Chapter 11.6, Problem 10PPB

Interpretation Introduction

Interpretation:

The pressure of the gas calculated using the ideal gas equation and the van der Waals equation has to be explained.

Concept introduction:

Ideal gas is the most usually used form of the ideal gas equation, which describes the relationship among the four variables $\text{P, V, n, and T}\text{.}$  An ideal gas is a hypothetical sample of gas whose pressure-volume-temperature behavior is predicted accurately by the ideal gas equation.

$\text{PV = nRT}$

Van der Waals equation:

We find the pressure by first solving algebraically for $\text{P}\text{.}$

$\text{P=}\frac{\text{nRT}}{\text{V-nb}}\text{-}\frac{{\text{an}}^{\text{2}}}{{\text{V}}^{\text{2}}}$

Answer to Problem 10PPB

(a)

The pressure of the gas calculated using the ideal gas equation is $\text{1}\text{.3}\text{atm}$

(b)

The pressure of the gas calculated using the van der Waals equation is $\text{1}\text{.3}\text{atm}$

Explanation of Solution

(a)

To determine: The pressure of the gas calculated using the ideal gas equation

The pressure of the gas calculated using the ideal gas equation is calculated below

Given,

$\begin{array}{l}\text{T}\text{=}\text{273}\text{+}\text{32}\text{K}\text{,}\text{n}\text{=}\text{0}\text{.35}\text{mol}\\ \text{V=}\text{6}\text{.50}\text{L}\end{array}$

Where

$\text{P}\text{=}\frac{\text{nRT}}{\text{V}}\text{=}\frac{\text{(0}\text{.35}\text{mol)}\left(\frac{\text{0}\text{.08206}\text{L}\text{.}\text{atm}}{\text{K}\text{.}\text{mol}}\right)\text{(305}\text{K)}}{\text{(6}\text{.50}\text{L)}}\text{=}\text{1}\text{.3}\text{atm}$

The pressure of the gas calculated using the ideal gas equation is $\text{1}\text{.3}\text{atm}$

(b)

To determine: The pressure of the gas calculated using the van der Waals equation

The pressure of the gas calculated using the van der Waals equation is calculated below

Given,

$\begin{array}{l}\text{T}\text{=}305\text{K}\text{,}\text{a}\text{=}\text{4}\text{.17}\text{atm}\text{.}{\text{L/mol}}^{\text{2}}\text{,}\\ \text{b}\text{=}\text{0}\text{.0371}\text{L/mol}\end{array}$

Where

Evaluating the correction terms in the van der Waals equation, we get

$\begin{array}{l}\frac{{\text{an}}^{\text{2}}}{{\text{V}}^{\text{2}}}\text{=}\frac{{\text{(0}\text{.35}\text{mol)}}^{\text{2}}\left(\frac{\text{4}\text{.17}\text{atm}\text{.}{\text{L}}^{\text{2}}}{{\text{mol}}^{\text{2}}}\right)}{{\text{(6}\text{.50}\text{L)}}^{\text{2}}}\text{=}\text{0}\text{.012}\text{atm}\\ \\ \text{nb}\text{=}\text{(0}\text{.35}\text{mol)}\left(\frac{\text{0}\text{.0371}\text{L}}{\text{mol}}\right)\text{=}\text{0}\text{.0129}\text{L}\end{array}$

$\begin{array}{l}\text{(P+}\text{0}\text{.012}\text{atm)(6}\text{.50L-0}\text{.0129}\text{L)}\text{=}\left(\frac{\text{0}\text{.08206}\text{L}\text{.}\text{atm}}{\text{K}\text{.}\text{mol}}\right)\text{(305}\text{K)}\\ \text{P = }\text{1}\text{.3}\text{atm}\end{array}$

The pressure of the gas calculated using the van der Waals equation is $\text{1}\text{.3}\text{atm}$

Conclusion

The pressure of the gas calculated using the ideal gas equation and the van der Waals equation was explained.