3. Liquefied air (assuming it is only O2 and N2) is kept in a container where pressure and temperature can be controlled.
a. What is the maximum temperature at which the mixture exists as a liquid? If this temperature is kept constant, what is the minimum pressure to keep the air in the liquid phase?
b. Because of a sudden drop in pressure to 2 bar, all the liquid evaporates. Calculate the volume of one mole of this vapor at -100oC.
To answer these questions, we need to understand the behavior of gases under different conditions using principles from thermodynamics.
a. To determine the maximum temperature at which the mixture exists as a liquid, we need to reference a phase diagram specific to the mixture of oxygen (O2) and nitrogen (N2). The phase diagram shows the boundary between the liquid and gas phases for different temperatures and pressures.
Now, if we assume that the mixture contains only O2 and N2, we can refer to the phase diagram of pure nitrogen to estimate the behavior of the mixture. Nitrogen (N2) has a boiling point of -195.8°C at atmospheric pressure (1 bar), and it boils at lower temperatures with reduced pressure.
To keep the mixture in the liquid phase, we need to maintain a temperature below the boiling point for the given pressure. However, without the exact phase diagram for this specific mixture, we can make an estimation based on average boiling points.
Oxygen (O2) has a higher boiling point than nitrogen (N2) at atmospheric pressure, which is -183°C. Considering that the mixture contains both O2 and N2, we can reasonably assume the maximum temperature for the liquid phase to be around -190 to -180°C.
To find the minimum pressure to keep the air in the liquid phase at a constant temperature, we would again need to refer to the phase diagram of the specific mixture of O2 and N2. The phase diagram illustrates the pressure and temperature at which the transition from liquid to gas occurs. Unfortunately, without specific data, it is challenging to provide an accurate answer.
b. Now let's calculate the volume of one mole of this vapor at -100°C after the sudden drop in pressure to 2 bar. To determine the volume, we need to use the ideal gas law equation: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.
First, convert the given temperature from Celsius to Kelvin:
T = -100°C + 273.15 = 173.15 K
Now, rearrange the ideal gas law equation to solve for volume (V):
V = (nRT) / P
Assuming we have one mole of the vapor, n = 1. The ideal gas constant, R, is approximately 0.0821 L·atm/(K·mol). However, we need to convert the pressure from bar (b) to atmospheric pressure (atm) since the ideal gas constant's unit is in atm:
1 atm = 1.01325 bar
So, the pressure in units of atm is 2 bar / 1.01325 = 1.97 atm.
Substitute the values into the equation:
V = (1 mole * 0.0821 L·atm/(K·mol) * 173.15 K) / 1.97 atm
Now, calculate the volume:
V = 0.9076 L
Therefore, the volume of one mole of vapor at -100°C and 2 bar pressure is approximately 0.9076 liters.