Chapter 1, Problem 1C.8AE

(i)

Interpretation Introduction

Interpretation: The temperature and pressure at which $1.0\text{mol}$ of ${\text{NH}}_{\text{3}}$ will be in states that corresponds to $1.0\text{mol}$ of ${\text{H}}_2$ at $1.0\text{atm}$ and $25°\text{C}$.

Concept introduction: The principle of corresponding states, the real gases that have the same reduced pressure, temperature and volume are said to correspond.

Answer to Problem 1C.8AE

The temperature and pressure at which $1.0\text{mol}$ of ${\text{NH}}_3$ is in corresponding the state of ${\text{H}}_2$ is $\underset{\_}{3637.335K}$ and $\underset{\_}{8.6814atm}$, respectively.

Explanation of Solution

The pressure of ${\text{H}}_2$ is $1\text{atm}$.  The temperature of ${\text{H}}_2$ is $25°\text{C}$.

The conversion of given temperature $\left(°\text{C}\right)$ into K is shown below.

$\begin{array}{c}\text{K}=°\text{C+273}\\ =25°\text{C}+273\\ =298\text{K}\end{array}$

The reduced pressure and temperature is given by,

    $p_t=\frac{p}{p_c}$ (1)

  $T_t=\frac{T}{T_c}$                                                                                                            (2)

Where,

• Ø  $T_c$ is the critical temperature.
• Ø  $p_c$ is the critical pressure.

The critical temperature of ${\text{H}}_2$ is $33.23\text{K}$ and the critical pressure of ${\text{H}}_2$ is $12.8\text{atm}$.

Using equation (1), the reduced pressure is,

    $\begin{array}{c}{p}_t=\frac{p}{p_c}\\ =\frac{1.0\text{atm}}{12.8\text{atm}}\\ =0.078\end{array}$

Hence, the reduced pressure of ${\text{H}}_2$ is $0.078$.

Using equation (2), the reduced temperature is,

    $\begin{array}{c}{T}_t=\frac{T}{T_c}\\ =\frac{298.15\text{K}}{33.23\text{K}}\\ =8.97\end{array}$

Hence, the reduced temperature of ${\text{H}}_2$ is $8.97$.

The critical temperature of ${\text{NH}}_3$ is $405.5\text{K}$ and the critical pressure of ${\text{NH}}_3$ is $111.3\text{atm}$.

According to the principle of corresponding states, the value of reduced pressure for ${\text{NH}}_3$ is same as that of ${\text{H}}_2$.

The pressure at which $1.0\text{mol}$ of ${\text{NH}}_3$ is in corresponding the state of ${\text{H}}_2$ can be calculated using the equation shown below.

    $\begin{array}{c}p=p_t\times p_c\\ =0.078\times 111.3\text{atm}\\ =\underset{\_}{8.6814atm}\end{array}$

Hence, the pressure of $1.0\text{mol}$ of ${\text{NH}}_3$ is $\underset{\_}{8.6814atm}$.

The temperature at which $1.0\text{mol}$ of ${\text{NH}}_3$ is in corresponding the state of ${\text{H}}_2$ can be calculated using the equation shown below,

    $\begin{array}{c}T=T_t\times T_c\\ =8.97\times 405.5\text{K}\\ =\underset{\_}{3637.335K}\end{array}$

Hence, the temperature of $1.0\text{mol}$ of ${\text{NH}}_3$ is $\underset{\_}{3637.335K}$.

(ii)

Interpretation Introduction

Interpretation: The temperature and pressure at which $1.0\text{mol}$ of $\text{Xe}$ will be in states that corresponds to $1.0\text{mol}$ of ${\text{H}}_2$ at $1.0\text{atm}$ and $25°\text{C}$.

Concept introduction: The principle of corresponding states, the real gases that have the same reduced pressure, temperature and volume are said to correspond.

Answer to Problem 1C.8AE

The temperature and pressure at which $1.0\text{mol}$ of $\text{Xe}$ is in corresponding the state of ${\text{H}}_2$ is $\underset{\_}{2599.0575K}$ and $\underset{\_}{4.524atm}$, respectively.

Explanation of Solution

The pressure of ${\text{H}}_2$ is $1\text{atm}$.  The temperature of ${\text{H}}_2$ is $25°\text{C}$.

The conversion of given temperature $\left(°\text{C}\right)$ into K is shown below.

$\begin{array}{c}\text{K}=°\text{C+273}\\ =25°\text{C}+273\\ =298\text{K}\end{array}$

The principle of corresponding states, the real gases that have the same reduced pressure, temperature and volume are said to correspond.  The reduced pressure and temperature is given by,

    $p_t=\frac{p}{p_c}$ (1)

  $T_t=\frac{T}{T_c}$                                                                                                            (2)

Where,

• Ø  $T_c$ is the critical temperature.
• Ø  $p_c$ is the critical pressure.

The critical temperature of ${\text{H}}_2$ is $33.23\text{K}$ and the critical pressure of ${\text{H}}_2$ is $12.8\text{atm}$.

