{"id":2306,"date":"2023-10-02T12:05:59","date_gmt":"2023-10-02T12:05:59","guid":{"rendered":"http:\/\/localhost\/ecole9ja\/?p=2306"},"modified":"2023-10-02T12:15:13","modified_gmt":"2023-10-02T12:15:13","slug":"week-8-ss1-second-term-chemistry-notes","status":"publish","type":"post","link":"https:\/\/ecolebooks.com\/nigeria\/posts\/week-8-ss1-second-term-chemistry-notes\/","title":{"rendered":"Week 8 &#8211; SS1 Second Term Chemistry Notes"},"content":{"rendered":"<p>\u00a0<br \/>\n\u00a0<strong>WEEK  8<br \/>\n<\/strong><strong>GAS LAWS<br \/>\n<\/strong>Behaviour of gases is expected to differ from that of solids and liquids. This was investigated by many early scientists e.g. Boyle, Charles, Graham and Dalton, Avogadro&#8217;s. They studied the physical behaviour of gases. Gay-Lussac: he studied the chemical behaviour.<br \/>\n<strong>Boyle&#8217;s Law<br \/>\n<\/strong><\/p>\n<ul>\n<li>By <strong>Robert Boyle in 1662<\/strong>\n\t\t\t<\/li>\n<li>States that the volume of a given mass of a gas is inversely proportional to its pressure, provided temperature remains constant.\n<\/li>\n<li>Law is about the relationship between volume (V) and pressure (P)\n<\/li>\n<\/ul>\n<p>As volume of a gas increases, the pressure decreases and vice versa<br \/>\n<strong>Mathematical expression<br \/>\n<\/strong>V \u03b1 1\/p<br \/>\nV \u03b1 k\/p<br \/>\nK = PV<br \/>\nV = volume<br \/>\nP = pressure<br \/>\nK = constant<br \/>\nFor initial and final pressure and volume, we have<br \/>\nP<sub>1<\/sub>V<sub>1<\/sub> = P<sub>2<\/sub>V<sub>2<\/sub><br \/>\n\t\tAccording to the kinetic theory,the gas pressure is caused by molecular collisions with the walls of the container.Therefore, the larger the number of molecules per unit volume,the larger the number of collisions and the higher the pressure.Reducing the volume of a gas container will increase the collisions on the walls of the container per unit time and consequently the pressure of the gas will incr<strong><br \/>\n\t\t\t\t<\/strong><br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<strong>Using kinetic theory to explain Boyle&#8217;s Law<br \/>\n<\/strong><br \/>\n\u00a0<img decoding=\"async\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100223_1205_Week8SS1Se1.png\" alt=\"\"\/><strong><br \/>\n\t\t\t<\/strong><strong>NOTE:<br \/>\n<\/strong><strong>In stage 2:<br \/>\n<\/strong><\/p>\n<ul>\n<li><strong>The piston is kept stationary by placing a heavy weight on it.<br \/>\n<\/strong><\/li>\n<li><strong>The space occupied by the molecules is constant i.e. volume is constant.<br \/>\n<\/strong><\/li>\n<li><strong>The gas exerts constant pressure.<br \/>\n<\/strong><\/li>\n<\/ul>\n<p><strong>In stage II:<br \/>\n<\/strong><\/p>\n<ul>\n<li><strong>Weight of piston is replaced with a lighter one, so the piston moves up.<br \/>\n<\/strong><\/li>\n<li><strong>The space occupied by the gas is doubled (increased volume).<br \/>\n<\/strong><\/li>\n<li><strong>The pressure exerted by the gas is halved (pressure increases).<br \/>\n<\/strong><\/li>\n<\/ul>\n<p>\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<strong>In stage III:<br \/>\n<\/strong><\/p>\n<ul>\n<li>Weight of the piston is now replaced with a heavier one, so the piston moved down.