Gas State Question: Composition of Two Temperature Jacketed 1L Bulbs

[AlchemistK]

Homework Statement

Two temperature jacketed 1L bulbs are connected by stopcocks. Bulb A has a mix of H2O (g), CO2 (g) and N2 (g) at 25°C and at a pressure of 564 mm Hg. The second bulb is empty and at a temperature of -70°C. Once the stopcocks are opened what would be the composition in the two bulbs? ( CO2 sublimes at -78°C and N2 boils at -196°C)

Homework Equations

The Attempt at a Solution

As soon, as the stopcocks are opened and then later the apparatus is allowed to achieve equilibrium(given pressure will be 219 mm Hg), a part of the gasses in the first bulb would transfer into the second and since the temperature of the second bulb is -70°C, H2O would freeze into a solid. So bulb A would contain N2(g) and CO2(g) while bulb B would contain N2 (g), CO2(g) and H2O(s).

But the book has the answer as :A contains CO2 (g),N2(g) and H2O(s) and B contains CO2(g) and N2 (g).
Why is that true? Also, how can we calculate the number of H2O moles in the system?

 

[anathema]

I would like to clarify a few things about this problem. First, it is important to note that the temperature and pressure given for bulb A are not the same as the temperature and pressure at equilibrium. Therefore, we cannot assume that the pressure in bulb A will remain at 564 mm Hg when the stopcocks are opened.

Assuming that the apparatus is allowed to reach equilibrium at a pressure of 219 mm Hg, we can use the ideal gas law to calculate the number of moles of each gas in the system. The ideal gas law is given by PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin.

First, we need to calculate the volume of each bulb. Since both bulbs are 1L, the volume will remain the same. However, the temperature of bulb A will change to -70°C (203 K) and the pressure will change to 219 mm Hg. Therefore, the number of moles of each gas in bulb A can be calculated as follows:

n = (PV)/(RT) = (219 mm Hg * 1 L)/(0.08206 L*atm/mol*K * 203 K) = 0.0129 moles

Since CO2 sublimes at -78°C (195 K), we can assume that all of the CO2 in bulb A will transfer to bulb B. Therefore, bulb B will contain 0.0129 moles of CO2 and 0.0129 moles of N2.

In bulb A, we will have 0.0129 moles of N2 and 0.0129 moles of H2O. However, since the temperature is now -70°C, the H2O will freeze into a solid. Therefore, bulb A will contain 0.0129 moles of N2 and 0 moles of H2O.

In conclusion, the composition in the two bulbs at equilibrium will be as follows:

Bulb A: CO2 (g), N2 (g), and H2O (s)
Bulb B: CO2 (g) and N2 (g)

To calculate the number of moles of H2O in the system, we can use the ideal gas law again. Since the temperature and pressure are the same in both