| 2E-1 : | Equilibrium Pressure When Two Gases Are Mixed | 5 pts | |||||||
| Two storage tanks are connected and their contents mixed. Tank A initially contains 5 kg of nitrogen gas at 150 oC and 150 KPa. Tank B initially contains 2 kg of oxygen gas at 100oC and 75 KPa. Can either gas be treated as an ideal gas? If both tanks eventually cool to room temperature (20oC) after mixing, what will the final pressure be at equilibrium? | ![]() |
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| Read : | The key here is to assume that both the pure gases and the final mixture of gases behave as ideal gases. We can immediately verify that the pure gases in their initial state are ideal gases, but we cannot verify that the final mixture of gases is ideal until we solve the problem and determine the final pressure. | ||||||||
| Choose the contents of both tanks as the system. The fact that links the initial and final states of this system is that the total number of moles in the system does not change. This is a closed system. | |||||||||
| Given: | mN2 = | 5 |
kg | Find: | |||||
| TA = | 150 |
oC = | 423.15 | K | Pequ = | ??? |
kPa | ||
| PA = | 150 |
kPa | |||||||
| mO2 = | 2 |
kg | |||||||
| TB = | 100 |
oC = | 373.15 | K | |||||
| PB = | 75 |
kPa | |||||||
| Tequ = | 20 |
oC = | 293.15 | K | |||||
| Assumptions : | |||||||||
| - Both pure gases, as well as the final mixture, behave as ideal gases. | |||||||||
| Solution : | A diatomic gas can be considered ideal when the following criterion is satisfied: | Eqn 1 | |||||||
| VN2 = | 23.45 |
L/mol | Where: | R = | 8.314 |
J/mol K | |||
| VO2 = | 41.36 |
L/mol | |||||||
| Since both molar volumes are much greater than 5 L/mole, it is safe to consider both gases to be ideal gases. | |||||||||
| The key to solving the problem is to ASSUME that the equilibrium mixture will be an ideal gas: | ![]() |
Eqn 2 | |||||||
| Let's begin by determining how many moles of gas are initially in each tank. | Eqn 3 | ||||||||
| Then we can determine the total moles of gas in the system : | Eqn 4 | ||||||||
| MWN2 = | 28.01 |
g/mol | nN2 = | 178.5 |
mole N2 | ||||
| MWO2 = | 32.00 |
g/mol | nO2 = | 62.5 |
mole O2 | ||||
| ntotal = | 241.0 |
mole total | |||||||
| The total number of moles in the system does not changes as the gases mix ! | |||||||||
| The system, consisting of both tanks, is closed. | |||||||||
| Next, we can use the IG EOS to determine the volume of each tank and then the total volume of the system. | |||||||||
| Eqn 5 | Where (for ideal gases) : |
Eqn 6 | |||||||
| and : | |||||||||
| VA = | 4.19 |
m3 | Eqn 7 | ||||||
| VB = | 2.59 |
m3 | |||||||
| Vtot = | 6.77 |
m3 | |||||||
| From the IG EOS we derive the following equation for the equilibrium pressure : | ![]() |
Eqn 8 | |||||||
| Answer : | Peq = | 86.7 |
kPa | ||||||
| Verify : | Now, calculate the molar volume at the equilibrium state, just to be sure it is still safe to treat the gas as an ideal gas ! | ||||||||
| Veq = | 28.10 |
L/mol | |||||||
| Since Veq is greater than 5 L/mole, we were justified in using the ideal gas EOS for the equilibrium state as well. | |||||||||


