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ιδανικα αερια και μεταβολες

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ιδανικα αερια και μεταβολες

  1. 1. eal Gas Law In studying the behavior of gases in conf ined spaces, it is useful to limit ourselves to the study of ideal gases. Ideal gases are theoretical models of real gases, w hich utilize a number of basic assumptions to simplify their study. These assumptions include treating the gas as being comprised of many particles w hich move randomly in a container. The particles are, on average, far apart f rom one another, and they do not exert forces upon one another unless they come in contact in an elastic collision. Under normal conditions such as standard temperature and pressure, most gases behave in a manner quite similar to an ideal gas. Heavy gases as w ell as gases at very low temperatures or very high pressures are not as w ell modeled by an ideal gas. The Ideal Gas Law relates the pressure, volume, number of particles, and temperature of an ideal gas in a single equation, and can be w ritten in a number of dif ferent ways. In this equation, P is the pressure of the gas (in Pascals), V is the volume of the gas (in cubic meters), n is the number of moles of gas, N is the number of molecules of gas, R is the universal gas constant equal to 8.31 J/mol·K (w hich is also 0.08206 L·atm/mol·K), kB is Boltzmann’s Constant (1.38×10-23 J/K), and T is the temperature, in Kelvins. To convert f rom molecules to moles, you can use Avogadro’s Number (N0=6.02×10-23 molecules/mole): Note that a Pascal multiplied by a cubic meter is a new ton-meter, or Joule. As w ell, Boltzmann’s constant is the ideal gas law constant divided by Avogadro’s number. Question: How many moles of an ideal gas are equivalent to 3.01×1024 molecules? Answer: Question: Find the number of molecules in 0.4 moles of an ideal gas. Answer: Question: How many moles of gas are present in a 0.3 m3 bottle of carbon dioxide held at a temperature of 320K and a pressure of 1×106 Pascals? Answer: Question: A cubic meter of carbon dioxide gas at room temperature (300K) and atmospheric pressure (101,325 Pa) is compressed into a volume of 0.1 m3 and held at a temperature of 260K. What is the pressure of the compressed carbon dioxide? Answer: Since the number of moles of gas is constant, you can simplify the ideal gas equation into the combined gas law by setting the initial pressure, volume, and temperature relationship equal to the f inal pressure, volume, and temperature relationship. Since you know all the quantities in this equation except for the f inal pressure, you can solve for the f inal pressure directly.
  2. 2. Question: One mole of helium gas is placed inside a balloon. What is the pressure inside the balloon w hen the balloon rises to a point in the atmosphere w here the temperature is -12°C and the volume of the balloon is 0.25 cubic meters? Answer: First you must convert the temperature f rom degrees Celsius to Kelvins. Next, you can use the ideal gas law to solve for the pressure inside the balloon. It’s also quite straightf orw ard to f ind the total internal energy of an ideal gas. Recall that the average kinetic energy of the particles of an ideal gas are described by the formula: The total internal energy of an ideal gas can be found by multiplying the average kinetic energy of the gas’s particles by the number of particles (N) in the gas. Therefore, the internal energy of the gas can be calculated using: Question: Find the internal energy of 5 moles of oxygen at a temperature of 300K.
  3. 3. Answer: Question: What is the temperature of 20 moles of argon w ith a total internal energy of 100 kJ? Answer: Thermodynamics Thermodynamics, which began as an ef fort to increase the ef ficiency of steam engines in the early 1800s, can be thought of as the study of the relationship betw een heat transferred to or f rom an object, and the w ork done on or by an object. Both heat and w ork deal w ith the transfer of energy, but heat in- volves energy transfer due to a temperature dif ference. The zeroth law of thermodynamics (don’t blame me, I didn’t name it!) states that if object A is in thermal equilibrium w ith object B, and object B is in thermal equilibrium w ith object C, then objects A and C must be in thermal equilibrium w ith each other. This law is so intuitive it almost doesn’t need stating, but in def ining proof s of the 1st and 2nd law s of thermodynamics, scientists realized they needed this law specifically stated to complete their proofs. The first law of thermodynamics is really a restatement of the law of conservation of energy. Specifically, it states that the change in the internal energy of a closed system is equal to the heat added to the system plus the w ork done on the system, and is w ritten as: In this equation it is important to note the sign conventions, where a positive value for heat, Q, represents heat added to the system, and a positive value for w ork, W, indicates w ork done on the gas. If energy w ere being pulled f rom the system, as in heat taken f rom the system or w ork done by the system, those quantities w ould be negative. In most cases, you’ll utilize the f irst law of thermodynamics to analyze the behavior of ideal gases, w hich can be streamlined by analyzing the def inition of w ork on a gas. If w ork is force multiplied by displacement, and pressure is force over area, force can be replaced w ith pressure multiplied by area. The area multiplied by the displacement gives you the change in volume of the gas. Due to the sign convention that w ork done on the gas is positive (corresponding to a decrease in volume), you can w rite work as W=-PΔV. Question: Five thousand joules of heat is added to a closed system, w hich then does 3000 joules of w ork. What is the net change in the internal energy of the system? Answer: Question: A liquid is changed to a gas at atmospheric pressure (101,325 Pa). The volume of the liquid w as 5×10- 6 m3. The volume of gas is 5×10-3 m3. How much w ork w as done in the process? Answer:
