Question

Liquid nitrogen is an excellent bath for keeping temperatures around \(77^{\circ} \mathrm{K}\), its normal boiling point. What pressure would you need to maintain over the liquid nitrogen if you wanted to set the bath temperature at \(85^{\circ} \mathrm{K} ?\) Heat of vaporization is about \(5560(\mathrm{~J} / \mathrm{mole})\).

Short answer

To maintain a liquid nitrogen bath at \(85^{\circ} \mathrm{K}\), a pressure of approximately \(2.07 \times 10^5 \, \mathrm{Pa}\) is required.

Step-by-step solution

Identify the given information

We are given: - Normal boiling point, \(T_0 = 77 \, \mathrm{K}\), - Heat of vaporization, \(\Delta H_v = 5560 \, \frac{\mathrm{J}}{\mathrm{mol}}\), - Desired temperature, \(T = 85 \, \mathrm{K}\), - Ideal gas constant, \(R = 8.314 \, \frac{\mathrm{J}}{\mathrm{mol} \cdot \mathrm{K}}\). We need to find the pressure \(P\) at the desired temperature.

Find vapor pressure at the normal boiling point

At the normal boiling point, the vapor pressure is 1 atm (101325 Pa). Hence, the known vapor pressure \(P_0 = 1 \, \mathrm{atm}\), or \(1 \times 10^5 \, \mathrm{Pa}\), since 1 atm is approximately equal to \(10^5 \, \mathrm{Pa}\).

Substitute the given values into the Clausius-Clapeyron Equation

Now we can substitute the given values into the Clausius-Clapeyron Equation: \[P = (1 \times 10^5 \, \mathrm{Pa}) \times e^{(-\dfrac{5560 \, \frac{\mathrm{J}}{\mathrm{mol}}}{8.314 \, \frac{\mathrm{J}}{\mathrm{mol} \cdot \mathrm{K}}})(\dfrac{1}{85 \, \mathrm{K}} - \dfrac{1}{77 \, \mathrm{K}})}\]

Calculate the pressure

Simplify the equation and solve for the pressure \(P\): \[P = (1 \times 10^5 \, \mathrm{Pa}) \times e^{(-\dfrac{5560}{8.314})(\dfrac{1}{85} - \dfrac{1}{77})}\] \[P \approx 2.07 \times 10^5 \, \mathrm{Pa}\] So, the pressure required to maintain the liquid nitrogen bath at 85 K is approximately \(2.07 \times 10^5 \, \mathrm{Pa}\).

Key concepts

Vapor Pressure

Understanding vapor pressure is fundamental when studying phase transitions, such as boiling. It represents the pressure exerted by a vapor in thermodynamic equilibrium with its condensed phases (solid or liquid) at a given temperature in a closed system. The vapor pressure increases as the temperature rises because more molecules have enough kinetic energy to escape from the liquid surface into the vapor phase.

Consider a liquid in a closed container. A dynamic equilibrium is established between the molecules leaving the liquid to become gas and the gas molecules returning to the liquid state. The pressure exerted by the gas phase on the liquid is what we refer to as the vapor pressure. If we increase the temperature, more molecules can escape the liquid, which increases the vapor pressure. This concept is crucial for explaining why we would need to apply more pressure over liquid nitrogen to raise its boiling point from 77 K to 85 K.

Heat of Vaporization

The heat of vaporization, \(\Delta H_v\), is the energy required to transform a quantity of a substance from a liquid to a gas at a constant pressure and temperature. It's essentially the amount of heat energy needed to break the intermolecular forces that keep the liquid molecules together without changing the temperature. During the phase change, the substance absorbs heat energy, which is then utilized to break these bonds.

High heat of vaporization signifies strong intermolecular forces within the liquid, requiring more energy to undergo phase transition. When we talk about liquid nitrogen, we say it has a heat of vaporization of 5560 J/mol, which is relatively high. This indicates the amount of energy needed to turn one mole of liquid nitrogen into gaseous nitrogen at the substance's normal boiling point.

Boiling Point

The boiling point of a substance is the temperature at which its vapor pressure is equal to the external pressure exerted on the liquid. When the ambient pressure changes, so does the boiling point. For instance, water boils at 100°C at sea level, where the atmospheric pressure is 1 atm. However, on top of a mountain, where the pressure is lower, water will boil at a lower temperature.

When examining boiling points, it's essential to understand that a liquid can boil when its vapor pressure equals the external pressure, not necessarily when it reaches a specific 'boiling point' temperature. Liquid nitrogen, which normally boils at 77 K, can have its boiling point altered by varying the pressure, as explored with the Clausius-Clapeyron Equation. When we increase the pressure, we raise the temperature at which nitrogen can boil, thus adjusting the system to reach the desired equilibrium at a higher temperature, such as 85 K.