Chapter 10, Problem 106AP

Interpretation Introduction

Interpretation:

The partial pressures of helium and neon are to be calculated with given volume, pressure, and temperature.

Concept Introduction:

The ideal gas equation elaborates the physical properties of gases by relating the pressure, volume, temperature, and number of moles with each other with the help of four gas laws the expression fortheideal gas equation is

$PV=nRT.$

Here, $P$ is pressure, $V$ is volume, $n$ is number of moles, $R$ is thegas constant, and $T$ is temperature.

The expression to calculate mole fraction $\left({\chi }_i\right)$ is

${\chi }_i=\frac{n_i}{n_{\text{total}}}.$

Here, $n_i$ is the number of moles of the component and $n_{\text{total}}$ is the total number of moles.

The relation between partial pressure $\left(P_i\right)$ and total pressure $\left(P_{\text{total}}\right)$ is

${\chi }_i=\frac{P_i}{P_{\text{total}}}.$

The conversion of temperature from degree Celsius to Kelvin can be done by using the formula given below:

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

Answer to Problem 106AP

Solution: $\text{0}\text{.16 atm}$ and $\text{2}\text{.0 atm}$

Explanation of Solution

Given information: For $\text{He}$,

Volume is $1.2\text{ L}$.

Pressure is $0.63\text{ atm}$.

Temperature is ${16}^o\text{C}$

For $\text{Ne}$,

Volume is $\text{3}\text{.4 L}$.

Pressure is $\text{2}\text{.8 atm}$.

Temperature is ${16}^o\text{C}$.

After the stopcock is opened, both the gases mix togethercompletely.

First, find the number of moles of each gas with the help of theideal gas equation as

$n=\frac{PV}{RT}.$…… (1)

Temperature is in Celsius. Convert it into Kelvin as

$\begin{array}{c}T=\left(16+273\right)\text{ K}\\ =289\text{ K}\text{.}\end{array}$

For $\text{He}$,

Substitute $0.63\text{ atm}$ for $P$, $1.2\text{ L}$ for $V$, $0.0821\text{ L}\text{.atm/mol-K}$ for $R,$ and $289\text{ K}$ for $T$ in equation (1):

$\begin{array}{c}n=\frac{0.63\cancel{\text{atm}}\times 1.2\cancel{\text{L}}}{0.0821\cancel{\text{L}}\text{.}\cancel{\text{atm}}\text{/mol-}\cancel{\text{K}}\times 289\cancel{\text{K}}}\\ =0.032\text{ mol He}\end{array}$ $\begin{array}{c}n=\frac{0.63\text{ atm}\times 1.2\text{ L}}{0.0821\text{ L}\text{.atm/mol-K}\times 289\text{ K}}\\ =0.032\text{ mol He}\end{array}$

For $\text{Ne}$,

Substitute $\text{2}\text{.8 atm}$ for $P$, $\text{3}\text{.4 L}$ for $V$, $0.0821\text{ L}\text{.atm/mol-K}$ for $R,$ and $289\text{ K}$ for $T$ in equation (1):

$\begin{array}{c}n=\frac{\text{2}\text{.8 }\cancel{\text{atm}}\times 3.4\cancel{\text{L}}}{0.0821\cancel{\text{L}}\text{.}\cancel{\text{atm}}\text{/mol-}\cancel{\text{K}}\times 289\cancel{\text{K}}}\\ =0.40\text{ mol Ne}\end{array}$ $\begin{array}{c}n=\frac{\text{2}\text{.8 atm}\times 3.4\text{ L}}{0.0821\text{ L}\text{.atm/mol-K}\times 289\text{ K}}\\ =0.40\text{ mol Ne}\end{array}$

Calculate the total number of moles as

$\text{Total moles}=\text{moles of He + moles of Ne}\text{.}$

Substitute $0.032\text{ mol}$ for moles of $\text{He}$ and $0.40\text{ mol}$ for moles of $\text{Ne}$ in the above expression as:

$\begin{array}{c}\text{Total moles}=0.032\text{ mol + }0.40\text{ mol}\\ =0.432\text{ mol}\text{.}\end{array}$

Now, rearrange theideal gas equation for $P$ as:

$P=\frac{nRT}{V}.$

Substitute $0.432\text{ mol}$ for $n$, $0.0821\text{ L-atm/mol-K}$ for $R$, $289\text{ K}$ for $T,$ and $\left(1.2\text{ +3}\text{.4}\right)\text{ L}$ for $V$ in the above reaction as:

$\begin{array}{c}P=\frac{0.432\cancel{\text{mol}}\times 0.0821\cancel{\text{L}}\text{-atm/mol-}\cancel{\text{K}}\times 289\cancel{\text{K}}}{\left(1.2\text{ +3}\text{.4}\right)\cancel{\text{L}}}\\ =2.2\text{ atm}\end{array}$ $\begin{array}{c}P=\frac{0.432\text{ mol}\times 0.0821\text{ L-atm/mol-K}\times 289\text{ K}}{\left(1.2\text{ +3}\text{.4}\right)\text{ L}}\\ =2.2\text{ atm}\end{array}$

The formula to calculate the mole fraction is

${\chi }_i=\frac{n_i}{n_{\text{total}}}.$ …… (2)

Now, calculate the mole fractions of each component.

For $\text{He}$,

Substitute $\text{0}\text{.032 mol}$ for $n_i$ and $0.432\text{ mol}$ for $n_{\text{total}}$ in the above expression as

${\chi }_i=\frac{n_i}{n_{\text{total}}}.$

$\begin{array}{c}{\chi }_{\text{He}}=\frac{0.032\cancel{\text{mol}}}{0.432\cancel{\text{mol}}}\\ =\mathrm{0.0741.}\end{array}$

For $\text{Ne}$,

Substitute $\text{0}\text{.40 mol}$ for $n_i$ and $0.432\text{ mol}$ for $n_{\text{total}}$ in equation (2) as

${\chi }_i=\frac{n_i}{n_{\text{total}}}$

$\begin{array}{c}{\chi }_{\text{He}}=\frac{0.40\cancel{\text{mol}}}{0.432\cancel{\text{mol}}}\\ =\mathrm{0.926.}\end{array}$

The relation between partial pressure and total pressure is

$P_i={\chi }_i{P}_T.$ …… (3)

For $\text{He}$,

Substitute $0.926$ for ${\chi }_i$ and $2.2\text{ atm}$ for $P_T$ in equation (3) as

$P_i={\chi }_i{P}_T$

$\begin{array}{c}{P}_{\text{He}}=\left(0.926\right)\left(2.2\text{ atm}\right)\\ =2.0\text{ atm}\text{.}\end{array}$

For $\text{Ne}$,

Substitute $0.0741$ for ${\chi }_i$ and $2.2\text{ atm}$ for $P_T$ in equation (3) as

$P_i={\chi }_i{P}_T$

$\begin{array}{c}{P}_{\text{Ne}}=\left(0.0741\right)\left(2.2\text{ atm}\right)\\ =0.16\text{ atm}\text{.}\end{array}$

Conclusion

The partial pressures of helium and neon are $\text{0}\text{.16 atm}$ and $\text{2}\text{.0 atm}$, respectively.