Chapter 5, Problem 5.40QP

Interpretation Introduction

Interpretation:

The volume of gas at ${\text{150}}^{\text{o}}\text{C}$ has to be calculated.

Explanation of Solution

Concept introduction:

Ideal gas: A hypothetical gas whose molecules occupy negligible space and have no interactions, and which consequently obeys the gas law exactly.

Ideal gas equation: An equation describes the relationship among the four variables P, V, T, and n.

$PV=nRT$

The unit of each term in the equation is,

Where, R is proportionality constant called gas constant $\left(atm.L/mol.K\right)$

V is volume $\left(L\right)$

n is number of moles $\left(moles\right)$

P is Pressure $\left(atm\right)$

T is temperature $\left(K\right)$

Standard temperature and pressure: The conditions ${\text{0}}^{\text{o}}\text{C}$ and $\text{1}\text{atm}$ are called standard atmospheric conditions.

The ideal gas equation is, $PV=nRT-----\left(1\right)$

Where, R is proportionality constant called gas constant $\left(atm.L/mol.K\right)$

V is volume $\left(L\right)$

N is number of moles $\left(moles\right)$

P is Pressure $\left(atm\right)$

T is temperature $\left(K\right)$

Given:

At STP,

Volume, ${\text{V}}_{\text{1}}$=$\text{488}\text{mL}\text{=}\text{488}{\text{×10}}^{\text{-3}}\text{L}$;

Pressure, ${\text{P}}_{\text{1}}$= $\text{1}\text{atm}$

Temperature, ${\text{T}}_{\text{1}}$ = $0^{\text{o}}\text{C}\text{=}{\text{0}}^{\text{o}}\text{C}\text{+}\text{273}\text{=}273\text{K}$

At non-standard conditions,

Volume, ${\text{V}}_{\text{2}}$ =$\text{?}$;

Pressure, ${\text{P}}_{\text{2}}$ = $\text{22}\text{.5}\text{atm}$

Temperature, ${\text{T}}_{\text{2}}$ = ${\text{150}}^{\text{o}}\text{C}\text{=}{\text{150}}^{\text{o}}\text{C}\text{+}\text{273}\text{=}42\text{3K}$

Substituting the known values into the equation below as follows,

${\text{P}}_{\text{1}}{\text{V}}_{\text{1}}\text{=}{\text{nRT}}_{\text{1}}\text{and}{\text{P}}_{\text{2}}{\text{V}}_{\text{2}}\text{=}{\text{nRT}}_{\text{2}}$

By combing both mentioned above equation,

$\begin{array}{l}{\text{P}}_{\text{1}}{\text{V}}_{\text{1}}\text{=}{\text{nRT}}_{\text{1}}\text{and}{\text{P}}_{\text{2}}{\text{V}}_{\text{2}}\text{=}{\text{nRT}}_{\text{2}}\\ \\ \frac{{\text{P}}_{\text{1}}{\text{V}}_{\text{1}}}{{\text{P}}_{\text{2}}{\text{V}}_{\text{2}}}=\frac{\cancel{\text{nR}}{\text{T}}_{\text{1}}}{\cancel{\text{nR}}{\text{T}}_{\text{2}}}\\ \\ \frac{{\text{P}}_{\text{1}}{\text{V}}_{\text{1}}}{{\text{P}}_{\text{2}}{\text{V}}_{\text{2}}}=\frac{{\text{T}}_{\text{1}}}{{\text{T}}_{\text{2}}}\end{array}$

On substituting,

$\begin{array}{l}\frac{{\text{P}}_{\text{1}}{\text{V}}_{\text{1}}}{{\text{P}}_{\text{2}}{\text{V}}_{\text{2}}}\text{=}\frac{{\text{T}}_{\text{1}}}{{\text{T}}_{\text{2}}}\\ \\ \frac{\left(\text{1}\cancel{\text{atm}}\right)\left(\text{488}{\text{×10}}^{\text{-3}}\text{L}\right)}{\left(\text{22}\text{.5}\cancel{\text{atm}}\right)\left({\text{V}}_{\text{2}}\right)}\text{=}\frac{\text{273}\cancel{\text{K}}}{\text{423}\cancel{\text{K}}}=0.70449\\ \\ \frac{1}{{\text{V}}_{\text{2}}}=\frac{\left(0.34539\right)\left(\text{22}\text{.5}\right)}{\left(\text{488}{\text{×10}}^{\text{-3}}\text{L}\right)}\\ \\ {\text{V}}_{\text{2}}=\frac{\left(\text{488}{\text{×10}}^{\text{-3}}\text{L}\right)}{\left(0.34539\right)\left(\text{22}\text{.5}\right)}\\ \\ =\frac{\text{488}{\text{×10}}^{\text{-3}}\text{L}}{14.52127}=33.6\times 10^{-3}L\end{array}$

Therefore, the volume obtained is, $33.6\times 10^{-3}L$

Note: As the pressure is increased, the volume decreases. The obtained value of volume is less than the initial volume. Hence, the obtained value is reliable.