Presentation on theme: "Properties of Gases Compiled by Mr. Walia Gas form is an essential form of the matter O 2, O 3, N 2, CO 2 are essential gases in the atmosphere. O 2,"— Presentation transcript:
Properties of Gases Compiled by Mr. Walia
Gas form is an essential form of the matter O 2, O 3, N 2, CO 2 are essential gases in the atmosphere. O 2, and N 2 provide the plants essential compounds for photosynthesis. O 3 protects us from the harmful solar radiations.
The proportion of these gases in the atmosphere is measured using balloons by meteorologists everyday.
11 of the periodical table elements are in the gas form.
Many inorganic and organic compounds are gases: NO 2, SO 2, N 2 O, CH 4, C 2 H 6, C 3 H 8, CH 3 NH 2
Pressure of the Gases Pressure (P) is the force exerted on a surface divided by the area of the surface
Properties of Pressure Pressure increases as more gas is added Conclusion - Pressure (P) is directly proportional to moles of gas (n) Pressure due to a gas is the same in all directions whereas pressure due to weight is directional Pressure unit in SI is Pascal:
Theres another way to measure the gas pressure: Barometer An apparatus used to measure pressure; derived from the Greek "baros" meaning "weight. Created by Evangelista Torricelli in Gas pressure can be measured by relating to the atmospheric pressure.
At sea level and 0°C this height is 760 mmHg and the pressure supporting this height is called 1 atmosphere. Torricelli inverted a tube filled with mercury into a dish until the force of the Hg inside the tube balanced the force of the atmosphere on the surface of the liquid outside the tube. The hight of the mercury in the tube is a measure of the atmospheric pressure.
Atmospheric Pressure is measured by different units: Atmospheric pressure is equal to 760 mmHg and is called 1 atm. 1 atm = 760 mmHg 1 mmHg = 1 torr so 1atm = 760 torr 1 atm = kPa 1 bar = 10 5 Pa so 1 atm = 1 bar
Change in average atmospheric pressure with altitude
Why Mercury? In theory, any liquid can be used in abarometer. Mercury is so dense that can form a usable height in the tube. A similar barometer made of water, in comparison, would have to be more than 34 feet (100 meters) high. Mercury also has a low vapor pressure, meaning it does not evaporate very easily. Water has a greater vapor pressure. Because of this, the pressure exerted by water vapor at the top of the barometer would affect the level of the mercury in the tube and the barometric reading.
Aneroid barometer A major disadvantage of the mercury barometer is its bulkiness and fragility. The long glass tube can break easily, and mercury levels may be difficult to read under unsteady conditions, as on board a ship at sea. To resolve these difficulties, the French physicist Lucien Vidie invented the aneroid ("without liquid") barometer in 1843.
Aneroid Barometer An aneroid barometer is a container that holds a sealed chamber from which some air has been removed, creating a partial vacuum. An elastic disk covering the chamber is connected to a needle or pointer on the surface of the container by a chain, lever, and springs. As atmospheric pressure increases or decreases, the elastic disk contracts or expands, causing the pointer to move accordingly.
This type of aneroid barometer has a pointer that moves from left to right in a semicircular motion over a dial, reflecting low or high pressure. The simple clock-like aneroid barometer hanging on the wall of many homes operates on this basis.
Open- tube manometer The open-tube manometer is another device that can be used to measure pressure. The open-tube manometer is used to measure the pressure of a gas in a container.
Atmospheric pressure pushes on the mercury from one direction, and the gas in the container pushes from the other direction. In a manometer, since the gas in the bulb is pushing more than the atmospheric pressure, you add the atmospheric pressure to the height difference: gas pressure = atmospheric pressure + h h is the difference in mercury levels. Gas pressure will be in units of torr or mmHg.
Ideal Gas Laws There are some laws that explain (a) the relationship between the pressure and volume of the gas at a fixed temperature, (b) the relationship between the volume and temperature of it in a fixed pressure and, (c) the relationship between the pressure and temperature of a gas in a fixed volume.
Boyle's Law (1662) The relationship between the pressure and the volume of a given sample of gas at fixed temperature. A sample of gas compresses if the external pressure applied to it increases and the product PV is constant. The Pressure (P) of a gas is inversely proportional to Volume (V) at constant Temperature (T) and moles of gas (n).
Boyle's law, stated in mathematical terms for a gas whose pressure and volume is measured at two different pressure/volume states at a constant temperature is then, P 1 V 1 = P 2 V 2
Boyles Mathematical Law: What if we had a change in conditions since PV = k P 1 V 1 = P 2 V 2 Eg: A gas has a volume of 3.0 L at 2 atm. What is its volume at 4 atm?
