Question

If liquid nitrogen is boiled slowly-that is, reversiblyto transform it into nitrogen gas at a pressure \(P=100.0 \mathrm{kPa}\), its entropy increases by \(\Delta S=72.1 \mathrm{~J} /(\mathrm{mol} \mathrm{K}) .\) The latent heat of vaporization of nitrogen at its boiling temperature at this pressure is \(L_{\text {vap }}=5.568 \mathrm{~kJ} / \mathrm{mol}\). Using these data, calculate the boiling temperature of nitrogen at this pressure.

Short answer

Answer: The boiling temperature of nitrogen at the given pressure is approximately 77.2 K.

Step-by-step solution

Rewrite the formula for Temperature

Given the formula \(\Delta S = \frac{L_{\text {vap}}}{T}\), we can rewrite the formula to find \(T\) by multiplying both sides with \(T\) and dividing by \(\Delta S\): \(T = \frac{L_{\text {vap}}}{\Delta S}\)

Substitute the values

Now, we can substitute the given values into the formula: \(T = \frac{5.568 \mathrm{~kJ} / \mathrm{mol}}{72.1 \mathrm{~J} /(\mathrm{mol} \mathrm{K})}\) First, we should convert the latent heat of vaporization from kJ to J. Since 1 kJ = 1000 J: \(L_{\text {vap}} = 5.568 \mathrm{~kJ} \times \frac{1000 \mathrm{~J}}{1 \mathrm{~kJ}} = 5568 \mathrm{~J} / \mathrm{mol}\) Now, substitute the converted value: \(T = \frac{5568 \mathrm{~J} / \mathrm{mol}}{72.1 \mathrm{~J} /(\mathrm{mol} \mathrm{K})}\)

Calculate the boiling temperature

We can calculate the boiling temperature by dividing \(L_{\text {vap}}\) by \(\Delta S\): \(T = \frac{5568 \mathrm{~J} / \mathrm{mol}}{72.1 \mathrm{~J} /(\mathrm{mol} \mathrm{K})} = \frac{5568}{72.1} \mathrm{K} \approx 77.2 \mathrm{K}\) The boiling temperature of nitrogen at the given pressure is approximately 77.2 K.

Key concepts

Entropy

Entropy is a key concept in thermodynamics, often associated with the level of disorder or randomness in a system. It has a direct connection to the energy transformations that occur during various processes, like the phase transition from liquid to gas. When a liquid is transformed into a gas, the entropy increases because molecules in the gaseous state are more spread out and have more freedom to move. In the context of the exercise, nitrogen gas being formed from liquid nitrogen involves an increase in entropy, quantified as \(\Delta S = 72.1 \, \mathrm{J}/(\mathrm{mol} \, \mathrm{K})\). Understanding entropy helps us predict how energy will be distributed in a system. If you think about energy as being more usable when organized, then higher entropy means less usability in that sense. But it's necessary for processes like vaporization, where liquids gain enough energy for molecules to break free as gas. This kind of insight is crucial in fields across chemistry, physics, and engineering.

Latent heat of vaporization

Latent heat of vaporization is the amount of heat required to convert a unit mass of a liquid into gas at constant temperature and pressure. For nitrogen, this value is significant because it tells us how much energy is needed to overcome the forces keeping the nitrogen molecules in liquid form. In the problem, the latent heat of vaporization \(L_{\text{vap}}\) is given as \(5.568 \, \mathrm{kJ}/\mathrm{mol}\). To fully understand the process of vaporization, consider that this heat does not raise the temperature of the substance. Instead, it goes into changing the phase of the substance, from liquid to gas. It's a crucial element in calculating another fundamental property, like the boiling temperature, using the relation \(T = \frac{L_{\text{vap}}}{\Delta S}\). When considering practical applications, the latent heat of vaporization is vital in designing systems like refrigeration, where phase changes are a core part of heat transfer mechanisms.

Boiling temperature

The boiling temperature is the temperature at which a substance changes from a liquid to a gaseous state. At this point, the vapor pressure of the liquid equals the atmospheric pressure. In the given problem, calculating the boiling temperature of nitrogen required understanding the relationships between latent heat and entropy changes. Using the formula \(T = \frac{L_{\text{vap}}}{\Delta S}\), we deduced that the boiling temperature of nitrogen is approximately 77.2 K at 100.0 kPa pressure. Boiling temperature is pivotal in many scientific and industrial processes. Knowing when a substance will boil allows scientists and engineers to design equipment that can safely handle phase transitions. This applies not only to studying gases but also to liquids under different pressures and temperatures, informing practices in fields like chemical engineering and materials science.

Nitrogen gas

Nitrogen gas is a diatomic molecule, forming approximately 78% of the Earth's atmosphere, and has various industrial and scientific uses. In its gaseous state, nitrogen is chemically inert under most conditions due to its strong triple bond. This makes it valuable as a protective atmosphere in food packaging, electronics manufacturing, and other sensitive processes. The transition from liquid nitrogen, which is often used for cooling and freezing, to nitrogen gas involves the concepts we've discussed of latent heat and entropy. Understanding the properties of nitrogen gas, including its boiling temperature, helps scientists and engineers use this substance effectively. For example, in cryogenics, knowing the exact boiling point ensures optimal performance in applications where maintaining very low temperatures is necessary.