Chapter 14.7, Problem 113RP

(a)

To determine

The molar analysis of the saturated air.

Answer to Problem 113RP

The molar analysis of water is $0.0313$, nitrogen is $0.7566$, oxygen is $0.2025$ and argon is $0.0097$.

Explanation of Solution

Express the pressure of air.

$\begin{array}{c}{P}_{\text{air}}=P-P_v\\ =P-P_{\text{sat@25°C}}\end{array}$ (I)

Here, the saturation pressure at temperature of $\text{25°C}$ is $P_{\text{sat@25°C}}$, vapor pressure is $P_v$ and atmospheric pressure is $P$.

Express the molar fraction of water.

$\begin{array}{c}{y}_{{\text{H}}_{\text{2}}\text{O}}=\frac{P_v}{P}\\ =\frac{P_{\text{sat@25°C}}}{P}\end{array}$ (II)

Here, pressure of water is $P_{{\text{H}}_{\text{2}}\text{O}}$.

Express the molar fraction of nitrogen.

$\begin{array}{c}{y}_{{\text{N}}_{\text{2}}}=\frac{P_{{\text{N}}_{\text{2}}}}{P}\\ =\frac{y_{{\text{N}}_{\text{2,dry}}}{P}_{\text{air}}}{P}\end{array}$ (III)

Here, pressure of nitrogen is $P_{{\text{N}}_{\text{2}}}$ and mass fraction of dry nitrogen is $y_{{\text{N}}_{\text{2,dry}}}$.

Express the molar fraction of oxygen.

$\begin{array}{c}{y}_{{\text{N}}_{\text{2}}}=\frac{P_{{\text{O}}_{\text{2}}}}{P}\\ =\frac{y_{{\text{O}}_{\text{2,dry}}}{P}_{\text{air}}}{P}\end{array}$ (IV)

Here, pressure of oxygen is $P_{{\text{O}}_{\text{2}}}$ and mass fraction of dry oxygen is $y_{{\text{O}}_{\text{2,dry}}}$.

Express the molar fraction of argon.

$\begin{array}{c}{y}_{\text{Ar}}=\frac{P_{\text{Ar}}}{P}\\ =\frac{y_{{\text{Ar}}_{\text{,dry}}}{P}_{\text{air}}}{P}\end{array}$ (V)

Here, pressure of oxygen is $P_{{\text{O}}_{\text{2}}}$ and mass fraction of dry oxygen is $y_{{\text{O}}_{\text{2,dry}}}$.

Conclusion:

Refer Table A-4, “saturated water-temperature table”, and write the saturation pressure at temperature of $\text{25°C}$.

$P_{\text{sat@25°C}}=3.1698\text{kPa}$

Perform unit conversion of atmospheric pressure from $\text{atm}\text{to}\text{kPa}$.

$\begin{array}{c}P=1\text{atm}\\ =1\text{atm}\left[\frac{101.325\text{kPa}}{\text{atm}}\right]\\ =101.325\text{kPa}\end{array}$

Substitute $101.325\text{kPa}$ for $P$ and $3.1698\text{kPa}$ for $P_{\text{sat@25°C}}$ in Equation (I).

$\begin{array}{c}{P}_{\text{air}}=101.325\text{kPa}-3.1698\text{kPa}\\ =98.155\text{kPa}\end{array}$

Substitute $101.325\text{kPa}$ for $P$ and $3.1698\text{kPa}$ for $P_{\text{sat@25°C}}$ in Equation (II).

$\begin{array}{c}{y}_{{\text{H}}_{\text{2}}\text{O}}=\frac{3.1698\text{kPa}}{101.325\text{kPa}}\\ =0.0313\end{array}$

Hence, the molar analysis of water is $0.0313$.

Substitute $0.781$ for $y_{{\text{N}}_{\text{2}},\text{dry}}$, $98.155\text{kPa}$ for $P_{\text{air}}$ and $101.325\text{kPa}$ for $P$ in Equation (III).

$\begin{array}{c}{y}_{{\text{N}}_{\text{2}}}=\frac{\left(0.781\right)\left(98.155\text{kPa}\right)}{101.325\text{kPa}}\\ =0.7566\end{array}$

Hence, the molar analysis of nitrogen is $0.7566$.

