Ideal Gas Lawpreap

Avogadro's Hypothesis • Avogadro's Hypothesis – Equal volumes of gases at the same temperature and pressure contain equal numbers of particles. – 1 mole of any gas @ STP has 6.02 x 1023molecules – 22.4 L = 1 mole @ STP V1 = V2 n1

n2

Where V = volume and n = moles

Sample Problem Avogadro • A student placed 4.2 moles of a gas in a container, measuring the volume to be 1.25 L. The student then added more gas particles increasing the number of moles to 8.6. What is the new volume of the gas? V1 = V2 n1 n2 1.25 = V2 4.2 8.6 V2 = 2.6 Liters

Gay-Lussac’s Law of Combining Gases • If temperature & pressure are constant, constant the volumes of gases can be given as small, whole # ratios. – Coefficients in balanced equations can be used to figure out the relationships between liters of gas.

• 2H2 + O2  2H2O • 1 L of O2 will combine with 2L of H2 to make 2L of H2O.

Graham’s Law • Graham’s Law of Effusion: Gases at the same temperature & pressure have rates of effusion inversely proportional to the square roots of their molar masses. • Bigger gas molecules move slower through openings based on mass.

Ideal Gas Law • Ideal Gas Law – Describes the behavior of an ideal gas. PV = nRT P = Pressure ( in atmospheres usually) V = volume (in Liters) n = moles R = 0.0821 L atm/mole K OR 8.31 kPa L/mol K T = Temperature (in Kelvin)

Sample Problem - Ideal • How many moles of oxygen will occupy a volume of 2.5 Liters at 1.2 atm and 25 °C? – n=? – V = 2.5 L – P = 1.2 atm – T = 298 K (273.15 + 25) – R = 0.0821 L atm/mole K PV = nRT (1.2)(2.5) = n (0.0821)(298) n= (1.2)(2.5) = 0.12 moles (0.0821)(298)

Your Turn! • How many liters of hydrogen will occupy a container of 6.32 moles of gas at 2.45 atm and 53.0 °C? – n= 6.32 – V=? – P = 2.45 atm – T = 326.15 K (273.15 + 53) – R = 0.0821 L atm/mole K PV = nRT (2.45)V = (6.32)(0.0821)(326.15) V = (6.32)(0.0821)(326.15) = 69.1 Liters

Another One! • How many moles of C2H4 are in a 15.0 L tank at 4.40 atm and 305 K? •n=? • P = 4.40 atm • V=15.0 L • T= 305 K • R= 0.0821 L atm/mol K • PV=nRT so n= PV/RT • (4.40)(15.0) = 2.63 moles (0.0821)(305)

Same Problem…different Question • What is the density of C2H4 ? (D=m/v) 2.63 moles 28.06 g = 73.80 g 1 1 mole Density = 73.80 grams = 4.92 g/L 15.0 L

Check it Out! • Find the molar mass of an unknown gas that has a density of 2.58 g/L at STP? 2.58 g 22.4 L = 57.8 g/mole 1L 1 mole

Last One! • How many moles/liter of a gas at 3.5 kPa will you have at 27 °C? PV = nRT or n = P V RT 3.5 kPa = 0.0014 mol/L (300)(8.3) • If this substance is C2H4, what is its density? 0.00141 mol 28.02 g = 0.039 g/L 1L 1 mole

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