Scan_Doc0106 - However, m/V is the gas density d in units...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
In this case n = 0.189 mol, T = 125°C + 273 = 398 K, P R = 0.08206 L . atmlK . mol. Thus (0.189 mot)(0.08206 L . atrrfIK . mot)(398 K) V = = 2.47 L 2.50 atm 198 Chapter Five Gases mass Density = volume EXAMPLE 5.14 2.50 atm, ane This represents the volume of CO 2 produced under these conditions. SEE EXERCISES 5.69 AND 5.72 Molar Mass of a Gas One very important use of the ideal gas law is in the calculation of the molar mass (mo- lecular weight) of a gas from its measured density. To see the relationship between gas density and molar mass, consider that the number of moles of gas n can be expressed as grams of gas mass m n= - molar mass molar mass molar mass Substitution into the ideal gas equation gives nRT (m/molar mass)RT m(RT) P=--= =----- V V V(molar mass)
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: However, m/V is the gas density d in units of grams per liter. Thus dRT P=----molar mass or dRT Molar mass = p (5.1) Thus, if the density of a gas at a given temperature and pressure is known, its molar mas can be calculated. The density of a gas was measured at 1.50 atm and 27C and found to be 1.95 g/L. Cal-culate the molar mass of the gas. Solution Where are we going? To determine the molar- mass of the gas What do we know? ,/ P = 1.50 atm ,/ T = 27C + 273 = 300. K ,/ d = 1.95 g/L What information do we need? dRT ,/ Molar mass = -P ,/ R = 0.08206 L . atmlK . mol...
View Full Document

This note was uploaded on 12/13/2010 for the course CHEM 2301 taught by Professor Bill during the Spring '10 term at South Texas College.

Ask a homework question - tutors are online