Chapter 6, Problem 6.128QE

(a)

Interpretation Introduction

Interpretation:

The pressure of $30.33\text{ mol}$ hydrogen at $240°\text{C}$ in a $2.44\mathrm{L}$ container has to be calculated.

Concept Introduction:

Van der Waals equation is given as follows:

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

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 6.128QE

The pressure of $30.33\text{ mol}$ hydrogen at $240°\text{C}$ in a $2.44\mathrm{L}$ container is $44.054\text{ atm}$.

Explanation of Solution

The formula to convert degree Celsius to kelvin is as follows:

  $T\left(\text{K}\right)=T\left(°\text{C}\right)+273\text{ K}$        (1)

Here,

$T\left(\text{K}\right)$ denotes the temperature in kelvins.

$T\left(°\text{C}\right)$ denotes the temperature in Celsius.

Substitute $240°\text{C}$ for $T\left(°\text{C}\right)$ in equation (1).

  $\begin{array}{c}T\left(\text{K}\right)=240°\text{C}+273\text{ K}\\ =513\text{ K}\end{array}$

Van der Waals equation is given as follows:

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

Here,

$a$ is a constant that gives the strength of the attractive force.

$b$ gives the measure of size of the gas molecules.

Rearrange the expression (2) to obtain the formula of pressure as follows:

  $P=\frac{nRT}{\left(V-nb\right)}-\frac{a{n}^2}{V^2}$        (3)

Refer to table 6.4 for the values of Van der Waals constants of hydrogen.

Substitute $513\text{ K}$ for $T$ , $\text{0}\text{.244 atm}\cdot {\text{L}}^{\text{2}}{\text{/mol}}^{\text{2}}$ for $a$, $0.0266\text{ L/mol}$ for $b$, $2.44\mathrm{L}$ for $V$, $30.33\text{ mol}$ for $n$ and $0.08206\text{ L}\cdot \text{atm/mol}\cdot \text{K}$ for $R$ in equation (3).

  $\begin{array}{c}P=\left[\begin{array}{c}\frac{\left(30.33\text{ mol}\right)\left(0.08206\text{ L}\cdot \text{atm/mol}\cdot \text{K}\right)\left(513\text{ K}\right)}{\left(2.44\mathrm{L}-\left(30.33\text{ mol}\right)\left(0.0266\text{ L/mol}\right)\right)}-\\ \frac{\left(\text{0}{\text{.244 atm L}}^{\text{2}}{\text{/mol}}^{\text{2}}\right){\left(30.33\text{ mol}\right)}^2}{{\left(\text{2}\text{.44 L}\right)}^2}\end{array}\right]\\ =744\text{atm}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The pressure of $30.33\text{ mol}$ methane at $240°\text{C}$ in a $2.44\mathrm{L}$ container has to be calculated.

Concept Introduction:

Refer to part (a).

Answer to Problem 6.128QE

The pressure of $30.33\text{ mol}$ methane at $240°\text{C}$ in a $2.44\mathrm{L}$ container is $770.5015\text{ atm}$.

Explanation of Solution

Van der Waals equation is given as follows:

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

Here,

$a$ is a constant that gives the strength of the attractive force.

$b$ gives the measure of size of the gas molecules.

Rearrange the expression (2) to obtain the formula of pressure as follows:

  $P=\frac{nRT}{\left(V-nb\right)}-\frac{a{n}^2}{V^2}$        (3)

Refer to table 6.4 for the values of Van der Waals constants of methane.

Substitute $513\text{ K}$ for $T$, $\text{2}{\text{.25 atm L}}^{\text{2}}{\text{/mol}}^{\text{2}}$ for $a$, $0.0428\text{ L/mol}$ for $b$, $2.44\mathrm{L}$ for $V$, $30.33\text{ mol}$ for $n$ and $0.08206\text{ L}\cdot \text{atm/mol}\cdot \text{K}$ for $R$ in equation (3).

  $\begin{array}{c}P=\left[\begin{array}{c}\frac{\left(30.33\text{ mol}\right)\left(0.08206\text{ L}\cdot \text{atm/mol}\cdot \text{K}\right)\left(513\text{ K}\right)}{\left(2.44\mathrm{L}-\left(\left(30.33\text{ mol}\right)\left(0.0428\text{ L/mol}\right)\right)\right)}-\\ \frac{\left(\text{2}{\text{.25 atm L}}^{\text{2}}{\text{/mol}}^{\text{2}}\right){\left(30.33\text{ mol}\right)}^2}{{\left(2.44\mathrm{L}\right)}^2}\end{array}\right]\\ =1118.161\text{ atm}-347.654\text{ atm}\\ =770.50\text{ atm}\end{array}$

(c)

Interpretation Introduction

Interpretation:

The reason behind the differences in the pressures observed for the two gases has to be determined.

Concept Introduction:

Refer to part (a).

Explanation of Solution

The value of Van der Waals constants $a$ for hydrogen and methane is $\text{0}{\text{.244 atm L}}^{\text{2}}{\text{/mol}}^{\text{2}}$ and $\text{2}{\text{.25 atm L}}^{\text{2}}{\text{/mol}}^{\text{2}}$ respectively. Greater the value of $a$ greater is the contribution of pressure as per the Van der Waals equation. Hence the pressure has been found to be greater in methane than the hydrogen.