)

Interpretation Introduction

Interpretation: The equation that can be used to calculate the new volume of the balloon needs to be determined.

Concept Introduction: The combined gas law is combination of three gas laws that is Boyle’s law, Charles’ law and Gay-Lussac’s law. According to this law the ratio of temperature to the product of pressure and volume is a constant.

Explanation of Solution

The three gas laws are explained as follows:

Boyle’s law:

This law explains the change in pressure of the gas with volume of the container at constant temperature. As volume of the container decreases, pressure tends to increase.

This is mathematically represented as follows:

  $P_1{V}_1=P_2{V}_2$

Here, P₁ is initial pressure, P₂ is final pressure, V₁ is initial volume and V₂ is final volume.

Charles’s law:

At constant pressure, volume and temperature of the gas are directly proportional to each other.

This is mathematically represented as follows:

  $\frac{V_1}{T_1}=\frac{V_2}{T_2}$

Here, V₁ is initial volume, V₂ is final volume, T₁ is initial temperature and T₂ is final temperature.

Gay-Lussac’s law:

At constant volume, pressure and temperature of the gas are directly proportional to each other.

This is mathematically represented as follows:

  $\frac{P_1}{T_1}=\frac{P_2}{T_2}$

Here, T₁ is initial volume, T₂ is final volume, P₁ is initial pressure and P₂ is final pressure.

When the three gases are combined, the combined gas law will be:

  $\frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2}$

Thus, the new volume can be calculated using the following equation:

  $V_2=\frac{P_1{V}_1{T}_2}{P_2{T}_1}$

Here, P₁, V₁, T₁, P₂, V₂ and T₂ is initial pressure, initial volume, initial temperature, final pressure, final volume and final temperature respectively.

(c)

Interpretation Introduction

Interpretation: The volume of the balloon needs to be determined at 20 ^oC and 0.95 atm.

Concept Introduction: According to Gay-Lussac’s law, at constant volume, pressure and temperature of the gas are directly proportional to each other. This is mathematically represented as follows:

  $\frac{P_1}{T_1}=\frac{P_2}{T_2}$

Here, T₁ is initial volume, T₂ is final volume, P₁ is initial pressure and P₂ is final pressure.

Explanation of Solution

When the three gases are combined, the combined gas law will be:

  $\frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2}$

Thus, the new volume can be calculated using the following equation:

  $V_2=\frac{P_1{V}_1{T}_2}{P_2{T}_1}$

The volume and the pressure of the helium gas at 25 ^oC is 1.5 L and 1.0 atm respectively. The new pressure and temperature of the helium gas is 0.95 atm and 20 ^oC respectively.

The initial and final temperature in K can be calculated as follows:

  $\begin{array}{l}{T}_1={25}^{\text{o}}\text{C}+273.15=298.15\text{ K}\\ T_2={20}^{\text{o}}\text{C}+273.15=293.15\text{ K}\end{array}$

Putting all the values,

  $\begin{array}{l}{V}_2=\frac{P_1{V}_1{T}_2}{P_2{T}_1}\\ =\frac{\left(1\text{ atm}\right)\left(1.5\text{ L}\right)\left(293.15\text{ K}\right)}{\left(0.95\text{ atm}\right)\left(298.15\text{ K}\right)}\\ =1.55\text{ L}\end{array}$