Chapter 5, Problem 5.125P

(a)

Interpretation Introduction

Interpretation:

The pressure of tanks of ${\text{H}}_{\text{2}}$ and of ${\text{O}}_{\text{2}}$ at $23.8°\text{C}$ by the ideal gas equation is to be calculated.

Concept introduction:

According to Boyle’s law, the volume occupied by the gas is inversely proportional to the pressure at the constant temperature.

The relationship between pressure and volume can be expressed as follows,

$PV=\text{constant}$

Here, $P$ is the pressure and $V$ is the volume.

According to Charles's law, the volume occupied by the gas is directly proportional to the temperature at the constant pressure.

The relationship between pressure and temperature can be expressed as follows,

$V\alpha T$

Here, $T$ is the temperature and $V$ is the volume.

According to Avogadro’s law, the volume occupied by the gas is directly proportional to the mole of the gas at the constant pressure and temperature.

The relationship between volume and mole can be expressed as follows,

$\text{V }\alpha n$

Here, $n$ is the mole of the gas and $V$ is the volume.

The ideal gas equation can be expressed as follows,

$PV=nRT$

Here, $P$ is the pressure, $V$ is the volume, $T$ is the temperature, $n$ is the mole of the gas and $R$ is the gas constant.

Answer to Problem 5.125P

The pressure of tanks by the ideal gas equation of ${\text{H}}_{\text{2}}$ is $34.1\text{ atm}$ and ${\text{O}}_{\text{2}}$ is $17.1\text{ atm}$

Explanation of Solution

The equation for the reaction of ${\text{H}}_{\text{2}}$ with ${\text{O}}_{\text{2}}$ is as follows:

${\text{2H}}_{\text{2}}\left(g\right)+{\text{ O}}_{\text{2}}\left(g\right)\to 2{\text{H}}_{\text{2}}\text{O}\left(g\right)$ (1)

From the equation (1), two moles of the ${\text{H}}_{\text{2}}\text{O}$ will be formed one mole of ${\text{O}}_{\text{2}}$ . Thus, the moles of ${\text{O}}_{\text{2}}$ is calculated from ${\text{H}}_{\text{2}}\text{O}$ as follows:

${\text{Moles of O}}_{\text{2}}=\left[\left({\text{Moles of H}}_{\text{2}}\text{O}\right)\left(\frac{{\text{1 mol O}}_{\text{2}}}{{\text{2 mol H}}_{\text{2}}\text{O}}\right)\right]$ (2)

Substitute the value $\text{28 mol}$ for moles of ${\text{H}}_{\text{2}}\text{O}$ in the equation (2).

$\begin{array}{c}{\text{Moles of O}}_{\text{2}}=\left[\left(28\text{ mol}\right)\left(\frac{{\text{1 mol O}}_{\text{2}}}{{\text{2 mol H}}_{\text{2}}\text{O}}\right)\right]\\ =14\text{ mol}\end{array}$

From the equation (1), two moles of the ${\text{H}}_{\text{2}}\text{O}$ will be formed two moles of ${\text{H}}_{\text{2}}$ . Thus, the moles of ${\text{H}}_{\text{2}}$ is calculated from ${\text{H}}_{\text{2}}\text{O}$ as follows:

${\text{Moles of H}}_{\text{2}}=\left[\left({\text{Moles of H}}_{\text{2}}\text{O}\right)\left(\frac{{\text{2 mol H}}_{\text{2}}}{{\text{2 mol H}}_{\text{2}}\text{O}}\right)\right]$ (3)

Substitute the value $\text{28 mol}$ for moles of ${\text{H}}_{\text{2}}\text{O}$ in the equation (3).

$\begin{array}{c}{\text{Moles of H}}_{\text{2}}=\left[\left(28\text{ mol}\right)\left(\frac{{\text{2 mol H}}_{\text{2}}}{{\text{2 mol H}}_{\text{2}}\text{O}}\right)\right]\\ =28\text{ mol}\end{array}$

The formula to convert $°\text{C}$ to Kelvin is:

$T_{\text{K}}=T_{\text{°C}}+273.15$ (4)

Substitute $\text{23}\text{.8 }°\text{C}$ for $T_{\text{°C}}$ in equation (4)

$\begin{array}{c}{T}_{\text{K}}=23.8+273.15\\ =296.95\text{ K}\end{array}$

The expression to calculate the pressure of ${\text{H}}_{\text{2}}$ is as follows:

$P_{\text{IGL}}V=nRT$ (5)

Here, $P$ is the pressure, $V$ is the volume, $T$ is the temperature, $n$ is the mole of ${\text{H}}_{\text{2}}$ and $R$ is the gas constant.

