Chapter 5, Problem 59E

Interpretation Introduction

Interpretation: For the given data, the final pressure and temperature in the container should be determined.

Concept introduction:

By combining the three gaseous laws namely Boyle’s law, Charles’s law and Avogadro’s law a combined gaseous equation is obtained. This combined gaseous equation is called Ideal gas law.

According to ideal gas law,

                                             $\text{PV=nRT}$

Where,

          P = pressure in atmospheres

          V= volumes in liters

          n = number of moles

          R =universal gas constant ( $\text{0}\text{.08206L×atm/K×mol}$)

          T = temperature in kelvins

By knowing any three of these properties, the state of a gas can be simply identified with applying the ideal gas equation. For a gas at two conditions, the unknown variable can be determined by knowing the variables that change and remain constant and can be generated an equation for unknown variable from ideal gas equation.

Answer to Problem 59E

Answer

The pressure inside the container if it is heated to $45°C=12.8atm$

The temperature inside the container if the excreted pressure be $6.50atm$ $=161K$

The temperature inside the container if the excreted pressure be $25.0atm=620.0K$

Explanation of Solution

Explanation

According to ideal gas equation,

$PV=nRT$

By rearranging the above equation,

$\frac{\text{PV}}{\text{nT}}\text{=R}$

Since R is a gas constant, and at constant n and V for a gas at two conditions the equation can be written as:

$\frac{P_1{}_{}}{T_1}=VnR=\frac{P_2}{T_2}or\frac{P_1{}_{}}{T_1}=\frac{P_2}{T_2}$ (1)

At constant volume and number of moles, for finding the equation for final pressure the above equation (1) becomes,

$P_2=\frac{P_1{T}_2}{T_1}$ (2)

From the ideal gas equation, the equation for final pressure of gas for a gas at two conditions can be derived by knowing initial pressure ( ${\text{P}}_{\text{1}}$), temperature ( ${\text{T}}_{\text{1}}$) and final temperature ( ${\text{T}}_{\text{2}}$). It is the ratio of product of initial pressure and final temperature to the initial temperature.

The given data and its values.

$\begin{array}{l}{\text{P}}_{\text{1}}\text{=11}\text{.0 atm}\\ {\text{T}}_{\text{1}}\text{=0°C=273K}\\ \text{since,1K=°C+273}\\ \text{=0°C+273}\\ \text{=273K}\\ {\text{T}}_{\text{2}}\text{=45°C=318K}\\ \text{since,1K=°C+273}\\ \text{=45°C+273}\\ \text{=318K}\\ \\ \end{array}$

To find out the final pressure of gas, it is needed to take and write the given data and substitute their values in the equation (2). For two conditions problem, units for P and V just needed to be the same units and it is not needed to convert the standard units. But in the case of pressure, it must be converted to the Kelvin.

The final pressure of gas can be determined by substituting the given values to the equation (2) that derived from ideal gas law.

The given data into the equation (2) to get the final pressure of gas,

$\begin{array}{l}\\ {\text{P}}_{\text{2}}\text{=}\frac{\text{11}\text{.0}\text{atm×318K}}{\text{273K}}\\ \text{=12}\text{.8}\text{atm}\end{array}$

Derive the equation for final temperature of gas from ideal gas equation for a gas at two conditions

According to ideal gas equation,

$PV=nRT$

By rearranging the above equation,

$\frac{\text{PV}}{\text{nT}}\text{=R}$

Since R is a gas constant, and at constant n and V for a gas at two conditions the equation can be written as:

$\frac{P_1{}_{}}{T_1}=VnR=\frac{P_2}{T_2}or\frac{P_1{}_{}}{T_1}=\frac{P_2}{T_2}$ (1)

At constant volume and number of moles, for finding the equation for final temperature the above equation (1) becomes,

                                             $T_2=\frac{T_1{P}_2}{P_1}$ (2)

From the ideal gas equation, the equation for final pressure of gas for a gas at two conditions can be derived by knowing initial pressure ( ${\text{P}}_{\text{1}}$), temperature ( ${\text{T}}_{\text{1}}$) and final pressure ( ${\text{P}}_{\text{2}}$). It is the ratio of product of initial temperature and final pressure to the initial pressure.

The given data and its values to determine the temperature inside the container if the excreted pressure be $6.50atm$

$\begin{array}{l}{\text{P}}_{\text{1}}\text{=11}\text{.0}\text{atm}\\ {\text{T}}_{\text{1}}\text{=0°C=273K}\\ \text{since,1K=°C+273}\\ \text{=0°C+273}\\ \text{=273K}\\ {\text{P}}_{\text{2}}\text{=}\text{6}\text{.50}\text{atm}\\ \\ \end{array}$

To find out the final temperature of gas, it is needed to take and write the given data and substitute their values in the equation (2). For two conditions problem, units for P and V just needed to be the same units and it is not needed to convert the standard units. But in the case of pressure, it must be converted to the Kelvin.

The given data into the equation (2) to get the final temperature of gas.

$\begin{array}{l}\\ {\text{T}}_{\text{2}}\text{=}\frac{\text{273 K×6}\text{.50 atm}}{\text{11}\text{.0atm}}\\ \text{=161 K}\end{array}$

The final temperature of gas can be determined by substituting the given values to the equation (2) that derived from ideal gas law.

The given data and its values to determine the temperature inside the container if the excreted pressure is $25.0atm$

$\begin{array}{l}{\text{P}}_{\text{1}}\text{=11}\text{.0}\text{atm}\\ {\text{T}}_{\text{1}}\text{=0°C=273K}\\ \text{since,1K=°C+273}\\ \text{=0°C+273}\\ \text{=273K}\\ {\text{P}}_{\text{2}}\text{=}\text{25}\text{atm}\\ \\ \end{array}$

To find out the final temperature of gas, it is needed to take and write the given data and substitute their values in the equation (2). For two conditions problem, units for P and V just needed to be the same units and it is not needed to convert the standard units. But in the case of pressure, it must be converted to the Kelvin.

Substitute the given data into the equation (2) to get the final temperature of gas.

                                    $\begin{array}{l}\\ {\text{T}}_{\text{2}}\text{=}\frac{\text{273 K×25}\text{.0 atm}}{\text{11}\text{.0atm}}\\ \text{=620}\text{. K}\end{array}$

The final temperature of gas can be determined by substituting the given values to the equation (2) that derived from ideal gas law.

Conclusion

Conclusion

The final temperature and pressure in the container is measured by ideal gas equation.