\n<\/li>\n<li>The space occupied by the gas is halved (volume decreases)\n<\/li>\n<li>The pressure exerted by the gas is doubled (pressure increases)\n<\/li>\n<\/ul>\n<p>Therefore, at constant temperature; as the volume of a gas decreases, the pressure the gas exerts increases. Example, Ababio page 31, Fig. 5.15<br \/>\n<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100223_1205_Week8SS1Se2.png\" alt=\"\"\/>V\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<br \/>\n<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100223_1205_Week8SS1Se3.png\" alt=\"\"\/><\/p>\n<p>\u00a0<br \/>\n\u00a0<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100223_1205_Week8SS1Se4.png\" alt=\"\"\/>  0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<strong><sup>1<\/sup>\/<sub>P<\/sub><\/strong>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<br \/>\n<img decoding=\"async\" align=\"right\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100223_1205_Week8SS1Se5.gif\" alt=\"\"\/><br \/>\n\t\t<strong>Calculation Involving Boyle&#8217;s Law<br \/>\n<\/strong><strong>The volume of a gas means nothing unless the conditions under which it was measured are known.<\/strong><br \/>\n\t\t\u00a0 Tips for working with gas laws: <\/p>\n<ul>\n<li>\n<div><strong>All gas calculations must use Kelvin temperatures.<\/strong>\u00a0\n<\/div>\n<\/li>\n<li>\n<div>The conditions <strong>0 <sup>o<\/sup>C<\/strong> and <strong>1 atm<\/strong> are referred to as <strong>standard temperature and pressure<\/strong> &#8211; <strong>STP.<\/strong>\u00a0\n<\/div>\n<\/li>\n<li>\n<div>The volume occupied by one mole of a gas at STP, 22.4 liters, is referred to as <strong>molar volume<\/strong>.\u00a0\n<\/div>\n<\/li>\n<li>\n<div>Read the problem to see what conditions change. \u00a0\n<\/div>\n<\/li>\n<li>\n<div>Decide which gas law to use and write its equation. \u00a0\n<\/div>\n<\/li>\n<li>\n<div>Reread the problem to see what question is asked. \u00a0\n<\/div>\n<\/li>\n<li>\n<div>If needed, manipulate the gas law equation. \u00a0\n<\/div>\n<\/li>\n<li>\n<div>Plug numbers and units into the equation. \u00a0\n<\/div>\n<\/li>\n<li>\n<div>Pickup your calculator and punch buttons. \u00a0\n<\/div>\n<\/li>\n<li>\n<div>Write the answer to the problem, <em>don&#8217;t forget significant figures<\/em>, and circle it.\n<\/div>\n<\/li>\n<\/ul>\n<p>\u00a0 Example 1<strong><br \/>\n\t\t\t<\/strong>A sample of gas occupied 390cm3 ata pressure of 760mmHg.What volume will the gas occupy at 780mmHg, if the temperature remains constant?<br \/>\nSolution<br \/>\nP<sub>1<\/sub>V<sub>1<\/sub>=P<sub>2<\/sub>V<sub>2<\/sub>  (T constant)<br \/>\nP<sub>1<\/sub>=760mmHg<br \/>\nV<sub>1<\/sub>=390cm<sup>3<\/sup><br \/>\n\t\tP<sub>2<\/sub>=789mmHg<br \/>\nV<sub>2<\/sub>=?<br \/>\nP<sub>1<\/sub>V<sub>1<\/sub>=P<sub>2<\/sub>V<sub>2<\/sub>  (Boyle&#8217;s law)<br \/>\nV2=P<sub>1<\/sub>V<sub>1<\/sub>\/P<sub>2<\/sub><br \/>\n\t\t=760mmHg x 390cm<sup>3<\/sup>\/780mmHg<br \/>\nV2=380 cm<sup>3<\/sup><\/p>\n<p>\u00a0<\/p>\n<ol>\n<li>\n<div>375cm<sup>3<\/sup> of gas has a pressure of 770mmHg, Find its volume if the pressure is reduced to 750mmHg\n\t\t\t\t\t<\/div>\n<p><strong>Ans = 385cm<sup>3<\/sup><br \/>\n\t\t\t\t\t\t<\/strong><br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<strong>ABSOLUTE TEMPERATURE<br \/>\n<\/strong>The temperature at which the volume of a gas would be theoretically reduced to O. This temperature is 0C or 273K.<br \/>\n<strong>Note: <\/strong>practically, it is not possible as all gases liquefy above this temperature but it is significant because it is the lowest possible temperature that can be reached.