Chapter 1, Problem 1C.8BE

(i)

Interpretation Introduction

Interpretation: The temperature and pressure at which $1.0\text{mol}$ of ${\text{H}}_{\text{2}}\text{O}$ will be in states that corresponds to $1.0\text{mol}$ of ${\text{N}}_2$ at $1.0\text{atm}$ and $25°\text{C}$ has to be calculated.

Concept introduction: According to 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.8BE

The temperature and pressure at which $1.0\text{mol}$ of ${\text{H}}_{\text{2}}\text{O}$ corresponds to the state of $1.0\text{mol}$ ${\text{N}}_2$ at $1.0\text{atm}$ and $25°\text{C}$ is $\underset{\_}{5754.738K}$ and $\underset{\_}{1.72atm}$, respectively.

Explanation of Solution

The pressure of ${\text{N}}_2$ is $1\text{atm}$ and temperature of ${\text{N}}_2$ 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.
• Ø  $p$ is the pressure.
• Ø  $T$ is the temperature.

The critical temperature of ${\text{N}}_2$ is $33.54\text{K}$ and the critical pressure of ${\text{N}}_2$ is $126.3\text{atm}$.

Substitute the critical pressure of ${\text{N}}_2$ in equation (1).

    $\begin{array}{c}{p}_t=\frac{1.0\text{atm}}{126.3\text{atm}}\\ =0.0079\end{array}$

Hence, the reduced pressure of ${\text{N}}_2$ is $\underset{\_}{0.0079}$.

Substitute the critical temperature of ${\text{N}}_2$ in equation (2).

    $\begin{array}{c}{T}_t=\frac{298.15\text{K}}{33.54\text{K}}\\ =8.889\end{array}$

Hence, the reduced temperature of ${\text{N}}_2$ is $\underset{\_}{8.889}$.

The critical temperature of ${\text{H}}_{\text{2}}\text{O}$ is $647.4\text{K}$ and the critical pressure of ${\text{H}}_{\text{2}}\text{O}$ is $218.3\text{atm}$.

According to the principle of corresponding states, the value of reduced pressure for ${\text{H}}_{\text{2}}\text{O}$ is same as that of ${\text{N}}_2$.  Thus, the reduced pressure for ${\text{H}}_{\text{2}}\text{O}$ is $0.0079$.

The pressure at which $1.0\text{mol}$ of ${\text{H}}_{\text{2}}\text{O}$ is in corresponding the state of ${\text{N}}_2$ can be calculated using the equation shown below.

    $p=p_t\times p_c$                                                                                                     (3)

Substitute the reduced pressure and critical pressure of ${\text{H}}_{\text{2}}\text{O}$ in equation (3).

    $\begin{array}{c}p=0.0079\times 218.3\text{atm}\\ =\underset{\_}{1.72atm}\end{array}$

Hence, the pressure of $1.0\text{mol}$ of ${\text{H}}_{\text{2}}\text{O}$ is $\underset{\_}{1.72atm}$.

The temperature at which $1.0\text{mol}$ of ${\text{H}}_{\text{2}}\text{O}$ is in corresponding the state of ${\text{H}}_2$ can be calculated using the equation shown below,

    $T=T_t\times T_c$                                                                                                      (4)

Substitute the reduced temperature and critical temperature of ${\text{H}}_{\text{2}}\text{O}$ in equation (4).

    $\begin{array}{c}T=8.889\times 647.4\text{K}\\ =\underset{\_}{5754.738K}\end{array}$

Hence, the temperature of $1.0\text{mol}$ of ${\text{H}}_{\text{2}}\text{O}$ is $\underset{\_}{5754.738K}$.

(ii)

Interpretation Introduction

Interpretation: The temperature and pressure at which $1.0\text{mol}$ of ${\text{CO}}_{\text{2}}$ will be in states that corresponds to $1.0\text{mol}$ of ${\text{N}}_2$ at $1.0\text{atm}$ and $25°\text{C}$.

Concept introduction: According to 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.8BE

The temperature and pressure at which $1.0\text{mol}$ of ${\text{CO}}_{\text{2}}$ is in corresponding the state of ${\text{N}}_2$ is $\underset{\_}{2704.0338K}$ and $\underset{\_}{0.57591atm}$, respectively.

Explanation of Solution

The pressure of ${\text{N}}_2$ is $1\text{atm}$ and temperature of ${\text{N}}_2$ 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.
• Ø  $p$ is the pressure.
• Ø  $T$ is the temperature.

The critical temperature of ${\text{N}}_2$ is $33.54\text{K}$ and the critical pressure of ${\text{N}}_2$ is $126.3\text{atm}$.

Substitute the critical pressure of ${\text{N}}_2$ in equation (1).