  4. 4. Pressure-Volume Diagrams (PV diagrams) are useful tools for visualizing the thermodynamic processes of gases. These diagrams show pressure on the y-axis, and volume on the x-axis, and are used to describe the changes undergone by a set amount of gas. Because the amount of gas remains constant, a PV diagram not only tells you pressure and volume, but can also be used to determine the temperature of a gas w hen combined w ith the ideal gas law . A Sample PV diagram is shown below, showing two states of the gas, state A and state B. In transitioning f rom state A to state B, the volume of the gas increases, while the pressure of the gas decreases. In transitioning f rom state B to state A, the volume of the gas decreases, while the pressure increases. Because the w ork done on the gas is given by W=-PΔV, you can f ind the w ork done on the gas graphically f rom the PV diagram by taking the area under the curve. Because of the positive/negative sign convention, as the volume of gas expands the gas does w ork (W is negative), and as the gas compresses, work is done on the gas (W is positive). Question: Using the PV diagram at right, f ind the amount of w ork required to transition f rom state A to B, and then the amount of w ork required to transition f rom state B to state C. Answer: The amount of w ork in moving f rom state A to B is equal to the area under the graph for that transition. Since there is no area under the straight line, no w ork was done. The w ork in moving f rom state B to state C can be found by taking the area under the line in the PV diagram. Note that the w ork is negative, indicating the gas did w ork, which correlates with the gas expanding.
  5. 5. In exploring ideal gas state changes, there are a number of state changes in w hich one of the characteristics of the gas or process remain constant, and are illustrated on the PV diagram below . The types of processes include:  Adiabatic -- Heat (Q) isn’t transf erred into or out of the system  Isobaric -- Pressure (P) remains constant  Isochoric -- Volume (V) remains constant  Isothermal -- Temperature (T) remains constant In an adiabatic process, heat f low (Q) is zero. Applying the f irst law of thermodynamics, if ΔU=Q+W, and Q is 0, the change in internal energy of the gas must be equal to the w ork done on the gas ( ΔU=W). In an isobaric process, pressure of the gas remains constant. Because pressure is constant, the PV diagram for an isobaric process shows a horizontal line. Further, applying this to the ideal gas law , you f ind that V/T must remain constant for the process. In an isochoric process, the volume of the gas remains constant. The PV diagram for an isochoric process is a vertical line. Because W=-PΔV, and ΔV=0, the w ork done on the gas is zero. This is also ref lected graphically in the PV diagram. Work can be found by taking the area under the PV graph, but the area under a vertical line is zero. Applying this to the ideal gas law , you f ind that P/T must remain constant for an isochoric process. In an isothermal process, temperature of the gas remains constant. Lines on a PV diagram describing any process held at constant temperature are therefore called isotherms. In an isothermal process, the product of the pressure and the volume of the gas remains constant. Further, because temperature is constant, the internal energy of the gas must remain constant. Question: An ideal gas undergoes an adiabatic expansion, doing 2000 joules of w ork. How much does the gas’s internal energy change? Answer: Since the process is adiabatic, Q=0, therefore:
  6. 6. Question: Heat is removed f rom an ideal gas as its pressure drops f rom 200 kPa to 100 kPa. The gas then expands f rom a volume of 0.05 m3 to 0.1 m3 as show n in the PV diagram below . If curve AC represents an isotherm, f ind the w ork done by the gas and the heat added to the gas. Answer: The w ork done by the gas in moving f rom A to B is zero, as the area under the graph is zero. In moving f rom B to C, how ever, the w ork done by the gas can be found by taking the area under the graph. The negative sign indicates that 5000 joules of w ork was done by the gas. Since AC is on an isotherm, the temperature of the gas must remain constant, theref ore the gas’s internal energy must remain constant. Know ing that ΔU=Q+W, if ΔU=0, then Q must be equal to -W, therefore 5000 joules must have been added to the gas. The second law of thermodynamics can be stated in a variety of w ays. One statement of this law says that heat f low s naturally f rom a w armer object to a colder object, and cannot f low f rom a colder object to a w armer object w ithout doing w ork on the system. This can be observed quite easily in everyday circumstances. For example, your cold spoon contacting your hot soup never results in your soup becoming hotter and your spoon becoming colder. The second law of thermodynamics also limits the ef f iciency of any heat engine, and proves that it is not possible to make a 100 percent ef f icient heat engine, even if f riction w ere completely eliminated. Another statement of this law says that the level of entropy, or disorder, in a closed system can only increase or remain the same. This means that your desk w ill never naturally become more organized w ithout doing w ork. It also means that you can’t drop a handf ul of plastic building blocks and observe them spontaneously land in an impressive model of a medieval castle. Unfortunately, it even means that no matter how many times Humpty Dumpty falls of f his w all, all his pieces on the ground w ill never end up more organized af ter he hits the ground compared to before his balance failed him. The f inal law of thermodynamics, the third law of thermodynamics, also known as Ne rnst’s Theorem af ter its discoverer, Walter Nernst, states that no material can ever be cooled to absolute zero (although materials can get aw fully close!)

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