1. The apparatus shown helow consists of three 1.00T bulhs connected by stopcock
ID: 104102 • Letter: 1
Question
1. The apparatus shown helow consists of three 1.00T bulhs connected by stopcocks and held at the temperatures shown. Bulb A contains a mixture of HO-CO), and N, at 25 and a total pressure of 564 mmHg. Bulhs B and C are empty. CO2 sublimes a -78°C and NZ boils at-196. a. The stopcock hetween A and B is opened. The pressure in A and B is now 219 mmHa. What do bulbs A and B contain? (5 points) b. How manymeof H.O are in the systemS points) c. Both stopcocks are opened. The pressure throughout the system is 33.5mmHg What do hulbs A, B and C contain? 10 points) d. How many moles of N2 are in the system? (10 points) e. How many moles of CO are in the system (10 points) 2. The heal combuslion ol Butane As equal to the heat gencrated by the cooling of 17.2 g of water from 100C" to 20C (the specific heat for water is s 4.186 Jig.C) Calculate the mass of Butane that needs to be combusted to obtain 350 kJ of heat. (30 points) 3. The electron in a hydrogen atom falls from an excited energy level tote ground state in two steps, causing the emission of photons with wavelengths of 2624 nm and 97.2 nnm. What are the quantum numhers of the inal excited and the inteTmediate energy levels? (30 points)Explanation / Answer
After you open the stopcock you will have in bulb A CO2 and N2 , the water vapor will condense into liquid and then into solid water so in bulb B you will have
CO2, N2 and H2O (solid)
Part B
So lets calculate the number of gas moles that exists originally
at the beginning you have 564 mm Hg, change this to atm by dividing by 760
564 / 760 = 0.7421atm , you have 1 liter volume, apply ideal gas equation
Pv = nRT, p is pressure, v is volume, R is gas constant , T is temperature
n = PV / RT = 0.742 * 1 / (0.082*298) = 0.0303 moles of total gas
Now calculate the moles of gas after the solidification of water
As we said in part A, we have to calculate the number of moles in each bulb according to the temperature for each one
pressure of 219 mmHg is 0.288 atm for bulb A T = 298 Kelvin
n = PV / RT = 0.288 * 1 / (0.082*298) = 0.0117 moles of gas in A
For bulb B, T = -70 = -70 + 273 = 203 K
n = PV / RT = 0.288 * 1 / (0.082*203) = 0.0173 moles
moles after the expansion = 0.0117 + 0.0173 = 0.029 moles
moles of water = gas moles before expansion - gas moles after expansion = 0.0303 - 0.029 = 0.0013 moles of water
Part C. according to the statement at -190C the CO2 becomes solid but it is not enough to bring N2 to liquid phase so
Bulb A = N2 (g)
Bulb B = N2 (g) and H2O (s)
Bulb C = N2 (g) and CO2 (s)
Part D , calculate moles of N2 , we just need to make calculations for the three bulbs, at this point only N2 is in gas phase
33.5 mmHg is 0.044 atm
Bulb A : n = PV / RT = 0.044 * 1 / (0.082*298) = 0.0018 moles in A
Bulb B : n = PV / RT = 0.044 * 1 / (0.082*203) = 0.00264 moles in B
Bulb C: -190 + 273 = 83 K ; n = PV / RT = 0.044 * 1 / (0.082*83) = 0.00646 moles in C
total moles of N2 = 0.0018 + 0.00264 + 0.00646 = 0.0109 moles of N2
Moles of CO2 is just the difference from the gas moles in part B and part D
0.029 - 0.0109 = 0.0181 moles of CO2
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