<br \/>\n<strong>Temperature Conversion<br \/>\n<\/strong>0<sup>0<\/sup>C =273K, -273<sup>0<\/sup>C =0K (no degree sign)<br \/>\nC =Celsius or centigrade<br \/>\nK = Kelvin<\/p>\n<p>\u00a0     To convert:<br \/>\nCelsius to Kelvin =K=<sup>0<\/sup>C + 273<br \/>\nKelvin to Celsius = <sup>0<\/sup>C = K-273<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/p>\n<p>\u00a0<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100223_1205_Week8SS1Se6.png\" alt=\"\"\/><img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100223_1205_Week8SS1Se7.png\" alt=\"\"\/><img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100223_1205_Week8SS1Se8.png\" alt=\"\"\/><img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100223_1205_Week8SS1Se9.png\" alt=\"\"\/>\u00a0\u00a0\u00a0\u00a0          100<sup>0<\/sup>C\u00a0\u00a0\u00a0\u00a0                              373K<\/p>\n<p>\u00a0<br \/>\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<br \/>\n\u00a0<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100223_1205_Week8SS1Se10.png\" alt=\"\"\/><img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100223_1205_Week8SS1Se11.png\" alt=\"\"\/>0<sup>0<\/sup>C\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0273K<\/p>\n<p>\u00a0<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100223_1205_Week8SS1Se12.png\" alt=\"\"\/><img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100223_1205_Week8SS1Se13.png\" alt=\"\"\/>                          -273<sup>O<\/sup>C                                             0K\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<br \/>\nCELSIUS SCALE\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0KELVIN SCALE<\/p>\n<p>\u00a0<\/li>\n<\/ol>\n<p><strong>CHARLE&#8217;S LAW\u00a0\u00a0\u00a0\u00a0<\/strong><br \/>\n\t\tThe volume of a given mass of a gas is directly proportional to its absolute temperature provided that pressure remains constant.<br \/>\n<strong>MATHEMATICAL EXPRESSION<\/strong><\/p>\n<p>\u00a0<br \/>\n\u00a0<br \/>\n\u00a0V \u03b1 T<br \/>\nV \u03b1 KT<br \/>\nK= V<br \/>\n      T<br \/>\nV = Volume, T = Temperature (in Kelvin), K= constant<br \/>\nFor more than one gas, we have<br \/>\nV<sub>1 <\/sub><sub>=<\/sub>V<sub>2<br \/>\n<\/sub>T<sub>1     <\/sub>T<sub>2                      <\/sub><br \/>\n\t\tV<sub>2= <\/sub> V<sub>1<\/sub> T<sub>2<\/sub><br \/>\n\t\t          T<sub>1<br \/>\n<\/sub>The volume of a gas is zero at a temperature of -273<sup>0<\/sup>C which is zero Kelvin.<br \/>\nCharle&#8217;s laws explains that the behavior of gases at differenttemperature changes when the pressure is constant.Gases expand when heated.The rate of expansion or contraction is summaried as follows.At constant pressure,a gas increases by 1\/273 of its volume for each Celsius degree rise in temperature and this is true for all gases.For every one degrr centrigrade rise or fall in temperature.<\/p>\n<p>\u00a0The Kelvin  temperature  scale has -273oC as its starting point and it is called the absolute temperature scale.<br \/>\nHence,O<sup>0<\/sup>C=273K<br \/>\n-273=0<br \/>\nThe Kelvin  temperature  scale has -273oC as its starting point and it is called the absolute temperature<\/p>\n<p>\u00a0<br \/>\n\u00a0<br \/>\n\u00a0Charles&#8217; Law Problems<br \/>\n\t\tExample1<br \/>\nWhatvolume would be occupied by a given sample of gas at 45<sup>0<\/sup>C if it occupies 500cm<sup>3<\/sup> at O<sup>0<\/sup>c assuming the pressure is constant?<br \/>\nSolution<br \/>\nAcertain mass of gas occupies300cm3 at 35<sup>0<\/sup>C .At what temperature willit have its volume reduced by half,assuming the pressure remains constant.<br \/>\nSolution<br \/>\nV1\/T1=V2\/T2<br \/>\nV1=300cm<sup>3<\/sup><br \/>\n\t\tT1=35<sup>0<\/sup>C=(273+35)K=308K<br \/>\nV2=150cm<sup>3<\/sup><br \/>\n\t\tT2=?