# Volume Temperature Relationship: Charles s Law in Visual C# Scanner Data Matrix in Visual C# Volume Temperature Relationship: Charles s Law

Volume Temperature Relationship: Charles s Law
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Charles s law describes the volume and temperature relationship of a gas when the pressure and amount are constant If a sample of gas is heated, the volume must increase for the pressure to remain constant This is shown in Figure 84 Remember: In any gas law calculation, you must express the temperature in kelvin There is a direct relationship between the Kelvin temperature and the volume: as one increases, the other also increases Mathematically, Charles s law can be represented as: V/T = kc where kc is a constant and the temperature is expressed in kelvin Again, if there is a change from one set of volume temperature conditions to another, Charles s law can be expressed as: V1/T1 = V2/T2
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Volume pressure relationship for gases As the volume decreases, the number of collisions increase
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Volume temperature relationship for gases
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Pressure Temperature Relationship: Gay-Lussac s Law
Gay-Lussac s law describes the relationship between the pressure of a gas and its Kelvin temperature if the volume and amount are held constant Figure 85 represents the process of heating a given amount of gas at a constant volume As the gas is heated, the particles move with greater kinetic energy, striking the inside walls of the container more often and with greater force This causes the pressure of the gas to increase The relationship between the Kelvin temperature and the pressure is a direct one: P/T = kg or P1/T1 = P2/T2
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Figure 85 Pressure temperature relationship for gases As the temperature increases, the gas particles have greater kinetic energy (longer arrows) and collisions are more frequent and forceful
Combined Gas Law
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In the discussion of Boyle s, Charles s, and Gay-Lussac s laws we held two of the four variables constant, changed the third, and looked at its effect on the fourth variable If we keep the number of moles of gas constant that is, no gas can get in or out then we can combine these three gas laws into one, the combined gas law, which can be expressed as: (P1V1)/T1 = (P2V2)/T2
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Again, remember: In any gas law calculation, you must express the temperature in kelvin In this equation there are six unknowns; given any five, you should be able to solve for the sixth For example, suppose a 50-L bottle of gas with a pressure of 250 atm at 20 C is heated to 80 C We can calculate the new pressure using the combined gas law Before we start working mathematically, however, let s do some reasoning The volume of the bottle hasn t changed, and neither has the number of moles of gas inside Only the temperature
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and pressure have changed, so this is really a Gay-Lussac s law problem From Gay-Lussac s law you know that if you increase the temperature, the pressure should increase if the amount and volume are constant This means that when you calculate the new pressure, it should be greater than 250 atm; if it is less, you ve made an error Also, remember that the temperatures must be expressed in kelvin 20 C = 293 K (K = C + 273) and 80 C = 353 K We will be solving for P2, so we will take the combined gas law and rearrange for P2: (T2P1V1)/(T1V2) = P2 Substituting in the values: (353 K)(250 atm)(50 L)/(293 K)(50 L) = P2 30 atm = P2 The new pressure is greater than the original pressure, making the answer a reasonable one Note that all the units canceled except atm, which is the unit that you wanted Let s look at a situation in which two conditions change Suppose a balloon has a volume at sea level of 100 L at 7600 torr and 20 C (293 K) The balloon is released and rises to an altitude where the pressure is 4500 torr and the temperature is 10 C (263 K) You want to calculate the new volume of the balloon You know that you have to express the temperature in K in the calculations It is perfectly fine to leave the pressures in torr It really doesn t matter what pressure and volume units you use, as long as they are consistent in the problem The pressure is decreasing, so that should cause the volume to increase (Boyle s law) The temperature is decreasing, so that should cause the volume to decrease (Charles s law) Here you have two competing factors, so it is difficult to predict the end result You ll simply have to do the calculations and see Using the combined gas equation, solve for the new volume (V2): (P1V1)/T1 = (P2V2)/T2 (P1V1T2)/(P2T1) = V2 Now substitute the known quantities into the equation (You could substitute the knowns into the combined gas equation first, and then solve for the volume Do it whichever way is easier for you) (7600 torr)(100 L)(263 K)/(4500 torr)(293 K) = V2 152 L = V2 Note that the units canceled, leaving the desired volume unit of liters Overall, the volume did increase, so in this case the pressure decrease had a greater effect than the temperature decrease This seems reasonable, looking at the numbers There is a relatively small change in the Kelvin temperature (293 K versus 263 K) compared to a much larger change in the pressure (7600 torr versus 4500 torr)
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