Using equation (1), the reduced pressure is,

    $\begin{array}{c}{p}_t=\frac{p}{p_c}\\ =\frac{1.0\text{atm}}{12.8\text{atm}}\\ =0.078\end{array}$

Hence, the reduced pressure of ${\text{H}}_2$ is $0.078$.

Using equation (2), the reduced temperature is,

    $\begin{array}{c}{T}_t=\frac{T}{T_c}\\ =\frac{298.15\text{K}}{33.23\text{K}}\\ =8.97\end{array}$

Hence, the reduced temperature of ${\text{H}}_2$ is $8.97$.

The critical temperature of $\text{Xe}$ is $\text{289}\text{.75K}$ and the critical pressure of $\text{Xe}$ is $\text{58}\text{.0}\text{atm}$.

According to the principle of corresponding states, the value of reduced pressure for $\text{Xe}$ is same as that of ${\text{H}}_2$.

The pressure at which $1.0\text{mol}$ of $\text{Xe}$ is in corresponding the state of ${\text{H}}_2$ can be calculated using the equation shown below.

    $\begin{array}{c}p=p_t\times p_c\\ =0.078\times 58\text{atm}\\ =\underset{\_}{4.524atm}\end{array}$

Hence, the pressure of $1.0\text{mol}$ of $\text{Xe}$ is $\underset{\_}{4.524atm}$.

The temperature at which $1.0\text{mol}$ of $\text{Xe}$ is in corresponding the state of ${\text{H}}_2$ can be calculated using the equation shown below,

    $\begin{array}{c}T=T_t\times T_c\\ =8.97\times 289.75\text{K}\\ =\underset{\_}{2599.0575K}\end{array}$

Hence, the temperature of $1.0\text{mol}$ of $\text{Xe}$ is $\underset{\_}{2599.0575K}$.

(iii)

Interpretation Introduction

Interpretation: The temperature and pressure at which $1.0\text{mol}$ of $\text{He}$ will be in states that corresponds to $1.0\text{mol}$ of ${\text{H}}_2$ at $1.0\text{atm}$ and $25°\text{C}$.

Concept introduction: The principle of corresponding states, the real gases that have the same reduced pressure, temperature and volume are said to correspond.

Answer to Problem 1C.8AE

The temperature and pressure at which $1.0\text{mol}$ of $\text{He}$ is in corresponding the state of ${\text{H}}_2$ is $\underset{\_}{46.644K}$ and $\underset{\_}{0.17628atm}$, respectively.

Explanation of Solution

The pressure of ${\text{H}}_2$ is $1\text{atm}$.  The temperature of ${\text{H}}_2$ is $25°\text{C}$.

The conversion of given temperature $\left(°\text{C}\right)$ into K is shown below.

$\begin{array}{c}\text{K}=°\text{C+273}\\ =25°\text{C}+273\\ =298\text{K}\end{array}$

The principle of corresponding states, the real gases that have the same reduced pressure, temperature and volume are said to correspond.  The reduced pressure and temperature is given by,

    $p_t=\frac{p}{p_c}$ (1)

  $T_t=\frac{T}{T_c}$                                                                                                            (2)

Where,

• Ø  $T_c$ is the critical temperature.
• Ø  $p_c$ is the critical pressure.

The critical temperature of ${\text{H}}_2$ is $33.23\text{K}$ and the critical pressure of ${\text{H}}_2$ is $12.8\text{atm}$.

Using equation (1), the reduced pressure is,

    $\begin{array}{c}{p}_t=\frac{p}{p_c}\\ =\frac{1.0\text{atm}}{12.8\text{atm}}\\ =0.078\end{array}$

Hence, the reduced pressure of ${\text{H}}_2$ is $0.078$.

Using equation (2), the reduced temperature is,

    $\begin{array}{c}{T}_t=\frac{T}{T_c}\\ =\frac{298.15\text{K}}{33.23\text{K}}\\ =8.97\end{array}$

Hence, the reduced temperature of ${\text{H}}_2$ is $8.97$.

The critical temperature of $\text{He}$ is $5.2\text{K}$ and the critical pressure of $\text{He}$ is $2.26\text{atm}$.

According to the principle of corresponding states, the value of reduced pressure for $\text{He}$ is same as that of ${\text{H}}_2$.

The pressure at which $1.0\text{mol}$ of $\text{He}$ is in corresponding the state of ${\text{H}}_2$ can be calculated using the equation shown below.

    $\begin{array}{c}p=p_t\times p_c\\ =0.078\times 2.26\text{atm}\\ =\underset{\_}{0.17628atm}\end{array}$

Hence, the pressure of $1.0\text{mol}$ of $\text{He}$ is $\underset{\_}{0.17628atm}$.

The temperature at which $1.0\text{mol}$ of $\text{He}$ is in corresponding the state of ${\text{H}}_2$ can be calculated using the equation shown below,

    $\begin{array}{c}T=T_t\times T_c\\ =8.97\times 5.2\text{K}\\ =\underset{\_}{46.644K}\end{array}$

Hence, the temperature of $1.0\text{mol}$ of $\text{He}$ is $\underset{\_}{46.644K}$.