Lecture 20 - Gases Part II

Tuesday, April 2, 2024

9:00 AM

"We live submerged at the bottom of an ocean of elementary air, which is known by incontestable experiments to have weight" - Torricelli Class notes (1-19): https://bricejurban.github.io/CHEM101/ Assignments this week: ﷟HYPERLINK "https://boisestatecanvas.instructure.com/courses/28698/assignments/1014190"HW 13 Chemistry of Gases Due Sunday Review Answer Key for Midterm 3 Read Chapter 7 (I may make a short reading quiz on this) Reminders: Midterm 3 grades are entered on canvas and published on Gradescope. An answer key is outside SCNC 336. If you notice any errors in grading, please make a regrade request through Gradescope. Statistics Average: 77 Median: 80 St. Dev: 24 A: 35 B: 25 C: 14 D: 7 F: 20 My CIC (EDUC 107) Hours: Friday 11AM - 1PM Office Hours (SCNC 314 or Zoom): ﷟HYPERLINK "https://calendly.com/bricejurban/office-hours"By appointment Today (4/2) Gases Part II Pressure Mixtures of Gases and Partial Pressures Kinetic Molecular Theory Gas Laws Looking Ahead Thursday (4/4) Finish discussion on Gases Phase Transitions Calorimetry Pressure Standard Pressure = 1 atm = 760 mmHg = 760 torr = 101.325 kPa = 14.7 psi = 29.92 inHg = 1.01325 bar Pressure is the next property we should discuss. Like temperature we have both an everyday understanding and a scientific definition. Commit to memory that pressure refers to the amount of force applied over an area. If we have more force and keep the area the same we have a greater pressure. We can also decrease the area the force is applied to also increase the pressure. Consider the action of a hammer on a nail. The force of the swing exerted on the small area of the nail causes it to be driven into a board. Or consider a submarine deep underwater withstanding the weight of the ocean above it through its thick walls. Now consider the atmosphere above us, also a fluid, pushing down upon us with 1 atmosphere of pressure at sea level. This pressure we experience everyday but don't notice it. Thus pressure is commonly thought of as how it compares to that at sea level, and this is historically how it's been measured. Barometers are devices that can measure subtle differences in air pressure. Watching a weather report we are constantly hearing about low and high pressure systems. The first barometer was invented in 1643 by Evangelista Torricelli after he interpreted the results of Gasparo Berti's experiment. In short: Berti's experiment consisted of filling with water a long tube that had both ends plugged, then standing the tube in a basin of water. The bottom end of the tube was opened, and water that had been inside of it poured out into the basin. However, only part of the water in the tube flowed out, and the level of the water inside the tube stayed at an exact level, which happened to be 10.3 m (34 ft) Gasparo_Berti_Experiment.jpg undefined It was traditionally thought that air did not have weight: that is, that the kilometers of air above the surface did not exert any weight on the bodies below it. Even Galileo had accepted the weightlessness of air as a simple truth. Torricelli questioned that assumption, and instead proposed that air had weight which held (or rather, pushed) up the column of water. He thought that the level the water stayed at (10.3 m) was reflective of the force of the air's weight pushing on it (specifically, pushing on the water in the basin and thus limiting how much water can fall from the tube into it). He viewed the barometer as a balance, an instrument for measurement. Untitled picture.png
The first barometer was invented in 1643 by Evangelista Torricelli after he interpreted the results of Gasparo Berti's experiment. In short: Berti's experiment consisted of filling with water a long tube that had both ends plugged, then standing the tube in a basin of water. The bottom end of the tube was opened, and water that had been inside of it poured out into the basin. However, only part of the water in the tube flowed out, and the level of the water inside the tube stayed at an exact level, which happened to be 10.3 m (34 ft) Gasparo_Berti_Experiment.jpg undefined It was traditionally thought that air did not have weight: that is, that the kilometers of air above the surface did not exert any weight on the bodies below it. Even Galileo had accepted the weightlessness of air as a simple truth. Torricelli questioned that assumption, and instead proposed that air had weight which held (or rather, pushed) up the column of water. He thought that the level the water stayed at (10.3 m) was reflective of the force of the air's weight pushing on it (specifically, pushing