1)determine which variables you have: P 1 = 2 atm V 1 = 3.0 L P 2 = 4 atm V 2 = ? P and V = Boyles Law 2)determine which law is being represented:
3) Rearrange the equation for the variable you dont know 4) Plug in the variables and chug it on a calculator: P 1 V 1 = V 2 P2P2 P2P2 (2.0 atm)(3.0L) = V 2 (4atm) V 2 = 1.5L
Example 1 At 0 °C and 5.00 atm, a given sample of a gas occupies 75.0 L. The gas is compressed to a final volume of 30.0 L at 0 °C. What is the final pressure? Answer: 12.5 atm
Charles' Law (1787) (It was developed by Guy Lussac in 1802) The volume of any gas increases directly with increasing temperature at constant pressure. If we plot a graph of the volume of a sample of gas versus the temperature at constant pressure we get something that looks like the following:
Absolute Temperature Experimental data show that 1°C increase in the temperature of an ideal gas would increase its volume as 1/273 of the volume at 0°C. So if the volume in 0°C was 273 mL, it would increase 1/273 x 273 mL = 1 mL at 1°C and the total volume will be 274 mL. A 10°C increase will increase the volume: 10 x 1/273 x 273 mL = 10 mL, the total volume: 283 mL. A 273°C increase will increase the volume: 273 x 1/273 x 273 mL = 273 mL, the total volume: 546 mL.
Although the the volume increases in a regular manner with increase in temperature, it is not directly proportional to the Celsius temperature. An increase in temperature from 1°C to 10 °C, eg., does not increase the volume 10- fold, but only from 274 mL to 283 mL. An absolute temperature scale, kelvin temperature, is defined in such a way the volume is directly proportional to kelvin temperature. A 2-fold increase in the absolute temperature would increase the volume the same: an increase from 273 K (0°C) to 546 K (273 °C) increases the volume from 273 mL to 546 mL. A Kelvin reading is (T) is obtained by adding 273 to the Celsius temperature (t): T = t +273
From the extrapolated line, we can determine the temperature at which an ideal gas would have a zero volume. Since ideal gases have infinitely small atoms the only contribution to the volume of a gas is the pressure exerted by the moving atoms bumping against the walls of the container. If no volume then there must be no kinetic energy left. Thus, absolute zero is the temperature at which all kinetic energy (motion) has been removed. NOTE: This does not mean all energy has been removed, merely all kinetic energy.
To avoid the need to know k, we use ratios. The ratio of V to T of an ideal gas at constant pressure is constant over all temperatures. Or...
A balloon racer uses the Charles law. When the air in the balloon gets warmer it expands and will become less dense and balloon floats in the air. When the air gets colder it will become more dense and it will come down in the air.
Example 2 A sample of a gas has a volume of 79.5 mL at 45 °C. What volume will the sample occupy at 0 °C when the pressure is held constant? Answer: 68.2 mL
Amonton's Law (1703) The Pressure of a gas is directly proportional to the Temperature (Kelvin) at constant V and n. Example 3 A 10.0 L container is filled with a gas to a pressure of 2.00 atm at 0 °C. At what temperature will the pressure inside the container be 2.50 atm? Answer: 341 K = 68 °C
Molar Volume Molar volume (Vm) is the volume that every mole of the material occupies (L. mol -1 ) Vm = V. occupied by the material/ No. of moles of it Vm = V / n Physical data show that molar volumes of gases are equal at the same pressure and temperatures. Vm for some gases at 0 °C and 1 atm: Argon Carbon dioxide Nitrogen Oxygen Hydrogen 22.43
Avogadro's Law (1811) The Volume of a gas is directly proportional to the moles of the gas, n at constant P and T. According to Avogadro, equal volumes of different (ideal) gases at the same temperature and pressure contain equal numbers of molecules (moles) of the different gases.
Ideal Gas Law If we take the three of the gas laws we've studied so far, we can combine them into a single law called the Ideal Gas law. This law covers the relationship between temperature, pressure, volume and number of moles of an Ideal gas. Avogadro's Law: V = k 1 n | T,P Boyle's Law: V = k 2 /P | T,n Charles' Law: V = k 3 T | n,P
After some consideration and algebra, we arrive at: V = k overall nT/P where k overall turns out to be the Ideal gas constant (or universal gas constant) We're more familiar with the equation written as: PV = nRT This is the Ideal Gas Law or the equation of state for an ideal gas. At ordinary conditions of temperature and pressure, most gases conform well to the behavior described by this equation. Deviations occur, however, under extreme conditions (low temperature and high pressure).
The molar volume of an ideal gas at STP (0 °C and 1 atm): L So we can calculate the ideal gas constant: R = PV / nT = (1 atm) ( L) / (1 mol) ( K) = = x 10 –2 L.atm / K.mol Another form of the equation of state for an ideal gas: Since: n = g / MW So: PV = (g/MW) RT
Example 4 The volume of a sample of gas is 462 mL at 35 °C and 1.15 atm. Calculate the volume of the sample at STP. Example 5 What is the density of NH 3 (g) at 100 °C and 1.15 atm?