<br \/>\n300cm<sup>3<\/sup>\/308K=150cm<sup>3<\/sup>\/T2<br \/>\nT2=308K X 150cm<sup>3<\/sup>\/300cm<sup>3<\/sup><br \/>\n\t\t154K =(-119<sup>0<\/sup>C)<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0\u00a0\u00a0\u00a0\u00a0<strong><sub><br \/>\n\t\t\t\t<\/sub><\/strong><br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<strong><sub>USING KINETIC THEORY TO EXPLAIN CHARLE&#8217;S LAW. EXAMPLE.<\/sub><\/strong><br \/>\n\t\t<img decoding=\"async\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100223_1205_Week8SS1Se14.png\" alt=\"\"\/><br \/>\n\t\t<strong>NOTE:<br \/>\n<\/strong>In stage:<br \/>\nThe gas is heated, molecules acquire more kinetic energy, move faster and collide more often with the walls of the vessel, and hence, pressure is exerted.<br \/>\nIn stage II:<br \/>\nThe temperature is increased through heating <strong>(T<sub>1 <\/sub>\u2013T<sub>2<\/sub>).<\/strong><br \/>\n\t\tThe pressure is still constant because the piston has been moved up but the volume occupied by the gas has increased.<br \/>\n<strong>Conclusion<\/strong>: As temperature increases, the volume increases and vice versa.<br \/>\n<strong>\u00a0\u00a0\u00a0\u00a0<br \/>\n<\/strong>\u00a0\u00a0\u00a0\u00a0<\/p>\n<p>\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<strong>Graphical illustration of Charles law.<br \/>\n<\/strong><img decoding=\"async\" align=\"right\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100223_1205_Week8SS1Se15.gif\" alt=\"\"\/><br \/>\n\t\t<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100223_1205_Week8SS1Se16.png\" alt=\"\"\/><img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100223_1205_Week8SS1Se17.png\" alt=\"\"\/><img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100223_1205_Week8SS1Se18.png\" alt=\"\"\/>Ababio page 81, fig 5.16 and page 83<br \/>\n<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100223_1205_Week8SS1Se19.png\" alt=\"\"\/>Vol (dm<sup>3<\/sup>)                                                                                Vol<br \/>\n<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100223_1205_Week8SS1Se20.png\" alt=\"\"\/><br \/>\n\t\t<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100223_1205_Week8SS1Se21.png\" alt=\"\"\/><img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100223_1205_Week8SS1Se22.png\" alt=\"\"\/><img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100223_1205_Week8SS1Se23.png\" alt=\"\"\/><img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100223_1205_Week8SS1Se24.png\" alt=\"\"\/><br \/>\n\t\t<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100223_1205_Week8SS1Se25.png\" alt=\"\"\/> -273                          O                                       Temp<sup>o<\/sup>C                                                                  K<br \/>\n\u00a0\u00a0\u00a0\u00a0<strong>Calculations involving Charles law<br \/>\n<\/strong><\/p>\n<ol>\n<li>At 17<sup>0<\/sup>C, a sample of hydrogen gas occupies 125cm<sup>3<\/sup>. What will the volume be at 100<sup>0<\/sup>C, if the pressure remains constant? Ans. = 161cm<sup>3<\/sup>\n\t\t\t<\/li>\n<li>To what temperature in Celsius must a gas be raised from 0<sup>0<\/sup>C in order to double its volume? <strong>Ans. = 273<sup>0<\/sup>C<\/strong>\n\t\t\t<\/li>\n<li>\n<div>20cm<sup>3<\/sup> of a gas at 55<sup>0<\/sup>C exerts 160mm Hg pressure. At the same pressure, calculate the volume of the temperature is doubled? <strong>Ans. = 23.35cm<sup>3<\/sup><\/strong>\n\t\t\t\t<\/div>\n<p><strong>GENERAL GAS EQUATION<br \/>\n<\/strong><strong>Boyle&#8217;s and Charles&#8217;s laws<\/strong> show that there is a relationship between the temperature, press and volume. The relationship is expressed by what is called <strong>general gas equation <\/strong>I.e.