Chapter 16, Problem 32SP

One kilomole of ideal gas occupies 22.4 ${\text{m}}^{\text{3}}$ at 0 °C and 1 atm. (a) What pressure is required to compress 1.00 kmol into a 5.00 ${\text{m}}^{\text{3}}$ container at 100 °C? (b) If 1.00 kmol was to be sealed in a 5.00 ${\text{m}}^{\text{3}}$ tank that could withstand a gauge pressure of only 3.00 atm, what would be the maximum temperature of the gas if the tank was not to burst?

To determine

The pressure required to compress $1.00\text{ kmol}$ into a $5.00{\text{m}}^3$ container at $100°\text{C}$ if $1.00\text{ kmol}$ ideal gas occupies $22.4{\text{ m}}^3$ at $0°\text{C}$ at $1\text{atm}$ pressure.

Answer to Problem 32SP

Solution:

$6.12\text{atm}$

Explanation of Solution

Given data:

The total initial volume of $1.00\text{ kmol}$ gas is $22.4{\text{ m}}^3$.

The initial temperature of $1.00\text{ kmol}$ gas is $0°\text{C}$.

The initial pressure of $1.00\text{ kmol}$ gas is $1\text{atm}$.

The final volume of $1.00\text{ kmol}$ gas is $5.00{\text{m}}^3$.

The final temperature of $1.00\text{ kmol}$ gas is $100°\text{C}$.

Formula used:

Write the ideal gas equation,

$PV=nRT$

Here, $P$ is the absolute pressure, $V$ is the volume of the gas, $n$ is the number of moles of the gas, and $T$ is the temperature of the gas.

Write the conversion relation for Celsius temperature to Kelvin temperature.

$T_{\text{K}}=T_{\text{°C}}+273.15$

Here, $T_{\text{K}}$ is the temperature in Kelvin and $T_{°\text{C}}$ is the temperature in degree centigrade.

Explanation:

Convert the temperature from $°\text{C}$ to $\text{K}$.

For $0°\text{C}$,

$\begin{array}{c}{T}_1=0+273.15\text{ K}\\ =273.15\text{ K}\end{array}$

For $100°\text{C}$,

$\begin{array}{c}{T}_2=100+273.15\text{ K}\\ =373.15\text{ K}\end{array}$

Recall the expression obtained from the ideal gas law.

$PV=nRT$

If all other variables are constant then the relation between pressure, volume, and temperature between two states of the gas is written as follows:

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

Here, $T_1$ is the initial temperature, $T_2$ is the final temperature of the gas, $V_1$ is the initial volume, $V_2$ is the final volume of the gas, $P_1$ is the initial pressure, and $P_2$ is the final pressure of the gas.

Rearrange the expression for final pressure.

$P_2=\frac{P_1{V}_1}{T_1}\left(\frac{T_2}{V_2}\right)$

Substitute $273.15\text{ K}$ for $T_1$, $1\text{atm}$ for $P_1$, $22.4{\text{ m}}^3$ for $V_1$, $5.00{\text{m}}^3$ for $V_2$, and $373.15\text{ K}$ for $T_2$,

$\begin{array}{c}{P}_2=\frac{\left(1\text{atm}\right)\left(22.4{\text{ m}}^3\right)}{273.15\text{ K}}\left(\frac{373.15\text{ K}}{5.00{\text{m}}^3}\right)\\ =6.12\text{ atm}\end{array}$

Conclusion:

Therefore, the final requiredpressure of $1.00\text{ kmol}$ the gas is $6.12\text{atm}$.

To determine

The maximum temperature that the tankwithstands at the pressure of $3\text{atm}$ and volume $5.00{\text{m}}^3$ when the $1.00\text{ kmol}$ ideal gas which hasinitially occupied $22.4{\text{ m}}^3$ volume at $0°\text{C}$ temperature, is sealed in the tank.

Answer to Problem 32SP

Solution:

$-29.26°\text{C}$

Explanation of Solution

Given data:

The total initial volume of $1.00\text{ kmol}$ gas is $22.4{\text{ m}}^3$.

The initial temperature of $1.00\text{ kmol}$ gas is $0°\text{C}$.

The initial pressure of $1.00\text{ kmol}$ gas is $1\text{atm}$.

The final volume of $1.00\text{ kmol}$ gas is $5.00{\text{m}}^3$.

The container can withstands the $3\text{atm}$ pressure only.

Formula used:

Write the ideal gas equation,

$PV=nRT$

Here, $P$ is the absolute pressure, $V$ is the volume of the gas, $n$ is the number of moles of the gas, and $T$ is the temperature of the gas.

Write the conversion relation for Celsius temperature to Kelvin temperature.

$T_{\text{K}}=T_{\text{°C}}+273.15$

Here, $T_{\text{K}}$ is the temperature in Kelvin and $T_{°\text{C}}$ is the temperature in degree centigrade.

Explanation:

Convert the temperature from $°\text{C}$ to $\text{K}$.

For $0°\text{C}$,

$\begin{array}{c}{T}_1=0+273.15\text{ K}\\ =273.15\text{ K}\end{array}$

For $100°\text{C}$.

$\begin{array}{c}{T}_2=100+273.15\text{ K}\\ =373.15\text{ K}\end{array}$

Calculate the total maximum pressure on the container that withstandsitself.

$P_{\max }=P_{atm}+P_{abs}$

Here, $P_{atm}$ is the atmospheric pressure and $P_{abs}$ is the absolute pressure.

Substitute $1\text{atm}$ for $P_{atm}$ and $3\text{atm}$ for $P_{abs}$

$\begin{array}{c}{P}_{\max }=1\text{atm}+3\text{atm}\\ =4\text{atm}\end{array}$

Recall the expression obtained from the ideal gas law.

$PV=nRT$

If all other variables are constant then the relation between pressure, volume, and temperature between twostates of the gas is written as follows:

$\frac{P_1{V}_1}{P_{\max }{V}_2}=\frac{T_1}{T_{\max }}$

Here, $T_1$ is the initial temperature, $T_{\max }$ is the final temperature of the gas, $V_1$ is the initial volume, $V_2$ is the final volume of the gas, $P_1$ is the initial pressure, and $P_{\max }$ is the final pressure of the gas.

Rearrange the expression for final temperature.

$T_{\max }=\frac{P_{\max }{V}_2}{P_1{V}_1}\left(T_1\right)$

Substitute $273.15\text{ K}$ for $T_1$, $1\text{atm}$ for $P_1$, $22.4{\text{ m}}^3$ for $V_1$, $5.00{\text{m}}^3$ for $V_2$, and $4\text{atm}$ for $P_{\max }$ .

$\begin{array}{c}{T}_{\max }=\frac{\left(4\text{atm}\right)\left(5.00{\text{m}}^3\right)}{\left(1\text{atm}\right)\left(22.4{\text{ m}}^3\right)}\left(273.15\text{ K}\right)\\ =243.88\text{ K}\end{array}$

Convert this temperature into degree Celsius.

For $243.88\text{ K}$.

$\begin{array}{c}{T}_{\text{°C}}=243.884-273.15\\ \approx -29.26°\text{C}\end{array}$

Conclusion:

Therefore, the maximumtemperature of $1.00\text{ kmol}$ the gas is $-29.26°\text{C}$.