Rearrange the equation (5) to calculate $P_{\text{IGL}}$ as follows:

$P_{\text{IGL}}=\frac{nRT}{V}$ (6)

Substitute the value $\text{20}\text{.0 L}$ for $V$, $\text{296}\text{.15 K}$ for $T$, $\text{28 mol}$ for $n$ and $0.0821\frac{\text{L}\cdot \text{atm}}{\text{mol}\cdot \text{K}}$ for $R$ in the equation (6).

$\begin{array}{c}{P}_{\text{IGL}}=\frac{\left(\text{28 mol}\right)\left(0.0821\frac{\text{L}\cdot \text{atm}}{\text{mol}\cdot \text{K}}\right)\left(296.15\text{ K}\right)}{\left(\text{20}\text{.0 L}\right)}\\ =34.131\text{ atm}\\ \approx 34.\text{1 atm}\end{array}$

The expression to calculate the moles of ${\text{O}}_{\text{2}}$ is as follows:

$PV=nRT$ (7)

Here, $P$ is the pressure, $V$ is the volume, $T$ is the temperature, $n$ is the mole of ${\text{O}}_{\text{2}}$ and $R$ is the gas constant.

Rearrange the equation (7) to calculate $n$ as follows:

$n=\frac{PV}{RT}$ (8)

Substitute the value $\text{20}\text{.0 L}$ for $V$, $\text{296}\text{.15 K}$ for $T$, $\text{14 mol}$ for $n$ and $0.0821\frac{\text{L}\cdot \text{atm}}{\text{mol}\cdot \text{K}}$ for $R$ in the equation (8).

$\begin{array}{c}{P}_{\text{IGL}}=\frac{\left(\text{14 mol}\right)\left(0.0821\frac{\text{L}\cdot \text{atm}}{\text{mol}\cdot \text{K}}\right)\left(296.15\text{ K}\right)}{\left(\text{20}\text{.0 L}\right)}\\ =17.0657\text{ atm}\\ \approx 17.\text{1 atm}\end{array}$

Conclusion

The pressure of tanks by the ideal gas equation of ${\text{H}}_{\text{2}}$ is $34.1\text{ atm}$ and ${\text{O}}_{\text{2}}$ is $17.1\text{ atm}$.

(b)

Interpretation Introduction

Interpretation:

The pressure of tanks of ${\text{H}}_{\text{2}}$ and of ${\text{O}}_{\text{2}}$ at $23.8°\text{C}$ by van der Waals equation is to be calculated.

Concept introduction:

The ideal gas equation can be expressed as follows,

$PV=nRT$

Here, $P$ is the pressure, $V$ is the volume, $T$ is the temperature, $n$ is the mole of the gas and $R$ is the gas constant.

The expression of the van der Waals equation is as follows:

$\left(P_{\text{VDW}}+\frac{n^{2}a}{V^2}\right)\left(V-nb\right)=nRT$ (8)

Here, $P$ is the pressure, $V$ is the volume, $T$ is the temperature, $n$ is the mole of the gas, $R$ is the gas constant, $a$ and $b$ are van der Waals constants.

Answer to Problem 5.125P

The pressure of tanks by van der Waals equation of ${\text{H}}_{\text{2}}$ is $35.0\text{ atm}$ and ${\text{O}}_{\text{2}}$ is $16.8\text{ atm}$.