<br \/>\nBoyle&#8217;s law \u2013 V \u03b1 1\/P<br \/>\n<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100223_1205_Week8SS1Se26.png\" alt=\"\"\/>Charles Law &#8211; V \u03b1 T\u00a0\u00a0\u00a0\u00a0<strong>i<\/strong><br \/>\n\t\t\t\tMultiply (i) x (ii)<br \/>\nV \u03b1 1 X V\u03b1 T<br \/>\nV \u03b1 (1\/P X T)<br \/>\nV \u03b1 T\/P<br \/>\nV= KT\/P<br \/>\nPV= KT<br \/>\nK= PV\/T<\/p>\n<p>\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0Formorethan one gas,<br \/>\nP<sub>1  <\/sub>V<sub>1  <\/sub>=  P<sub>2<\/sub>   V<sub>2<\/sub>V<sub>2<\/sub>=  P<sub>1<\/sub>  V<sub>1<\/sub>  T  <sub>2<\/sub><br \/>\n\t\t\t\t     T<sub>1\u00a0\u00a0\u00a0\u00a0<\/sub>T<sub>2 <\/sub>                     P<sub>2      <\/sub>T<sub>  1<\/sub>\n\t\t\t\t<\/li>\n<\/ol>\n<p>where P1= initial pressure<br \/>\nV1=initial volume<br \/>\nT1=\u00a0\u00a0\u00a0\u00a0Initial Kelvin temperature<br \/>\nP2=New pressure<br \/>\nV2=New volume<br \/>\nT2=New Kelvin temperature<\/p>\n<p>\u00a0<strong>STANDARD TEMPERATURE AND PRESSURE (S.T.P)<br \/>\n<\/strong><strong>It is the generally <\/strong>accepted standard temperature (0<sup>0<\/sup>C or 273K) and pressure (760mmHg or 1.01 x 10<sup>5<\/sup>Nm<sup>-2<\/sup>)<br \/>\n<strong>NOTE: if two chemists, one <\/strong>in a temperate country e.g. England and the other in tropical country e.g. Nigeria, should carry out investigations on the same gases. Their gas volumes would differ because of different in temperatures of the two countries. So, scientists decided to have standard temperature and pressure for calculations and experimentation.<br \/>\n<strong>CALCULATIONS<br \/>\n<\/strong><\/p>\n<ol>\n<li>At s.t.p, a certain mass of gas occupies a volume of 790cm<sup>3<\/sup>. Find the temperature at which the gas occupies 1000cm<sup>3<\/sup>. Ans. = 330.1K\n<\/li>\n<li>A given mass of a gas occupies 850cm<sup>3<\/sup> at 320K and 0.92 x 10<sup>5<\/sup>Nm<sup>-2<\/sup> pressures. Calculate the volume of the gas at s.t.p Ans. = 660.5cm<sup>3<\/sup>\n\t\t\t<\/li>\n<li>A sample of N<sub>2<\/sub> occupies a volume of 1dm<sup>3<\/sup> at 500k and 1.01 x 10<sup>5<\/sup> Nm<sup>-2<\/sup>. What will its volume be at 2.02 x 10<sup>5<\/sup>Nm<sup>-2<\/sup> and 400K?\n<\/li>\n<li>To what temperature must a given mass of N<sub>2<\/sub> at 0<sup>0<\/sup>C be heated so that both its volume and pressure will be doubled?\n<\/li>\n<li>130cm<sup>3<\/sup> of gas at 20<sup>o<\/sup>C exerts a pressure of 750mmHg. Calculate its pressure if its volume is increased to 150cm<sup>3<\/sup> at 35<sup>0<\/sup>C.\n<\/li>\n<li>Calculate the volume of hydrogen produced at s.t.p and r.t.p when 25g of zinc are added to excess dilute Hydrochloric acid at 31<sup>o<\/sup>Cand 778mmHg pressure\u00a0\u00a0\u00a0\u00a0 (H= 1, Zn= 65, Cl= 35.5,  GMV= 22.4dm<sup>3, <\/sup>24 dm<sup>3<\/sup> )\n<\/li>\n<li>\n<div><img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100223_1205_Week8SS1Se27.png\" alt=\"\"\/>The combustion of butane in oxygen (air) is represented in the equation below: 2C<sub>4<\/sub>H<sub>10<\/sub> + 130<sub>2<\/sub>                     10H<sub>2<\/sub>0 + 8CO<sub>2<\/sub>\n\t\t\t\t<\/div>\n<p>\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<strong>IDEAL GAS EQUATION<br \/>\n<\/strong>In all experimental works, measurements or calculation involving gases, four quantities are important-<strong>volume, pressure, temperature<\/strong>, <strong>and number of moles.