    $\begin{array}{c}{p}_t=\frac{1.0\text{atm}}{126.3\text{atm}}\\ =0.0079\end{array}$

Hence, the reduced pressure of ${\text{N}}_2$ is $\underset{\_}{0.0079}$.

Substitute the critical temperature of ${\text{N}}_2$ in equation (2).

    $\begin{array}{c}{T}_t=\frac{298.15\text{K}}{33.54\text{K}}\\ =8.889\end{array}$

Hence, the reduced temperature of ${\text{N}}_2$ is $\underset{\_}{8.889}$.

The critical temperature of ${\text{CO}}_{\text{2}}$ is $304.2\text{K}$ and the critical pressure of ${\text{CO}}_{\text{2}}$ is $72.9\text{atm}$.

According to the principle of corresponding states, the value of reduced pressure for ${\text{CO}}_{\text{2}}$ is same as that of ${\text{N}}_2$.  Thus, the reduced pressure for ${\text{CO}}_{\text{2}}$ is $0.0079$.

The pressure at which $1.0\text{mol}$ of ${\text{CO}}_{\text{2}}$ is in corresponding the state of ${\text{N}}_2$ can be calculated using the equation shown below.

    $p=p_t\times p_c$                                                                                                     (3)

Substitute the reduced pressure and critical pressure of ${\text{CO}}_{\text{2}}$ in equation (3).

    $\begin{array}{c}p=0.0079\times 72.9\text{atm}\\ =\underset{\_}{0.57591atm}\end{array}$

Hence, the pressure of $1.0\text{mol}$ of ${\text{CO}}_{\text{2}}$ is $\underset{\_}{0.57591atm}$.

The temperature at which $1.0\text{mol}$ of ${\text{CO}}_{\text{2}}$ is in corresponding the state of ${\text{N}}_2$ can be calculated using the equation shown below,

    $T=T_t\times T_c$                                                                                                      (4)

Substitute the reduced temperature and critical temperature of ${\text{CO}}_{\text{2}}$ in equation (4).

    $\begin{array}{c}T=8.889\times 304.2\text{K}\\ =\underset{\_}{2704.0338K}\end{array}$

Hence, the temperature of $1.0\text{mol}$ of ${\text{CO}}_{\text{2}}$ is $\underset{\_}{2704.0338K}$.

(iii)

Interpretation Introduction

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

Concept introduction: According to 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.8BE

The temperature at which $1.0\text{mol}$ of $\text{Ar}$ is in corresponding the state of ${\text{N}}_2$ is $\underset{\_}{1339.57K}$ and $\underset{\_}{0.3792atm}$, respectively.

Explanation of Solution

The pressure of ${\text{N}}_2$ is $1\text{atm}$ and temperature of ${\text{N}}_2$ 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.
• Ø  $p$ is the pressure.
• Ø  $T$ is the temperature.

The critical temperature of ${\text{N}}_2$ is $33.54\text{K}$ and the critical pressure of ${\text{N}}_2$ is $126.3\text{atm}$.

Substitute the critical pressure of ${\text{N}}_2$ in equation (1).

    $\begin{array}{c}{p}_t=\frac{1.0\text{atm}}{126.3\text{atm}}\\ =0.0079\end{array}$

Hence, the reduced pressure of ${\text{N}}_2$ is $\underset{\_}{0.0079}$.

Substitute the critical temperature of ${\text{N}}_2$ in equation (2).

    $\begin{array}{c}{T}_t=\frac{298.15\text{K}}{33.54\text{K}}\\ =8.889\end{array}$

Hence, the reduced temperature of ${\text{N}}_2$ is $\underset{\_}{8.889}$.

The critical temperature of $\text{Ar}$ is $150.7\text{K}$ and the critical pressure of $\text{Ar}$ is $48\text{atm}$.

According to the principle of corresponding states, the value of reduced pressure for $\text{Ar}$ is same as that of ${\text{N}}_2$.  Thus, the reduced pressure for $\text{Ar}$ is $0.0079$.

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

    $p=p_t\times p_c$                                                                                                    (3)

Substitute the reduced pressure and critical pressure of $\text{Ar}$ in equation (3).

    $\begin{array}{c}p=0.0079\times 48\text{atm}\\ =\underset{\_}{0.3792atm}\end{array}$

Hence, the pressure of $1.0\text{mol}$ of $\text{Ar}$ is $\underset{\_}{0.3792atm}$.

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

    $T=T_t\times T_c$                                                                                                     (4)

Substitute the reduced temperature and critical temperature of $\text{Ar}$ in equation (4).

    $\begin{array}{c}T=8.889\times 150.7\text{K}\\ =\underset{\_}{1339.57K}\end{array}$

Hence, the temperature of $1.0\text{mol}$ of $\text{Ar}$ is $\underset{\_}{1339.57K}$.