Lesson Notes By Weeks and Term - Senior Secondary School 1
GAS LAWS
GAS LAWS
SUBJECT: CHEMISTRY
CLASS: SS 1
DATE:
TERM: 2nd TERM
REFERENCE BOOKS
TOPIC: GAS LAWS CONTENT BOYLE’S LAW It states that the volume of a fixed mass of gas is inversely proportional to the pressure provided the temperature remains constant. Mathematically, V α 1/P V = k/P PV = k Hence, P1V1 = P2V2 Boyle’s law can be represented graphically as shown below. The graph shows that if the pressure is doubled, the volume is reduced to half its former value and if it is halved, the volume is doubled. EXPLANATION OF BOYLE’S LAW USING THE KINETIC THEORY When the volume of fixed mass of gas is decreased, the molecules of the gas will collide with each other more rapidly. This gives rise to an increase in pressure. However, if molecules are farther apart the number of collisions for unit time decreases, resulting in a decrease in pressure. CHARLES’ LAW Charles’ law states that the volume of a fixed mass of gas at constant pressure is directly proportional to its temperature in the Kelvin scale. Mathematically, V α T V = k/T V = k T Hence, V1 = V2 T1 T2 The graphical representation of Charles’ law is as shown below: EXPLANATION OF CHARLES’ LAW USING THE KINETIC THEORY When a given gas is heated at constant pressure, the molecules acquire more kinetic energy and move faster. They collide with one another and with the walls of the container more frequently. To maintain the same number of collisions on the walls of container (i.e. keep the pressure constant) the volume of the gas increases. CALCULATIONS BASED ON BOYLE’S AND CHARLES’ LAW Solution: V1 = 200cm3, P1 = 510mmHg, P2 = 780mmHg V2 = ? Using the expression for Boyle’s law: P1V1 = P2V2 V2 = P1V1 = 510mmHg x 200cm3 = 130.769 = 131 cm3 P2 780mmHg Solution: V1 = 300cm3, T1 = 35oC = (35 + 273)K = 308K, V2 = V1/2 = 300/2 = 150cm3, T2 = ? Using the formula for Charles’ law V1 = V2 T1 T2 T2 = V2T1 = 150cm3 x 308K = 154K V1 300cm3 EVALUATION GENERAL GAS EQUATION Boyle’s and Charles’ laws are combined into a single expression known as the general gas equation which can be expressed mathematically as P1V1 = P2V2 T1T2 IDEAL GAS EQUATION This equation states that for an ideal gas PV/T is a constant. That is, PV = R (R = molar gas constant) T PV = RT That is, for n mole of a gas, the equation becomes PV = nRT CALCULATIONS Solution: T1 = 273K, P1 = 1.01 x 105Nm-2, T2 = 25oC = (25 + 273) = 298K, P2 = 0.84 x 105Nm-2, V2 = 700cm3, V1 =? Using the general gas equation P1V1 = P2V2 T1 T2 V1 = P2V2T1 = 0.84 x 105Nm-2 x 700cm3 x 273K = 533.337 =533cm3 P1T2 1.01 x 105Nm-2 x 298K Solution: V = 6.5dm3, P = 3atm, T = 15oC = (15 + 273)K = 288K, n =? Using PV = nRT n = PV = 3atm x 6.5dm3= 0.8257 RT 0.082atmdm3K-1mol-1 x 288K Number of moles = 0.83 mole DALTON’S LAW OF PARTIAL PRESSURE This law state that in a mixture of gases which do not react chemically together, the total pressure exerted by the mixture of gases is equal to the sum of the partial pressure of the individual gases that make up the mixture. Mathematically, the law can be expressed as: Ptotal = PA + PB +PC........Pn Where Ptotal is the total pressure of the mixture and PA, PB, PC are the partial pressure exerted separately by the individual gases A, B, C that make up the mixture. The pressure each constituent gas exerts is called partial pressure and is expressed as Partial pressure of gas A (PA) = Number of moles of gas A x Ptotal Total number of moles of gas in mixture That is, PA = nA x Ptotal nA + nB + nC If the gas is collected over water, it is likely to be saturated with water vapour and the total pressure becomes Ptotal = Pgas + Pwater vapour Pgas = Ptotal – Pwater vapour CALCULATION ON THE LAW A gaseous mixture containing 64g of O2 and 70g of N2 exerts a total pressure of 1.8oatm. What is the partial pressure exerted by oxygen in the mixture? Solution: Molar mass of O2 = 16 x 2 = 32gmol-1 Molar mass of N2 = 14 x 2 = 28gmol-1 Number of mole of O2 = 64g = 2.0mole 32gmol-1 Number of mole of O2 = 70g = 2.5mole 28gmol-1 Total number of moles of gases in mixture = 2.0 + 2.5 = 4.5 mole Partial pressure of O2 = 2.0 x 1.80 = 0.80atm 4.5 GENERAL EVALUATION/REVISION READING ASSIGNMENT New School Chemistry for Senior Secondary School by O.Y. Ababio, Pg 78-85. WEEKEND ASSIGNMENT THEORY
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