<\/strong> The first three have been used in general gas equation. But the combination of four of them gives <strong>ideal gas equation<\/strong> as:<br \/>\n<strong>PV = nrt<br \/>\n<\/strong>Where<br \/>\nP = Pressure (in atm)<br \/>\nV = Volume (in dm<sup>3<\/sup>)<br \/>\nn = No. of moles<br \/>\nR = Gas constant (0.082 atm dm<sup>3 <\/sup>k<sup>-1 <\/sup>mol<sup>-1<\/sup>)<br \/>\nT = Temp (in K)\n<\/li>\n<\/ol>\n<p><strong>Gas density<\/strong> can also be calculated using the ideal-gas equation.<br \/>\nDensity is equal to mass divided by volume, d = m\/v.<br \/>\nThe ideal-gas equation can be arranged to give density in g\/L:<br \/>\n<img decoding=\"async\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100223_1205_Week8SS1Se28.gif\" alt=\"\"\/><br \/>\n\t\tThis equations shows the density of a gas depends on its pressure, molar mass, and temperature. The higher the molar mass and pressure, the greater the gas density; the higher the temperature, the less dense the gas.<br \/>\nEven though gases form homogeneous mixtures regardless of their identities, a less dense gas will lie above a more dense one if they are not physically mixed. The differences between the densities of hot and cold gases is responsible for CO<sub>2<\/sub> being able to keep oxygen from reaching combustible materials (thus acting as a fire extinguisher) and for many weather phenomena, such as the formation of large thunderhead clouds during thunderstorms.<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<\/p>\n<ol>\n<li>\n<div>Under a pressure of 3000, Nm<sup>-2<\/sup> a gas has a volume of 250cm<sup>3. <\/sup>What will its volume be if the pressure is changed to 100mmHgat the same\n<\/div>\n<p>Temperature?<br \/>\n\u00a0\u00a0\u00a0\u00a0Solution to NO.3\n<\/li>\n<\/ol>\n<p>Note: The pressure is not in the same unit. So conversion must be done first. 101325 Nm<sup>-2 <\/sup>= 760mmHg<br \/>\n<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100223_1205_Week8SS1Se29.png\" alt=\"\"\/>3000Nm<sup>-2 <\/sup>=760  \u00a0\u00a0\u00a0\u00a0x\u00a0\u00a0\u00a0\u00a03000= 22.gmmHg<br \/>\n101325<br \/>\nP<sub>1 =<\/sub>22.5mmHg<br \/>\n<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100223_1205_Week8SS1Se30.png\" alt=\"\"\/>Using\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 P<sub>1 <\/sub>V<sub>1  \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0= <\/sub>P<sub>2 <\/sub>V<sub>2<br \/>\n<\/sub>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0T<sub>1\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/sub>    T<sub>2<br \/>\n<\/sub>P<sub>2 =<\/sub> 100mmHg, V<sub>1 <\/sub>= 250cm<sup>3,<\/sup> V<sub>2 =<\/sub>?<br \/>\n<strong>Ans= 56.25cm<sup>3<br \/>\n<\/sup><\/strong><strong>Dalton&#8217;s Law of Partial Pressures<\/strong>, established by John Dalton, states that if there is a mixture of gases that do not react chemically together, then the total pressure exerted by the mixture is the sum of the partial pressures of the individual gases that make up the mixture i.e. <strong><br \/>\n\t\t\t<\/strong><strong>P<sub>Total =<\/sub>P<sub>A <\/sub>+ P<sub>B <\/sub>+ P<sub>C   <\/sub>+\u2026<\/strong><br \/>\n\t\tWhere<\/p>\n<p>\u00a0<br \/>\n\u00a0<strong>P<sub>Total =<\/sub> Total pressure of the mixture<br \/>\n<\/strong><strong>P<sub>A =<\/sub> Partial pressure of gas A<br \/>\n<\/strong><strong>P<sub>B <\/sub>= Partial pressure of gas B<br \/>\n<\/strong><strong>P<sub>C   <\/sub>= Partial pressure of gas C<br \/>\n<\/strong>Note:<\/p>\n<ol>\n<li>Gases A, B and C make up the mixture.