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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.



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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).

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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.

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