Substitute $0.209$ for $y_{{\text{O}}_{\text{2}},\text{dry}}$, $98.155\text{kPa}$ for $P_{\text{air}}$ and $101.325\text{kPa}$ for $P$ in Equation (IV).

$\begin{array}{c}{y}_{{\text{O}}_{\text{2}}}=\frac{\left(0.209\right)\left(98.155\text{kPa}\right)}{101.325\text{kPa}}\\ =0.2025\end{array}$

Hence, the molar analysis of oxygen is $0.2025$.

Substitute $0.01$ for $y_{\text{Ar},\text{dry}}$, $98.155\text{kPa}$ for $P_{\text{air}}$ and $101.325\text{kPa}$ for $P$ in Equation (V).

$\begin{array}{c}{y}_{\text{Ar}}=\frac{\left(0.01\right)\left(98.155\text{kPa}\right)}{101.325\text{kPa}}\\ =0.0097\end{array}$

Hence, the molar analysis of argon is $0.0097$.

(b)

To determine

The density of air before and after the process.

Answer to Problem 113RP

The density of air before and after the process is $\text{1}\text{.186}\text{kg}/{\text{m}}^{\text{3}}\text{and}\text{1}\text{.170}\text{kg}/{\text{m}}^{\text{3}}$ respectively.

Explanation of Solution

Express the molar mass of dry air.

$\begin{array}{c}{M}_{\text{dry}\text{air}}={\displaystyle \sum y_i{M}_i}\\ =y_{{\text{N}}_{2,\text{dry}}}{M}_{{\text{N}}_{\text{2}}}+y_{{\text{O}}_{2,\text{dry}}}{M}_{{\text{O}}_2}+y_{\text{Ar,dry}}{M}_{\text{Ar}}\end{array}$ (VI)

Here, total molar fraction is ${\displaystyle \sum y_i}$, total molar mass is ${\displaystyle \sum M_i}$, molar mass of nitrogen is $M_{{\text{N}}_{\text{2}}}$, molar mass of oxygen is $M_{{\text{O}}_{\text{2}}}$ and molar mass of argon is $M_{\text{Ar}}$.

Express the molar mass of saturated air.

$\begin{array}{c}{M}_{\text{sat}\text{air}}={\displaystyle \sum y_i{M}_i}\\ =y_{{\text{N}}_2}{M}_{{\text{N}}_{\text{2}}}+y_{{\text{O}}_2}{M}_{{\text{O}}_2}+y_{\text{Ar}}{M}_{\text{Ar}}+y_{{\text{H}}_{\text{2}}\text{O}}{M}_{{\text{H}}_{\text{2}}\text{O}}\end{array}$ (VII)

Here, molar mass of water vapor is ${\text{H}}_{\text{2}}\text{O}$.

Express the density of air before the process.

${\rho }_{\text{dry}\text{air}}=\frac{P}{\left(R_a/M_{\text{dry}\text{air}}\right)T}$ (VIII)

Here, gas constant of air is $R_a$.

Express the density of air after the process.

${\rho }_{\text{sat}\text{air}}=\frac{P}{\left(R_a/M_{\text{sat}\text{air}}\right)T}$ (IX)

Here, gas constant of air is $R_a$.

Conclusion:

Refer Table A-1, “molar mass, gas constant, and the critical point properties”, and write the molar masses.

$\begin{array}{l}{M}_{{\text{N}}_{\text{2}}}=\text{28}\text{kg}/\text{kmol}\\ M_{\text{Ar}}=\text{39}\text{.9}\text{kg}/\text{kmol}\\ M_{{\text{O}}_{\text{2}}}=\text{32}\text{kg}/\text{kmol}\\ M_{{\text{H}}_{\text{2}}\text{O}}=\text{18}\text{kg}/\text{kmol}\end{array}$

Refer Table A-1, “molar mass, gas constant, and the critical point properties”, and write the universal gas constant of air.

