Problem 9.112 The apparatus shown consists of three temperature-jacketed 1.460 L

Question

Problem 9.112

The apparatus shown consists of three temperature-jacketed 1.460 L bulbs connected by stopcocks. Bulb A contains a mixture of H2O(g), CO2(g), and N2(g) at 25C and a total pressure of 570 mm Hg . Bulb B is empty and is held at a temperature of -70C. Bulb C is also empty and is held at a temperature of -190C. The stopcocks are closed, and the volume of the lines connecting the bulbs is zero. CO2 sublimes at -78C, and N2 boils at -196C.

Part A

The stopcock between A and B is opened, and the system is allowed to come to equilibrium. The pressure in A and B is now 206 mm Hg . What do bulbs A and B contain?

The stopcock between and is opened, and the system is allowed to come to equilibrium. The pressure in and is now 206 . What do bulbs and contain? (multiple choice) (choose one)

4. Bulb A contains N2(g), and bulb B contains CO2(g) and H2O(s).

Part B

How many moles of H2O are in the system?

Express your answer using two significant figures.

Part C

Both stopcocks are opened, and the system is again allowed to come to equilibrium. The pressure throughout the system is 32.0 mm Hg . What do bulbs A, B, and C contain?

Both stopcocks are opened, and the system is again allowed to come to equilibrium. The pressure throughout the system is 32.0 . What do bulbs , , and contain? (multiple choice) (choose one)

4. Bulb A contains N2(g), bulb B contains N2(g) and H2O(s) and bulb C containsN2(g) and CO2(s).

Part D

How many moles of N2 are in the system?

Part E

How many moles of CO2 are in the system?

1. Bulb A contains CO2(g), and bulb B contains N2(g), CO2(g) and H2O(s). 2. Bulb A contains N2 (g), CO2(g) and H2O(g) and bulb B contains N2(g), CO2(g) and H2O(s). 3. Bulb A containsN2(g) and CO2(g), and bulb B contains .N2(g), CO2(g) and H2O(s).

4. Bulb A contains N2(g), and bulb B contains CO2(g) and H2O(s).

Part B

How many moles of H2O are in the system?

Express your answer using two significant figures.

Part C

Both stopcocks are opened, and the system is again allowed to come to equilibrium. The pressure throughout the system is 32.0 mm Hg . What do bulbs A, B, and C contain?

Both stopcocks are opened, and the system is again allowed to come to equilibrium. The pressure throughout the system is 32.0 . What do bulbs , , and contain? (multiple choice) (choose one)

1. Bulb A contains N2(g) and CO2(g), bulb B contains H2O(s) and bulb C contains CO2(s). 2. Bulb A contains H2O(g), N2(g), bulb B contains H2O(s)and bulb C contains N2(g) and CO2(s). 3. Bulb A contains N2(g), bulb B contains H2O(s) and bulb C contains CO2(s).

4. Bulb A contains N2(g), bulb B contains N2(g) and H2O(s) and bulb C containsN2(g) and CO2(s).

Part D

How many moles of N2 are in the system?

Part E

How many moles of CO2 are in the system?

25 OC 70 OC 190 oC

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)

First answer is A

Part B

So lets calculate the number of gas moles that exists originally

at the beginning you have 570 mm Hg, change this to atm by dividing by 760

570 / 760 = 0.75 atm , you have 1.46 liter volume, apply ideal gas equation

Pv = nRT, p is pressure, v is volume, R is gas constant , T is temperature

n = PV / RT = 0.75 * 1.46 / (0.082*298) = 0.04481 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 206 mmHg is 0.271 atm for bulb A T = 298 Kelvin

n = PV / RT = 0.271 * 1.46 / (0.082*298) = 0.01619 moles of gas in A

For bulb B, T = -70 = -70 + 273 = 203 K

n = PV / RT = 0.271 * 1.46 / (0.082*203) = 0.0237 moles

moles after the expansion = 0.01619 + 0.0237 = 0.0399 moles

moles of water = gas moles before expansion - gas moles after expansion =0.04481 - 0.0399 = 0.00491 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

32 mmHg is 0.0421 atm

Bulb A : n = PV / RT = 0.0421 * 1.46 / (0.082*298) = 0.0025 moles in A

Bulb B : n = PV / RT = 0.0421 * 1.46 / (0.082*203) = 0.0037 moles in B

Bulb C: -190 + 273 = 83 K ; n = PV / RT = 0.0421 * 1.46 / (0.082*83) = 0.009 moles in C

total moles of N2 = 0.0025 + 0.0037 + 0.009 = 0.0152 moles of N2

Moles of CO2 is just the difference from the gas moles in part B and part D

0.0399 - 0.0152 = 0.0247 moles of CO2

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