on the water in the basin and thus limiting how much water can fall from the tube into it). He viewed the barometer as a balance, an instrument for measurement. Torricelli was able to recreate Berti's experiment with mercury, which is about 14 times denser than water. Now only a tube 80 cm (0.8 m) was needed Untitled picture.png The mercury barometer's design gives rise to the expression of atmospheric pressure in the height of mercury in millimeters (mmHg) or centimeters (cmHg) or inches (inHg). Historically, to recognize Torricelli, we define the torr as 1 mmHg. The pressure is quoted as the level of the mercury's height in the vertical column. Typically, atmospheric pressure is measured between 26.5 inches (670 mm) and 31.5 inches (800 mm) of Hg. One atmosphere (1 atm) is equivalent to 29.92 inches (760 mm) of mercury. Torricelli observed that the height of the mercury in a barometer changed slightly each day concluding that this was due to the changing pressure in the atmosphere. Low pressures indicated a storm was likely the next day. He wrote that "we are submerged at the bottom of an ocean of elementary air, which is known by incontestable experiments to have weight." At the top of the mercury column is a void space known as a vacuum. Before this experiment, scientists questioned whether a vacuum could exist. 220px-MercuryBarometer.svg.png This is a depiction of a mercury thermometer.
°F = 9/5°C + 32 Standard Pressure = 1 atm = 760 mmHg = 760 torr = 101.325 kPa = 14.7 psi = 29.92 inHg = 1.01325 bar While travelling in Europe, you see a weather forecast of 20 C what is the temperature in F? Convert a pressure of 37.4 psi to atm. 3. What pressure in mmHg is equivalent to 1.350 atm? 4. Consider the image of a manometer below. Based on the image, how does the pressure of the gas in the manometer compare to the pressure of the atmosphere? Untitled picture.png Machine generated alternative text: Open end 26.4 cm Untitled picture.png Machine generated alternative text: The gas pressure is higher than the atmospheric pressure. The gas pressure is lower than the atmospheri pressure. The gas pressure is the same as the c atmospheric pressure. Untitled picture.png Machine generated alternative text: Consider the image of a mercury manometer below. Based on the image, if the pressure of the atmosphere is 760.0 mmHg, the pressure of the gas is Open end 26.4 cm Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings
Untitled picture.png Machine generated alternative text: Consider the image of a mercury manometer below. Based on the image, if the pressure of the atmosphere is 760.0 mmHg, the pressure of the gas is Open end 26.4 cm Mixtures of Gases Most gases don't exist as pure substances but rather as mixtures. When discussing a gas mixture we often make use of the mole fraction. Consider a gas sample composed of 15.0 g hydrogen (H2) and 15.0 g oxygen (O2). By mole, there is a lot more hydrogen than oxygen. Why? 15.0 g H2 ÷ 2.016 g/mol = 7.440 mol H2 15.0 g O2 ÷ 32.00 g/mol = 0.4688 mol O2 The mole fraction of H2 (χH2) is the mols of H2 divided by the total mols of gas: χH2 = 7.440 mol H2 / (7.440 mol H2 + 0.4688 mol O2) = 0.941 χO2 = 0.4688 mol O2 / (7.440 mol H2 + 0.4688 mol O2) = 0.059 χH2 + χO2 = 0.941 + 0.059 = 1 (Note the sum of all the mole fractions of a gas sample are equal to unity (1) Calculating the mole fraction is very useful for determining the individual gases contribution to the total pressure. If the total pressure in the gas sample is 3.00 atmospheres the partial pressures of the gases can be found by multiplying the mole fraction by the total pressure. Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings
If the total pressure in the gas sample is 3.00 atmospheres the partial pressures of the gases can be found by multiplying the mole fraction by the total pressure. P(H2) = 0.941 × 3.00 = 2.823 atm P(O2) = 0.059 × 3.00 = 0.177 atm P(H2) + P(O2) = 2.823 + 0.177 = 3.00 = Ptot (Note the sum of the partial pressures will be equal to the total partial. This assumption is known as Dalton's Law of Partial Pressures and has been proven experimentally. If we extend this to 3 or more gases in a sample it looks like: nA + nB + nc + . . . = ntot (nA = moles of gas component A) χA + χB + χC + . . . = 1 (χA = mole fraction gas component A) PA + PB + PC + . . . = Ptot (PA = partial pressure of gas component A) We can summarize these rules for one particular gas like this: PA/Ptot = nA/ntot PA/Ptot = χA PA = χA × Ptot This is very useful when we know the total pressure and have the amounts (mols) of each gas, but don't know the partial pressures. Here's a really common example of this law being applied: Consider the situation where we take a reactive metal and dissolve it in an acid. This will produce hydrogen gas which can then be collected in a container over water at 20 °C. An experimenter might want to determine what the partial pressure of the collected hydrogen would be in the container. ﷟HYPERLINK "https://www.youtube.com/watch?v=jg4c0rVSckg"Collecting A Gas Over Water Animation (video of how to collect a gas sample, but instead using KClO3 to produce oxygen gas). Collecting A Gas Over Water Animation Press enter to activate
