Introduction to Gases Ch 13 Suggested HW: 1, 3, 15, 16, 31, 39, 40, 49, 53, 55
About Gases • Gases are the most understood form of matter • Even though different gases have different chemical properties, they tend to exhibit similar physical properties • This situation arises because gas molecules expand to fill a given space, and are relatively far apart from one another • Thus, each molecule behaves as if the others are not there
The Kinetic Theory of Gases • Kinetic Theory tells us the following: • Gas molecules are ALWAYS in motion. They collide randomly with each other and with the walls of the container • All collisions involving gas particles are ELASTIC. This means they simply “bounce off” of one another. No energy is lost, and the speed of the particles does not change. • The distance between gas molecules is large. So, gas mainly consists of empty space. • Interactions between gas molecules are negligible (no attraction or repulsion). • The kinetic energy (speed) of gas molecules is proportional to temperature.
Pressure • The most readily measured properties of a gas are its temperature, volume, andpressure • Pressure describes the force that a gas exerts on an area, A. P = F/A • The image below shows gas molecules inside of a cubic container. The gas molecules strike against the walls of the container. These collisions are the source of the pressure.
Atmospheric Pressure • You and I are currently experiencing an attractive force that pulls us toward the center of the earth (gravity). • Gas molecules in the atmosphere also experience gravity. • Because of their small masses and thermal energies, gas molecules can somewhat counteract gravity, which is why gases don’t just sit on the surface • Nonetheless, gravity causes the gases in the atmosphere to “press down” on the surface. This is atmospheric pressure.
Atmospheric Pressure • To calculate atmospheric pressure, we calculate the mass of a 1m2 column of air extending through the entire atmosphere. • That column would have an approximate mass of 104 kg. • The force exerted on that 1m2 area would be: • F= ma = (104kg)(9.8 ms-2) = 105 N • Then, P = 105N/1m2 = 105N/m2 = 105Pa • SI unit of pressure is the Pascal(Pa). Related units are bar, mmHg, and atmospheres.
Atmospheric Pressure Closed, under vacuum • The image to the bottom left shows a barometer. This system is composed of a glass tube that is open on one end and closed on the other. The tube is under vacuum(no air in tube). • The tube is inverted into a dish that contains mercury. Some mercury flows into the because the atmospheric pressure forces it upward (pressure in tube is lower). • The flow stops when the atmospheric pressure is equal to the pressure inside the tube. • Pressure insided the tube is caused by the force (weight) of mercury. The height h is proportional to the atmospheric pressure. h Open-end, exposed to atmosphere
Units of Pressure 1 atm = 760 mm Hg
Gas Laws: Boyle’s Law • Robert Boyle was able to show that, at constant temperature, volume and pressure are inversely proportional. V α • In other words, if we decrease the volume of a gas, we increase its pressure. If we consider pressure to be caused by molecular collisions with the container walls, the figure below clearly indicates why this relationship exists. Large volume Small volume
Relationship between Volume and Temperature: Charles’s Law • The averagekinetic energy of a gas is given by: Where R is the gas constant, 8.314 J mol-1 K-1 (Since kinetic energy is equal to , we can determine the averagevelocity of the molecules. This will be discussed later.) • If we increase the temperature, the result will be molecules that are moving faster and causing higher energy collisions. This causes the gas to expand, increasing the volume.
Charles’s Law To the left, we have a small volume of air trapped by a plug. As we increase the temperature, the air expands (volume increases), moving the plug. To the right, we see the effect of pouring liquid nitrogen on a balloon. Lowering the temperature lowers the volume.
Charles’s Law • So how can we predict the change in volume with temperature? • Charles’s Law tells us that volume is directly proportional to temperature at constant pressure. • If V = mT (m is just a constant) • Then Vi/Ti = m, where i stands for ‘initial’. And… • Vf/Tf = m, where f stands for ‘final’. Thus… Charles’s Law
Example • Lets go back to this figure. At 0 oC, the volume of the air is 0.180 mL. Then, the water is raised to boiling temperature (100 oC), and the air expands. What is the new volume of air? • Here, we are looking for the final volume. We will use Charles’s Law and plug in the given information. USE KELVIN SCALE !!!