Explanation of Solution

Rearrange the equation (8) to calculate the pressure of $\left({\text{H}}_{\text{2}}\right)$$P_{\text{VDW}}$

$P_{\text{VDW}}=\frac{nRT}{\left(V-nb\right)}-\frac{n^{2}a}{V^2}$ (9)

Substitute the value $\text{20 L}$ for $V$, $\text{296}\text{.95 K}$ for $T$, $\text{28 mol}$ for $n$ , $0.244\frac{\text{atm}\cdot {\text{L}}^2}{{\text{mol}}^2}$ for $a$, $0.0266\frac{\text{L}}{\text{mol}}$ for $b$ and $0.0821\frac{\text{L}\cdot \text{atm}}{\text{mol}\cdot \text{K}}$ for $R$ in the equation (9).

$\begin{array}{c}{P}_{\text{VDW}}=\left[\begin{array}{l}\frac{\left(28\text{ mol}\right)\left(0.0821\frac{\text{L}\cdot \text{atm}}{\text{mol}\cdot \text{K}}\right)\left(296.15\text{ K}\right)}{\left(20\text{ L}-\left(28\text{ mol}\right)\left(0.0266\frac{\text{L}}{\text{mol}}\right)\right)}-\\ \frac{{\left(\text{28 mol}\right)}^2\left(0.244\frac{\text{atm}\cdot {\text{L}}^2}{{\text{mol}}^2}\right)}{{\left(\text{20 L}\right)}^2}\end{array}\right]\\ =34.978\text{ atm}\\ \approx 35\text{ atm}\end{array}$

Rearrange the equation (8) to calculate the pressure of $\left({\text{O}}_{\text{2}}\right)$$P_{\text{VDW}}$

$P_{\text{VDW}}=\frac{nRT}{\left(V-nb\right)}-\frac{n^{2}a}{V^2}$ (10)

Substitute the value $\text{20 L}$ for $V$, $\text{296}\text{.95 K}$ for $T$, $\text{14 mol}$ for $n$ , $0.244\frac{\text{atm}\cdot {\text{L}}^2}{{\text{mol}}^2}$ for $a$, $0.0266\frac{\text{L}}{\text{mol}}$ for $b$ and $0.0821\frac{\text{L}\cdot \text{atm}}{\text{mol}\cdot \text{K}}$ for $R$ in the equation (10).

$\begin{array}{c}{P}_{\text{VDW}}=\left[\begin{array}{l}\frac{\left(\text{14 mol}\right)\left(0.0821\frac{\text{L}\cdot \text{atm}}{\text{mol}\cdot \text{K}}\right)\left(296.15\text{ K}\right)}{\left(20\text{ L}-\left(\text{14 mol}\right)\left(0.0266\frac{\text{L}}{\text{mol}}\right)\right)}-\\ \frac{{\left(\text{14 mol}\right)}^2\left(0.244\frac{\text{atm}\cdot {\text{L}}^2}{{\text{mol}}^2}\right)}{{\left(\text{20 L}\right)}^2}\end{array}\right]\\ =16.7878\text{ atm}\\ \approx 16.8\text{ atm}\end{array}$

Conclusion

The pressure of tanks by van der Waals equation of ${\text{H}}_{\text{2}}$ is $35.0\text{ atm}$ and ${\text{O}}_{\text{2}}$ is $16.8\text{ atm}$.

(c)

Interpretation Introduction

Interpretation:

The result from the two equation is to be compared.

Concept introduction:

The ideal gas equation can be expressed as follows,

$PV=nRT$

Here, $P$ is the pressure, $V$ is the volume, $T$ is the temperature, $n$ is the mole of the gas and $R$ is the gas constant.

The expression of the van der Waals equation is as follows:

$\left(P_{\text{VDW}}+\frac{n^{2}a}{V^2}\right)\left(V-nb\right)=nRT$

Here, $P$ is the pressure, $V$ is the volume, $T$ is the temperature, $n$ is the mole of the gas, $R$ is the gas constant, $a$ and $b$ are van der Waals constants.

Answer to Problem 5.125P

The van der Waals value for hydrogen has a higher value as compared to the ideal gas law but the van der Waals value has a lower value as compared to the ideal gas law.

Explanation of Solution

For hydrogen, the van der Waals value for hydrogen has a higher value as compared to the ideal gas law. For oxygen, the van der Waals value has a lower value as compared to the ideal gas law.

Conclusion

The van der Waals value for hydrogen has a higher value as compared to the ideal gas law but the van der Waals value has a lower value as compared to the ideal gas law.