Chapter 5.3, Problem 5.5BFP

Interpretation Introduction

Interpretation:

The volume of air in the driver’s lung that occupies volume of $\text{2}\text{.2}\text{L}$ at $\text{0}\text{.980}\text{atm}$ and $28°\text{C}$ and changes to $\text{1}\text{.40}\text{atm}$ and $21°\text{C}$ is to be calculated.

Concept introduction:

Ideal gas law equation is as follows:

$PV=nRT$ (1)

Here,

$P$ is the pressure of the gas.

$V$ is the volume occupied by gas.

$n$ is the number of moles of gas.

$T$ is the temperature in Kelvin.

$R$ is the universal gas constant.

For two different values of pressure, temperature, volume and number of moles equation (1) can be modified as follows:

$\frac{P_1{V}_1}{n_1{T}_1}=\frac{P_2{V}_2}{n_2{T}_2}$ (2)

Here,

$P_1$ is the initial pressure of the gas.

$P_2$ is the final pressure of the gas.

$V_1$ is the initial volume of gas.

$V_2$ is the final volume of gas.

$n_1$ is the initial number of moles of gas.

$n_2$ is the final number of moles of gas.

For a fixed number of moles equation (2) can be written as follows:

$\frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2}$ (3)

The conversion factor to convert $°\text{C}$ to Kelvin is as follows:

$T\left(\text{K}\right)=T\left(°\text{C}\right)+273.15$ (4)

Answer to Problem 5.5BFP

The volume of air in the driver’s lung that occupies volume of $\text{2}\text{.2}\text{L}$ at $\text{0}\text{.980}\text{atm}$ and $28°\text{C}$ and changes to $\text{1}\text{.40}\text{atm}$ and $21°\text{C}$ is $\text{1}\text{.5}\text{L}$.

Explanation of Solution

The initial pressure $\left(P_1\right)$ is $\text{0}\text{.980}\text{atm}$.

The final pressure $\left(P_{\text{2}}\right)$ is $\text{1}\text{.40}\text{atm}$.

The initial volume $\left(V_1\right)$ is $\text{2}\text{.2}\text{L}$.

Initial temperature $\left(T_1\right)$ is $28°\text{C}$

Final temperature $\left(T_2\right)$ is $21°\text{C}$.

Substitute $28°\text{C}$ for $T\left(\text{°C}\right)$ in equation (4).

$\begin{array}{c}{T}_1\left(\text{K}\right)=28°\text{C}\text{+}\text{273}\text{.15}\\ =301.15\text{K}\end{array}$

Substitute $21°\text{C}$ for $T\left(\text{°C}\right)$ in equation (4).

$\begin{array}{c}{T}_1\left(\text{K}\right)=21°\text{C}\text{+}\text{273}\text{.15}\\ =294.15\text{K}\end{array}$

Rearrange equation (3) for $V_2$ as follows:

$V_2=\frac{P_1{T}_2{V}_1}{P_2{T}_1}$ (5)

Substitute $\text{2}\text{.2}\text{L}$ for $V_1$, $\text{0}\text{.980}\text{atm}$ for $P_1$, $\text{1}\text{.40}\text{atm}$ for $P_2$, $301.15K$ for $T_1$, $\text{294}\text{.15}\text{K}$ for $T_2$ in equation (5).

$\begin{array}{c}{V}_2\left(\text{L}\right)=\left(\text{2}\text{.2}\text{L}\right)\left(\frac{\left(\text{0}\text{.980}\text{atm}\right)\left(\text{294}\text{.15}\text{K}\right)}{\left(\text{1}\text{.40}\text{atm}\right)\left(\text{301}\text{.15}K\right)}\right)\\ =\text{1}\text{.5042}\text{L}\\ =\text{1}\text{.5}\text{L}\end{array}$

Conclusion

The volume of air in the driver’s lung that occupies volume of $\text{2}\text{.2}\text{L}$ at $\text{0}\text{.980}\text{atm}$ and $28°\text{C}$ and changes to $\text{1}\text{.40}\text{atm}$ and $21°\text{C}$ is $\text{1}\text{.5}\text{L}$.