Chapter 1, Problem 1A.1ST

Interpretation Introduction

Interpretation:

The molar volume of a perfect gas has to be calculated at $\text{0}\text{.0 °C and 1}\text{.000 atm}$.

Concept Introduction:

Ideal gas law:

The ideal gas law state that the relationship between pressure, temperature, volume and number of moles, it is

    $\begin{array}{l}\text{P}\text{V}\text{=}\text{n}\text{RT}\\ \text{Where,}\\ \text{P}\text{-presure}\\ \text{T}\text{-}\text{temperature}\\ \text{V-}\text{volume}\\ \text{n-}\text{number}\text{of}\text{mole}\\ \text{R-}\text{gas}\text{constant}\end{array}$

Explanation of Solution

Given:

Pressure $\text{=}\text{1}\text{.00}\text{atm}\text{=101}\text{.325}\text{kPa}$

Temperature $\text{=}\text{0}\text{.0°C}\text{=}\text{273}\text{.15}\text{K}$

Gas constant $\text{=}\text{8}{\text{.314dm}}^{\text{3}}{\text{ kPa K}}^{\text{-1}}{\text{ mol}}^{\text{-1}}$ 

Volume of 1 molar gas is,

    $\begin{array}{l}\text{V}\text{=}\frac{\text{nRT}}{\text{P}}\\ \text{=}\frac{\text{1×8}{\text{.314 dm}}^{\text{3}}{\text{ kPa K}}^{\text{-1}}{\text{ mol}}^{\text{-1}}\text{×273}\text{.15}\text{K}}{\text{101}\text{.325}\text{kPa}}\\ \text{=}\text{22}\text{.41}{\text{dm}}^{\text{3}}\end{array}$

Hence, the molar volume of a perfect gas is $\text{22}\text{.41}{\text{dm}}^{\text{3}}$.