Avogadro’s Law • Avogadro determined that equal volumes of gasesat a given pressure and temperature must contain equal numbers of molecules, and thus, equal moles 1L Cl2(g) 1L H2(g) 2L HCl(g) Volume is directly related to moles, and both are related to the stoichiometry 2L H2(g) 1L O2(g) 2L H2O(g)
The Ideal Gas Law • If we take the 3 previously discussed gas laws: • V α n (n = moles) Avogadro’s Law • V α Boyle’s Law • V α T Charles’s Law • We can combine these laws to obtain the IDEAL GAS LAW: PV = nRT Depending on the units of Pressure, we can use multiple values of R
Ideal Gases • Any gas that follows the ideal gas law is considered an ideal gas. • One mole of an ideal gas at 0 oC and 1 atmosphere of pressure occupies 22.4 L of space. The value of R is based on these values of n, T, P, and V. • The ideal gas law is valid only at low pressures • The conditions listed above (0 oC, 1 atm) are referred to as standard temperature and pressure(STP) • Note: The Ideal gas law is a theoretical approximation, and no gas follows this law exactly, but most gases are within a few percent of this approximation, so it is very useful.
Comparisons of Real Gases to the Ideal Gas Law Approximation at STP 22.42 L 22.41 L 22.4 L 22.31 L 22.06 L Molar Volume at STP (L) Cl2 CO2 Ideal Gas He H2 Deviations from the ideal gas law exist, but those deviations are reasonably small.
Example • The pressure in a 10.0 L gas cylinder containing N2(g) is 4.15 atmospheres at 20.0 oC.How many moles of N2(g)are there in the cylinder? *When dealing with ideal gas law questions, follow these steps: 1) Determine what it is you are solving for. 2) List the given information. Pay attention to the units of each parameter. Convert as needed, and MAKE SURE THAT THE TEMPERATURE IS IN KELVIN ! 3) Rearrange the ideal gas law equation accordingly to solve for the desired parameter.
Example, continued. The pressure in a 10.0 L gas cylinder containing N2(g) is 4.15 atmospheres at 20.0 oC.How many molesof N2(g) are there in the cylinder? We are solving for moles. V = 10.0 L P = 4.15 atm. T = 20. 0 oC 293 oK R = .0821 L•atm•mol-1• K-1 Rearrange the equation to solve for n. PV = nRT = 1.72 moles N2(g)
Example 2 PV = nRT • In carrying out the following reaction: Zn(s) + 2HCl (aq) ZnCl2 (aq) + H2 (g) • We obtain 0.0140 moles of H2 (g) at 2.25 atm and 25 oC. How many milliliters of H2(g) will we have? • We are solving for volume in mL. P = 2.25 atm T = 25 oC 298 oK n = .0140 mol R = 0.821 L•atm•mol-1• K-1 = 152 mL
Example 3 PV = nRT 2KClO3(s) 2KCl(s) + 3O2(g) • In the reaction above, 1.34 g of potassium chlorate is heated inside a container to yield oxygen gas and potassium chloride. The oxygen occupies 250 mL at 20.0 oC. What will the pressure of the gas be, in atmospheres? We are solving for pressure in atm. V = .250 L T = 20 oC 293 oK n = ? R = 0.821 L•atm•mol-1• K-1
Before we can find P, we must find n 2KClO3(s) 2KCl(s) + 3O2(g) 1.34 g .0164 mol .0109 mol
Root Mean Square Speed • Going back to the kinetic theory of gases, we found that the velocity of a gas molecule is proportional to its thermal energy: kinetic energy (per mole) = thermal energy • This equation is applied to moles of gas molecules. Thus, M is the molar mass in units of kg/mol. • Keep in mind, since the kinetic energy is in joules per mole(J), we express R in units of J mol-1 K-1 (8.314) • The speed in which the molecules of a gas travel can travel can be found by:
Example • Determine the thermal energy of a molecule of nitrogen gas at 20oC. Then, calculate the root mean square speed (RMS) • To find the RMS, we must first find the molar mass of the gas in kg/mol