Chapter 19, Problem 1DE

(a)

Interpretation Introduction

To determine: The pressure exerted by ${\text{N}}_2$ gas by ideal gas equation.

Answer to Problem 1DE

Solution: The pressure exerted by ${\text{N}}_2$ gas by ideal gas equation is $176.80\text{atm}$ .

Explanation of Solution

Given

The volume of the gas is $1100.0\text{L}$ .

The temperature of gas is $280°\text{C}$ .

The mass of ${\text{N}}_2$ gas is $120.0\text{kg}$ .

The molar mass of nitrogen gas is $28.0\text{g/mol}$ .

The conversion kilogram $\left(\text{kg}\right)$ into gram $\left(\text{g}\right)$ is done as,

$1\text{kg}={10}^3\text{g}$

Hence, the conversion $120\text{kg}$ into gram is,

$120\text{kg}=120\times {10}^3\text{g}$

The conversion of temperature from $°\text{C}$ into Kelvin is done as,

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

The conversion of temperature from $280°\text{C}$ into Kelvin is,

$\begin{array}{c}T\left(\text{K}\right)=\left(280+273\right)\text{K}\\ =553\text{K}\end{array}$

The pressure exerted by ${\text{N}}_2$ gas is calculated by the formula,

$P=\frac{mRT}{MV}$

Where,

• $m$ is the mass of the gas.
• $M$ is the molar mass of the gas.
• $P$ is the absolute pressure of the gas.
• $V$ is the volume of the gas.
• $T$ is the absolute temperature of the gas.
• $R$ is the gas constant $\left(0.08206\text{L}\cdot \text{atm/mol}\cdot \text{K}\right)$ .

Substitute the given values of $T,m,M,V$ and $R$ in above equation.

$\begin{array}{c}P=\frac{120\times {10}^3\text{g}\times 0.08206\text{L}\cdot \text{atm/mol}\cdot \text{K}\times 553\text{K}}{28.0\text{g/mol}\times 1100.0\text{L}}\\ =176.80\text{atm}\end{array}$

Conclusion

The pressure exerted by ${\text{N}}_2$ gas by ideal gas equation is $176.80\text{atm}$ .

(b)

Interpretation Introduction

To determine: The pressure exerted by ${\text{N}}_2$ gas by van der Waals equation.

Answer to Problem 1DE

Solution: The pressure exerted by ${\text{N}}_2$ gas by van der Waals equation is $187.32\text{atm}$ .

Explanation of Solution

Given

The mass of nitrogen gas is $120\text{kg}\left(=120\times {10}^3\text{g}\right)$ .

The number of moles of nitrogen gas is calculated by dividing the mass of nitrogen gas by its molar mass.

$\text{Moles}\text{of}{\text{N}}_2\text{gas}=\frac{\text{Mass}}{\text{Molar}\text{mass}}$

Substitute the value of mass and molar mass of nitrogen gas in the above equation.

$\begin{array}{c}\text{Moles}\text{of}{\text{N}}_2=\frac{120\times {10}^3\text{g}}{28.0\text{g/mol}}\\ =4.285\times {10}^3\text{mol}\end{array}$

Pressure exerted by nitrogen gas is calculated by van der Waals equation,

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

Where,

• $P$ is the pressure of the gas.
• $n$ is the number of moles of the gas.
• $V$ is the volume of gas.
• $R$ is the gas constant $\left(0.08206\text{L}\cdot \text{atm/mol}\cdot \text{K}\right)$ .
• $T$ is the temperature.
• $a$ is van der Waals constant $\left(1.39{\text{L}}^{\text{2}}\cdot {\text{atm/mol}}^{\text{2}}\right)$ . (refer to table 10.3)
  
• $b$ is van der Waals constant $\left(0.0391\text{L/mol}\right)$ .(refer to table 10.3)

Substitute the values of $n,V,T,R,a$ and $b$ in above equation.

$\begin{array}{c}\left[\begin{array}{c}\left(P+\frac{{\left(4.285\times {10}^3\text{mol}\right)}^2\times 1.39{\text{L}}^{\text{2}}\cdot {\text{atm/mol}}^{\text{2}}}{{\left(1100.0\text{L}\right)}^2}\right)\times \\ \left(1100.0\text{L}-4.285\times {10}^3\text{mol}\times 0.0391\text{L/mol}\right)\end{array}\right]=4.285\times {10}^3\text{mol}\times 0.08206\text{L}\cdot \text{atm/mol}\cdot \text{K}\times 553\text{K}\\ \left(P+21.09\text{atm}\right)\left(933\text{L}\right)=194450\text{L-atm}\\ P+21.09\text{atm}=\frac{194450\text{L-atm}}{933\text{L}}\\ P=208.41\text{atm}-21.09\text{atm}\\ =187.32\text{atm}\end{array}$

Conclusion

The pressure exerted by ${\text{N}}_2$ gas by van der Waals equation is $187.32\text{atm}$ .

(c)

Interpretation Introduction

To determine: The dominant correction in the manufacture of ammonia out of the given corrections of finite volume and the attractive interactions.

Answer to Problem 1DE

Solution: The correction of attractive interactions dominates in the manufacture of ammonia.

Explanation of Solution

In the manufacture of ammonia, due to the smaller size and high electronegativity difference of nitrogen and hydrogen the attractive interaction dominates. Nitrogen gas can easily abstract a proton from hydrogen atom due to its high electronegativity. Thereby, hydrogen is highly electropositive, it can easily donate its proton. Therefore, attractive interaction dominated over finite volume correction.

Conclusion

The correction of attractive interactions dominates in the manufacture of ammonia.