Chapter 1, Problem 1C.6E

(a)

n der Waal  ues in the perfect gas equation as follows,

to be detetrmined.

Interpretation Introduction

Interpretation:

The pressure exerted by 1.0 mol of $C_2{H}_6$ behaving as a perfect gas under the given conditions has to be determined.

Concept introduction:

Ideal gas equation,

$\text{PV}\text{=}\text{nRT}\boxed{\begin{array}{l}\text{Where,}\\ \text{P}\text{:}\text{Pressure}\\ \text{V:}\text{Volume}\\ \text{n:}\text{Number}\text{of}\text{moles}\\ \text{R:}\text{Gas}\text{constant}\\ \text{T}\text{:}\text{Temperature}\end{array}}$

Van der Waals equation:

The deviation from ideal gas law was accounted using two constants, a and b and was given by an equation called van der Waals equation which is similar to ideal gas law as follows,

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

  Where,

a and b are constants

P is the pressure

V is the volume

T is the temperature

R is molar gas constant

n is the mole

Using van der Waal’s equation, the pressure can be determined as,

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

Explanation of Solution

The pressure exerted by 1.0 mol of $C_2{H}_6$ behaving as a perfect gas under the given conditions can determined as follows,

1. (i) 273.15 K in $22.414d{m}^3$

Conversions:

$\begin{array}{l}1dm=10cm,1d{m}^3=1000c{m}^3\\ 22.414d{m}^3=22.414L\end{array}$

$\begin{array}{c}\text{n}\text{=}\text{1}\text{mol}\\ \text{R}\text{=}\text{0}\text{.0821}\text{L}\text{atm}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}\\ \text{T}\text{=}\text{273}\text{.15K}\\ \text{V}\text{=}\text{22}\text{.414}\text{L}\end{array}$

Substituting the known values in the perfect gas equation as follows,

$\begin{array}{c}P=\frac{nRT}{V}\\ =\frac{\left(1mol\right)\times \left(0.0821\right)\times \left(273.15K\right)}{22.414L}\\ =1atm\end{array}$

1. (ii) At 1000 K in $100c{m}^3$

$\text{100}{\text{cm}}^{\text{3}}\text{=}\text{100}\text{mL}\text{=}0.1L$

$\begin{array}{c}\text{n}\text{=}\text{1}\text{mol}\\ \text{R}\text{=}\text{0}\text{.0821}\text{L}\text{atm}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}\\ \text{T}\text{=}1000\text{K}\\ \text{V}\text{=}0.1\text{L}\end{array}$

Substituting the known values in the perfect gas equation as follows,

$\begin{array}{c}P=\frac{nRT}{V}\\ =\frac{\left(1mol\right)\times \left(0.0821\right)\times \left(1000K\right)}{0.1L}\\ =820atm\end{array}$

(b)

n der Waal  ues in the perfect gas equation as follows,

to be detetrmined.

Interpretation Introduction

Interpretation:

The pressure exerted by 1.0 mol of $C_2{H}_6$ behaving as a van der Waal gas under the given conditions has to be determined.

Concept introduction:

Ideal gas equation,

$\text{PV}\text{=}\text{nRT}\boxed{\begin{array}{l}\text{Where,}\\ \text{P}\text{:}\text{Pressure}\\ \text{V:}\text{Volume}\\ \text{n:}\text{Number}\text{of}\text{moles}\\ \text{R:}\text{Gas}\text{constant}\\ \text{T}\text{:}\text{Temperature}\end{array}}$

Van der Waals equation:

The deviation from ideal gas law was accounted using two constants, a and b and was given by an equation called van der Waals equation which is similar to ideal gas law as follows,

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

  Where,

a and b are constants

P is the pressure

V is the volume

T is the temperature

R is molar gas constant

n is the mole

Using van der Waal’s equation, the pressure can be determined as,

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

Explanation of Solution

The pressure exerted by 1.0 mol of $C_2{H}_6$ behaving as a van der Waal gas under the given conditions can be determined as follows,

1. (i) 273.15 K in $22.414d{m}^3$

Conversions:

$\begin{array}{l}1dm=10cm,1d{m}^3=1000c{m}^3\\ 22.414d{m}^3=22.414L\end{array}$

$\begin{array}{c}\text{n}\text{=}\text{1}\text{mol}\\ \text{R}\text{=}\text{0}\text{.0821}\text{L}\text{atm}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}\\ \text{T}\text{=}\text{273}\text{.15K}\\ \text{V}\text{=}\text{22}\text{.414}\text{L}\end{array}$

Using van der Waal’s equation, the pressure can be determined as,

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

$\begin{array}{l}a=5.507atmd{m}^{6}mo{l}^{-1}\\ b=6.51\times {10}^{-2}d{m}^{3}mo{l}^{-1}=6.51\times {10}^{-2}Lmo{l}^{-1}\end{array}$

Substituting the known values in the van der Waal gas equation as follows,

$\begin{array}{c}P=\text{}\frac{nRT}{V-nb}-\frac{n^{2}a}{V^2}\\ =\frac{\left(1mol\right)\left(0.0821\right)\left(273.15\right)}{22.414-\left(1mol\times 6.51\times {10}^{-2}\right)}-\frac{{\left(1\right)}^2\times 5.507atmd{m}^{6}mo{l}^{-1}}{{\left(22.414d{m}^3\right)}^2}\\ =0.99atm\end{array}$

1. (ii) At 1000 K in $100c{m}^3$

$\text{100}{\text{cm}}^{\text{3}}\text{=}\text{100}\text{mL}\text{=}0.1L$

$\begin{array}{c}\text{n}\text{=}\text{1}\text{mol}\\ \text{R}\text{=}\text{0}\text{.0821}\text{L}\text{atm}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}\\ \text{T}\text{=}1000\text{K}\\ \text{V}\text{=}0.1\text{L}\end{array}$

$\begin{array}{l}a=5.507atmd{m}^{6}mo{l}^{-1}\\ b=6.51\times {10}^{-2}d{m}^{3}mo{l}^{-1}=6.51\times {10}^{-2}Lmo{l}^{-1}\end{array}$

Substituting the known values in the van der Waal gas equation as follows,

$\begin{array}{c}P=\text{}\frac{nRT}{V-nb}-\frac{n^{2}a}{V^2}\\ =\frac{\left(1mol\right)\left(0.0821\right)\left(1000K\right)}{0.1L-\left(1mol\times 6.51\times {10}^{-2}\right)}-\frac{{\left(1\right)}^2\times 5.507atmd{m}^{6}mo{l}^{-1}}{{\left(0.1d{m}^3\right)}^2}\\ =1.8\times 10^{3}atm\end{array}$

n der Waal  ues in the perfect gas equation as follows,

to be detetrmined.

n der Waal  ues in the perfect gas equation as follows,

to be detetrmined.