Handout-09-KEY

Handout-09-KEY - Chemistry 111 KEY Fall, 2005 Prof. Simpson...

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Chemistry 111 KEY Fall, 2005 Prof. Simpson 1 Chemistry 111 Handout #9 – KEY Gas Laws, Kinetics, Thermodynamics & Thermochemistry Gas Laws. G1 . Oxygen in respiration therapy is stored in 43.0 L gas cylinders. At 20.0 o C, the pressure inside the tank is 1.50 x 10 2 atm. (a) If all of the gas was let out of the tank, how much volume would it occupy at 1.00 atm pressure and 20 o C? This corresponds to a change in pressure and volume at a constant temperature and number of moles, so the appropriate equation to use is P 1 V 1 = P 2 V 2 V 2 = (P 1 V 1 )/P 2 = (150 atm x 43.0 L) / (1.0 atm) = 6,450 L !! (b) What would the pressure inside the tank be if the temperature were doubled to 40.0 o C? Here, we are changing the pressure and temperature while keeping the volume and number of moles constant. Make sure you change to Kelvin, though! Although the temperature is doubled on the Celcius scale, it is not doubled on the Kelvin scale ! P 1 / T 1 = P 2 / T 2 P 2 = (P 1 T 2 ) / T 1 = (150 atm x 313 K) / (293 K) = 160.2 atm (c) A patient uses 30% of the oxygen in the tank. What is the new pressure (in atm)? Assume the temperature is 20.0 o C. Ok, so now we're changing the number of moles and the pressure while keeping the volume and temperature constant: P 1 / n 1 = P 2 / n 2 P 2 = P 1 x (n 2 / n 1 ) = 150 atm x (0.70) = 105 atm Notice that we didn't actually have to calculate the number of moles. We know that if the patient used 30% of the amount of the oxygen, there must be 70% left behind, and this corresponds to an n 2 / n 1 ratio of 0.700. (d) Sketch the Boltzmann distribution of speeds for the oxygen molecules in the tank for two different temperatures T 1 and T 2 . Make sure you clearly label which curve goes with which temperature, and which temperature is higher. This is a pretty straightforward sketching exercise … we saw this several times in class. # O T > w i t h g v e n s p d speed
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Chemistry 111 KEY Fall, 2005 Prof. Simpson 2 (e) What is the average kinetic energy per mole of the oxygen atoms in the tank originally (43.0 L, 20.0 o C, 1.50 x 10 2 atm)? The average kinetic energy per mole for a gas behaving ideally is given by the expression KE avg = 3/2 RT = (3/2) (8.314 J K -1 mol -1 ) (293 K) = 3.65 kJ mol -1 Unless you're told otherwise, you can assume the gases on this test behave ideally. (f) Does the average kinetic energy per mole increase with the process described in part (a)? What about in part (b)? What about in part (c)? Explain. In part (a), we changed the volume without changing the temperature or the number of moles. The pressure decreases, but not because the KE avg decreases. KE doesn't change. The pressure decreases because the surface area of the container increases, and therefore the same number of particles encounter the walls of the container less frequently, but with the same average force. In part (b), we changed the temperature. Clearly, from the expression in part (e), we can see that the KE will increase as the temperature increases. In part (c), we changed the number of moles of gas in the container without changing the temperature. While the total kinetic energy in the container will decrease in this process, the per mole of gas will not change.
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Handout-09-KEY - Chemistry 111 KEY Fall, 2005 Prof. Simpson...

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