$R_a=8.314\text{kPa}\cdot \text{m}/\text{kmol}\cdot \text{K}$

Substitute $0.781$ for $y_{{\text{N}}_{\text{2}},\text{dry}}$, $\text{28}\text{kg}/\text{kmol}$ for $M_{{\text{N}}_{\text{2}}}$, $\text{39}\text{.9}\text{kg}/\text{kmol}$ for $M_{\text{Ar}}$, $\text{32}\text{kg}/\text{kmol}$ for $M_{{\text{O}}_{\text{2}}}$, $0.209$ for $y_{{\text{O}}_{\text{2}},\text{dry}}$ and $0.01$ for $y_{\text{Ar},\text{dry}}$ in Equation (VI).

$\begin{array}{c}{M}_{\text{dry}\text{air}}=\left(0.781\right)\left(28\right)+\left(0.209\right)\left(32\right)+\left(0.01\right)\left(39.9\right)\\ =29\text{kg}/\text{kmol}\end{array}$

Substitute $0.7566$ for $y_{{\text{N}}_{\text{2}}}$, $\text{28}\text{kg}/\text{kmol}$ for $M_{{\text{N}}_{\text{2}}}$, $\text{39}\text{.9}\text{kg}/\text{kmol}$ for $M_{\text{Ar}}$, $\text{32}\text{kg}/\text{kmol}$ for $M_{{\text{O}}_{\text{2}}}$, $0.2025$ for $y_{{\text{O}}_{\text{2}}}$, $0.0097$ for $y_{\text{Ar}}$, $\text{18}\text{kg}/\text{kmol}$ for $M_{{\text{H}}_{\text{2}}\text{O}}$ and $0.0313$ for $y_{{\text{H}}_{\text{2}}\text{O}}$ in Equation (VII).

$\begin{array}{c}{M}_{\text{sat}\text{air}}=\left(0.7566\right)\left(28\right)+\left(0.2025\right)\left(32\right)+\left(0.0097\right)\left(39.9\right)+\left(0.0313\right)\left(18\right)\\ =28.62\text{kg}/\text{kmol}\end{array}$

Perform the unit conversion of temperature from $\text{°C}\text{to}\text{K}$.

$\begin{array}{c}T=25\text{°C}\\ =\left(25+273\right)\text{K}\\ =\text{298}\text{K}\end{array}$

Substitute $101.325\text{kPa}$ for $P$, $8.314\text{kPa}\cdot \text{m}/\text{kmol}\cdot \text{K}$ for $R_a$, $29\text{kg}/\text{kmol}$ for $M_{\text{dry}\text{air}}$ and $\text{298}\text{K}$ for $T$ in Equation (VIII).

$\begin{array}{c}{\rho }_{\text{dry}\text{air}}=\frac{101.325\text{kPa}}{\left(8.314\text{kPa}\cdot \text{m}/\text{kmol}\cdot \text{K}/29\text{kg}/\text{kmol}\right)\left(\text{298}\text{K}\right)}\\ =\text{1}\text{.186}\text{kg}/{\text{m}}^{\text{3}}\end{array}$ (X)

Hence, the density of air before the process is $\text{1}\text{.186}\text{kg}/{\text{m}}^{\text{3}}$.

Substitute $101.325\text{kPa}$ for $P$, $8.314\text{kPa}\cdot \text{m}/\text{kmol}\cdot \text{K}$ for $R_a$, $28.62\text{kg}/\text{kmol}$ for $M_{\text{sat}\text{air}}$ and $\text{298}\text{K}$ for $T$ in Equation (IX).

$\begin{array}{c}{\rho }_{\text{sat}\text{air}}=\frac{101.325\text{kPa}}{\left(8.314\text{kPa}\cdot \text{m}/\text{kmol}\cdot \text{K}/28.62\text{kg}/\text{kmol}\right)\left(\text{298}\text{K}\right)}\\ =\text{1}\text{.170}\text{kg}/{\text{m}}^{\text{3}}\end{array}$ (XI)

Hence, the density of air after the process is $\text{1}\text{.170}\text{kg}/{\text{m}}^{\text{3}}$.

From the result obtained in Equations (X) and (XI), It is obtained that the density of dry air is larger than that of saturated air as the molar mass of dry air being larger than that of water.