Collecting A Gas Over Water Animation Press enter to activate Consider this reaction collected over water: Zn (s) + 2HCl (aq) → ZnCl2 (aq) + H2(g) What is the partial pressure of H2 if the total pressure is 0.985 atm and the temperature of the water is 20.0 °C? As the container fills up with hydrogen gas, there will be contribution from water vapor H2O (g) Untitled picture.png Machine generated alternative text: Hydrogen plus water vapor Atmospheric pressure holds the water level up. phpaJ7HyG.png At 20.0 °C the partial pressure of water vapor is 17.5 mmHg 17.5 mmHg ÷ 760 mmHg/atm = 0.0230 atm P(H2) + P(H2O) = Ptot P(H2) + 0.0230 atm = 0.985 atm P(H2) = 0.985 atm - 0.0230 atm Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings
P(H2) = 0.985 atm - 0.0230 atm P(H2) = 0.962 atm Volumes of gases and the Kinetic Molecular Theory Gases are fun in that they take the shape of their container and can be easily compressed or expanded (unlike solids and liquids). One thing that is remarkable about gases is that equal amounts of gases, regardless of their molar mass, take up the same amount of space. This means that one mole of hydrogen takes up the same volume as one mole of oxygen. This is the Law of Amedeo Avogadro and has been proven experimentally. Avogadro used this law to help him determine the correct molar masses of the simplest gases. One useful conversion is that under standard temperature and pressure (STP) (273 K (0 C) and 1 atm) 1 mole of a gas = 22.4 L This is by no means an easily recognized truth and for a very long time most chemists did not accept this! Hydrogen is in fact smaller than oxygen, so why would it take up the same space? Well . . . the gas phase of matter is the state in which the particles are considered liberated and free from intermolecular forces. These particles are very far from each other, relatively, and thus do not interact much with each other especially under the conditions we generally observe gases. To really understand this, we need to consider the framework of Kinetic Molecular Theory that describes the behavior of gases in terms of the motion and interactions of their constituent molecules. It provides a microscopic explanation for the macroscopic properties of gases, such as pressure, temperature, and volume, and is based on several key postulates: Gas molecules are in constant, random motion: This postulate asserts that gas molecules are always moving in random directions with a variety of speeds. Their motion is straight-line, and they only change direction when they collide with each other or the walls of their container. The volume of individual gas molecules is negligible compared to the volume of their container: This means that the total volume occupied by all the gas molecules in a container is much less than the volume of the container itself. This is why gases are highly compressible. No intermolecular forces in ideal gases: KMT assumes that there are no forces of attraction or repulsion between the gas molecules in an ideal gas. This simplification is what makes a gas "ideal." In real gases, intermolecular forces do exist, but the ideal gas approximation is still useful under many conditions, particularly at high temperatures and low pressures. Elastic collisions: When gas molecules collide with each other or with the walls of their container, the collisions are perfectly elastic. This means that there is no net loss of kinetic energy from these collisions. The total kinetic energy of the molecules remains constant, assuming no external energy is added or removed. The average kinetic energy of gas molecules is proportional to the absolute temperature in K: This crucial postulate links the microscopic behavior of gas molecules with a macroscopic property, temperature. It states that the average kinetic energy of the molecules, which is a measure of their speed and mass, is directly proportional to the temperature of the gas in Kelvin. This implies that as the temperature increases, the average speed and kinetic energy of the gas molecules increase as well.