\n<\/li>\n<li>If a gas is collected by water, it is likely to be saturated with water vapor and the total pressure becomes: P<sub>total<\/sub>= P<sub>gas<\/sub> + P<sub>watervapour<\/sub>\n\t\t\t<\/li>\n<\/ol>\n<p>The pressure of the dry gas will now be:<br \/>\n<strong>P<sub>GAS =<\/sub> P<sub>TOTAL <\/sub>-P<sub>WATER<br \/>\n<\/sub><\/strong><br \/>\n\u00a0<img decoding=\"async\" align=\"right\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100223_1205_Week8SS1Se31.gif\" alt=\"\"\/>Dalton&#8217;s Law is helpful when collecting a gas &#8220;over water&#8221;. This diagram shows the collection of a gas by water displacement.<br \/>\nA collecting tube is filled with water and inverted in an open pan of water. Gas is then allowed to rise into the tube, displacing the water. By raising or lowering the collecting tube until the water levels inside and outside the tube are the same, the pressure inside the tube is exactly that of the atmospheric pressure.<br \/>\nA gas collected &#8220;over water&#8221; is a mixture of the gas and water vapor. Dalton&#8217;s law of partial pressures describes this situation as: <\/p>\n<p>\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0P<sub>total<\/sub> = P<sub>gas<\/sub> + P<sub>H2O<\/sub><br \/>\n\t\tCharts like this one are readily available that give water vapor pressure at any common temperature.<br \/>\n<strong>EXERCISES<\/strong><\/p>\n<ol>\n<li>\n<div>A certain mass of hydrogen gas collected over water of 6<sup>o <\/sup>C and 765mmHg pressure has a volume of 35cm<sup>3.  <\/sup>Calculate the volume when it is dry at s.t.p<strong>. (s.v.p. of water at 6<sup>o<\/sup>C=7mmHg).<br \/>\n<\/strong><\/div>\n<p><strong>Solution<br \/>\n<\/strong>To get the real pressure of H<sub>2<\/sub> gas i.e. when it is dry:<br \/>\n<strong>P<sub>DRY GAS <\/sub>=P<sub>TOTAL <\/sub>\u2013P<sub>WATER VAPOUR =<\/sub>765 -7 =758mmHg<br \/>\n<\/strong>Apply general gas equation to obtain the volume required in the question. <strong>Ans= 34.2cm<sup>3<\/sup><\/strong><sup><br \/>\n\t\t\t\t\t<\/sup><\/li>\n<\/ol>\n<p>      2 .272cm<sup>3 <\/sup>of CO<sub>2 <\/sub>were collected over water at 15<sup>o<\/sup>C and 782mmHg pressure. Calculate the \u00a0\u00a0\u00a0\u00a0Volume of the dry gas at s. t. p. (s.v.p of water at 15 0C= 12mmHg) \u00a0\u00a0\u00a0\u00a0<br \/>\n3 .A given amount of gas was collected over water at 302K where the water vapour pressure of was 4.0 KNm<sup>-2. <\/sup>Calculate the pressure of the dry gas if the atmospheric pressure at the same temperature was 101.3 KNm<sup>-2<\/sup><br \/>\n\t\t4 .200cm<sup>-2 <\/sup>of Nitrogen gas at a pressure of 500mmHg and 100cm<sup>3 <\/sup>of CO<sub>2 <\/sub>at a pressure of 50mmHg were introduced into a 150cm<sup>3 <\/sup>vessels. What is the total pressure in the vessel?<br \/>\n<strong>Solution for No. 4<br \/>\n<\/strong>For N<sub>2<br \/>\n<\/sub>V<sub>1<\/sub> =200 cm<sup>3<\/sup>, P<sub>1 <\/sub>=500 mmHg<\/p>\n<p>\u00a0<br \/>\n\u00a0<br \/>\n\u00a0 V2=150 cm<sup>3<\/sup>, P<sub>2 <\/sub>=? mmHg<br \/>\n<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100223_1205_Week8SS1Se32.png\" alt=\"\"\/>P<sub>2<\/sub>= P<sub>1<\/sub>V<sub>1 <\/sub>  = 666.67mmHg \u2013<strong>Boyles Law<\/strong><br \/>\n\t\t\u00a0\u00a0\u00a0\u00a0V<sub>2<br \/>\n<\/sub>For CO<sub>2<\/sub><br \/>\n\t\tV<sub>1 <\/sub>=10 cm<sup>3<\/sup>, P<sub>1<\/sub>=50 mmHg, V<sub>2<\/sub>=150 cm<sup>3<\/sup>, P<sub>2 <\/sub>mmHg=?<br \/>\nP<sub>2<\/sub>-33.33mmHg.<br \/>\nTotal pressure =666.67 +33.33 =700mmHg<\/p>\n<ol>\n<li>\n<div>A gas X was put in a 10dm<sup>3 <\/sup>vessel at 400K and 1.015 x 10<sup>5 <\/sup>Nm<sup>-2. <\/sup>. Another gas Y at 400K and 3.035 x 105 Nm<sup>-2<\/sup> is put in a 5dm3 vessels. What will be the total pressure if:\n<\/div>\n<ol>\n<li>X and Y are put in a 5dm<sup>3<\/sup> vessel at 400K?