Chapter 5, Problem 5.77PAE

(a)

Interpretation Introduction

To determine:

Calculate the pressure of $15.0\text{ g}$ of methane gas in a $\text{1}\text{.50 L}$ vessel at ${45}^{0}C$ using.

1. Apply ideal gas low

Explanation of Solution

No of moles of gas $=\frac{\text{mass of gas }}{{\text{molar mass of CH}}_4}$

$=\frac{15.0\text{ g}}{16.0\text{ g/m}}$

$\begin{array}{l}\left({\text{mass of CH}}_4\left(\text{given}\right)=15.0\text{ g}\right)\\ {\text{molar mass of CH}}_4=12+4=16.0\text{ g/mol}\end{array}$

$\begin{array}{c}V=1.5L\\ =1.5\times {10}^{-3}{m}^3\\ R=8.314J/mol.K\\ T={45}^{0}c=273+45=318K\\ PV=nRT\\ P=\frac{nRT}{V}\\ =\frac{\left(\frac{15}{16}\text{ mole}\right)\times \left(8.314Jmo{l}^{-1}{K}^{-1}\right)\times \left(318K\right)}{1.5\times {10}^{-3}}\\ =1652407.5Pascal\\ =16.30atm\end{array}$

(b)

Interpretation Introduction

To determine:

Calculate the pressure of $15.0g$ of methane gas in a $1.50L$ vessel at ${45.0}^{0}c$ using the van-der Waals equation using constant in table.

Explanation of Solution

$\begin{array}{l}\text{From the table a }\left(atm{L}^{2}mo{l}^{-2}\right)=2.253atm{L}^{2}mo{l}^{-2}\\ b\left(Lmo{l}^{-1}\right)=0.04278Lmo{l}^{-1}\\ R=0.08206Latmmo{l}^{-1}{K}^{-1}\\ V=1.50L\\ n=\left(\frac{15}{16}\right)=0.9375mol\\ T=273+45=318K\end{array}$

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

Putting the value of $a,b,n,R,T,V$ you get

$\begin{array}{l}=\left(P+\frac{\left(2.253atm{L}^{2}mo{l}^{-2}\right){\left(0.9375mol\right)}^2}{{\left(1.50L\right)}^2}\right)\times \left(1.5L-\left(0.9375mol\right)\times \left(0.4278Lmo{l}^{-1}\right)\right)\\ =\left(0.9375mol\times 0.08206Latmmo{l}^{-1}{K}^{-1}\right)\times \left(318K\right)\end{array}$

(Upon simplification)

$\begin{array}{l}P+\frac{2.253\times 0.9375}{{\left(1.50\right)}^2}atm=16.76atm\\ P=15.82atm\end{array}$

Hence $15.82atm$ (real behavior)

Conclusion

Ideal pressure is greater than real pressure (pressure calculated on the real behavior of gas).