The average kinetic energy of gas molecules is proportional to the absolute temperature in K: This crucial postulate links the microscopic behavior of gas molecules with a macroscopic property, temperature. It states that the average kinetic energy of the molecules, which is a measure of their speed and mass, is directly proportional to the temperature of the gas in Kelvin. This implies that as the temperature increases, the average speed and kinetic energy of the gas molecules increase as well. In a rigid container, how does the pressure change when the temperature increases? As the temperature increases, the motion of the particles increases, this causes more collisions With the walls of the container. This causes the pressure to increase. In a rigid container, how does the pressure change when the sample contains more moles of gas (at constant temperature)? Now that there are more particles to collide with the walls of the container, the pressure increases In a piston, how does the pressure change when the volume is decreased? There is less space for the gas particles to move around. This causes the pressure to increase Because there are more collisions with the walls of the container. In a balloon, how does the volume change when the temperature is increased at constant pressure? When we increase the temperature, the kinetic energy of the gases increase. This causes more collisions with the walls of the balloon (which can expand) Because the pressure is constant the volume must expand and the balloon increases in volume. Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings
Gas Laws The aforementioned discussion on the properties of gases and the kinetic molecular theory gives us the context to study how these properties relate. For example, how does an increase in pressure affect temperature and volume? Or, how does an increase in the amount (moles) of a gas affect its pressure in a container that is rigid? Or maybe, why does a balloon get larger and larger with altitude and eventually pop? These questions take us back to the early days of modern chemistry. There are a variety of laws that relate these individual properties and one law that governs them all. If you are to learn only one equation in this chapter it should be this one: PV= nRT (P = pressure in atm, V = volume in liters, n = amount in mols, R = gas constant (0.0821 atm·L/(mol·K)) and T = temp in K) This equation is known as the ideal gas law and relates all of the variables to a gas constant R. An ideal gas follows the postulates of the Kinetic Molecular Theory. (Note: Under conditions that are high pressure or low temperature (very cold e.g. 100 K) predicted properties will deviate from ideality) Before we get into this let's step back in time and consider the simplest relationships Pressure vs Volume (Boyle's Law) 1662 P V n T Initial 1.4 atm 250 mL Final ? 400 mL Sample of gas at 30 °C in a piston has an initial pressure of 1.4 atm and is then expanded by moving the piston Untitled picture.png As we expand the gas, there is more volume for the gas to take up. So its pressure decreases. Volume and pressure are inversely related Volume vs Temperature (Charles's Law) 1790's, attributed in 1802 P V n T Untitled picture.png Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings
P V n T Initial 1 L -80°C = Final ? 113°C = Closed balloon at atmospheric pressure in a dry ice bath has a volume of 1L is heated with steam. Untitled picture.png Pressure vs Temperature (Gay-Lussac's Law) ~1805 P V n T Initial 600 torr 0 °C = ? K Final ? 100°C = ? K Sample of neon in rigid 1.0 L container is equilibrated in an ice bath and has a pressure of 600 torr, is then placed in a boiling water bath Untitled picture.png Volume vs Amount (Avogadro's Law) ~1800 but recognized 1850 P V n T Initial 3 L 1.00 Final 12 L ? A balloon at constant atmospheric pressure and temperature is 3L in size and then blown up with more air to 12L. The initial mols of gas is 1.00 Untitled picture.png Combined Gas Law P V n T Initial 450 mmHg 600 mL 263 K Final 760 mmHg ? 273 K
A weather balloon's pressure and temperature changes as it descends. How does the volume change? Ideal Gas Law P V n T Initial ? 4.1 L 0.20 40°C = A sample of 0.20 mol of oxygen gas occupies a volume of 4.1 L at 40 °C. Extra Problems as needed

 

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