\n<\/li>\n<li>X and Y are put in a 15dm<sup>3<\/sup> vessel at 400K?\n<\/li>\n<li>X and Y are put in a 10dm<sup>3<\/sup> vessel at 400K?\n<\/li>\n<li>\n<div>X and Y are put in a 20dm<sup>3<\/sup> vessel at 400K?\n<\/div>\n<p><strong>SOLUTION<br \/>\n<\/strong>For X<\/p>\n<ol>\n<li>\n<div>P<sub>1<\/sub>= 1.015 x 105 Nm<sup>-2<\/sup>, V<sub>1<\/sub> = 10dm<sup>3<\/sup>, V<sub>2<\/sub> = 5dm<sup>3<\/sup>, P<sub>2<\/sub>=?\n<\/div>\n<p><img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100223_1205_Week8SS1Se33.png\" alt=\"\"\/>P<sub>2<\/sub> = P<sub>1<\/sub>V<sub>1<\/sub>\u00a0\u00a0\u00a0\u00a0=2.03 X 105 Nm<sup>-2<\/sup>.\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 NOTE: Temp is constant<br \/>\nV<sub>2 <\/sub>\n\t\t\t\t\t\t\t\t<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<p>\u00a0\u00a0\u00a0\u00a0For Y<br \/>\nIt already in 5 dm<sup>3<\/sup> and it is 3.035 x 10<sup>5 <\/sup>Nm<sup>-2<\/sup><br \/>\n\t\tP<sub> total<\/sub> = Px + Py<br \/>\n=2.03 x 10<sup>5<\/sup> + 3.035 10<sup>5<\/sup><br \/>\n\t\t=5.065 x 10<sup>5<\/sup> Nm<sup>-2<\/sup><br \/>\n\t\t<strong>EVALUATION<br \/>\n<\/strong><strong>EXERCISES<br \/>\n<\/strong>1.A certain amount of gas occupies 5.0dm<sup>3 <\/sup>at 2 atm and 10<sup> 0 <\/sup>C. Calculate the number of moles present (R=0.082atm dm<sup> 3 <\/sup>K<sup>&#8211; 1<\/sup>mol-<sup> 1)<\/sup>.<\/p>\n<ol>\n<li>2.0 moles of an Ideal Gas are at a temperature of -13<sup>0<\/sup>C and a pressure of 2\n<\/li>\n<\/ol>\n<p>\u00a0<br \/>\n\u00a0Atm. What volume in dm<sup>3 <\/sup>will the gas occupy at a temperature?<br \/>\n<strong>(R=0.082 atm dm<sup>3 <\/sup> K-<sup> 1 <\/sup> mol-<sup>1<\/sup>)<br \/>\n<\/strong><br \/>\n\u00a03.A given mass of nitrogen is 0.12dm<sup>3<\/sup> at 60C and 1.01 X 10<sup>5<\/sup> Nm<sup>-2<\/sup>. Find its pressure at the same temperature, if its volume is changed to 0.24dm<sup>3. <\/sup><br \/>\n\t\t\t4.A certain mass of gas occupies 600cm<sup>3<\/sup> and exerted 1.325 x 10<sup>5<\/sup> Nm<sup>-2<\/sup> pressures. At what pressure would the volume of the gas be halved?<br \/>\n\t\t\t<strong>Ans = 2.65 <\/strong>x <strong>10<sup>5<\/sup> Nm<sup>-2<\/sup><br \/>\n\t\t\t<\/strong><br \/>\n\u00a05.Convert the following Kelvin temperature to Celsius temperature<br \/>\n\t\t\tA. 405K    B. 298K<\/p>\n<p>\u00a06.Convert the following Celsius temperature to Kelvin temperature <strong><br \/>\n\t\t\t<\/strong>A. 0<sup>0<\/sup>C     B. -132<sup>0<\/sup>C<strong><br \/>\n\t\t\t<\/strong><br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<strong><br \/>\n\t\t\t<\/strong>\u00a0<\/p>\n","protected":false},"excerpt":{"rendered":"<p>\u00a0 \u00a0WEEK 8 GAS LAWS Behaviour of gases is expected to differ from that of&#8230;<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1,196],"tags":[],"class_list":["post-2306","post","type-post","status-publish","format-standard","hentry","category-posts","category-second-term-ss1-chemistry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/2306","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/comments?post=2306"}],"version-history":[{"count":1,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/2306\/revisions"}],"predecessor-version":[{"id":2307,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/2306\/revisions\/2307"}],"wp:attachment":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/media?parent=2306"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/categories?post=2306"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/tags?post=2306"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}