What is the entropy change when \(3.6 \mathrm{~g}\) of liquid water is completely converted into vapours at \(373 \mathrm{~K} ?\) The molar heat of vaporization is \(40.85 \mathrm{~kJ} / \mathrm{mol}\). (a) \(218.9 \mathrm{~J} / \mathrm{K}\) (b) \(2.189 \mathrm{~J} / \mathrm{K}\) (c) \(21.89 \mathrm{~J} / \mathrm{K}\) (d) \(0.2189 \mathrm{~J} / \mathrm{K}\)

Short Answer

Expert verified
The entropy change is \(218.9 \mathrm{~J} / \mathrm{K}\).

Step by step solution

01

Calculate the number of moles of water

Use the formula for the number of moles, which is mass (m) divided by molar mass (M). The molar mass of water (H2O) is approximately 18.015 g/mol. Hence, the number of moles of water (n) is given by: \( n = \frac{m}{M} = \frac{3.6 \mathrm{~g}}{18.015 \mathrm{~g/mol}} \).
02

Select the correct answer

From the calculated entropy change, select the answer that matches the calculated value.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Thermodynamics
Thermodynamics is a branch of physics concerned with heat and temperature and their relation to energy and work. It defines macroscopic variables, such as internal energy, entropy, and pressure, that partly describe a body of matter or radiation. It is concerned with the principles that govern the conversion of energy from one form to another and the direction in which processes can occur.

One of the key concepts in thermodynamics is the First Law, which states that energy cannot be created or destroyed, only transformed. Another crucial concept is the Second Law of thermodynamics, which introduces the notion of entropy. It states that the total entropy of an isolated system can never decrease over time, and is often interpreted as the degree of disorder or randomness in the system. This law predicts that certain processes, such as the spontaneous conversion of heat into work, are impossible, and it underpins the principle behind calculations of entropy change in physical chemistry.
Entropy in Physical Chemistry
In physical chemistry, entropy is a measure of the randomness or disorder of particles in a thermodynamic system. It is a central concept in determining the feasibility and spontaneity of chemical reactions and physical processes. Entropy can be represented by the symbol 'S' and has units of joules per kelvin (J/K).

When a substance undergoes a phase transition, such as from liquid to vapor, there is a change in entropy. The calculation of this entropy change (\( \triangle S \)) in a process where the temperature (T) is constant, often involves using the molar heat of vaporization (\( \triangle H_{vap} \)) and is given by the formula: \[ \triangle S = \frac{\triangle H_{vap}}{T} \].

The entropy change can be further understood by considering molecular order. As a substance vaporizes, the molecules spread out and move more freely, increasing the system’s disorder and, consequently, its entropy. This underlines why the entropy of vapor is greater than that of the corresponding liquid. Understanding entropy change calculations is essential for predicting whether a process will occur spontaneously and for evaluating the energy requirements of industrial processes.
Molar Heat of Vaporization
The molar heat of vaporization (\( \triangle H_{vap} \)) is the amount of energy required to convert one mole of a liquid into its vapor phase at a constant temperature and pressure. The units for molar heat of vaporization are typically joules per mole (J/mol) or kilojoules per mole (kJ/mol). It is an essential thermodynamic quantity that contributes to the understanding of phase transitions.

Knowing the molar heat of vaporization enables the calculation of entropy changes during the vaporization process. For example, in the original exercise, the provided molar heat of vaporization (40.85 kJ/mol) indicates how much energy is required to vaporize each mole of liquid water at the boiling point.

The molar heat of vaporization not only reflects the strength of intermolecular forces within the liquid but also can indicate the increase in entropy when the liquid turns into vapor. Substances with higher molar heats of vaporization require more energy to vaporize because their particles are typically more strongly attracted to one another in the liquid state. In contrast, when such a liquid vaporizes, the increase in disorder (entropy) is also more significant.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

With what minimum pressure must a given volume of an ideal gas \((\gamma=1.4)\), originally at \(400 \mathrm{~K}\) and \(100 \mathrm{kPa}\), be adiabatically compressed in order to raise its temperature up to \(700 \mathrm{~K}\) ? (a) \(708.9 \mathrm{kPa}\) (b) \(362.5 \mathrm{kPa}\) (c) \(1450 \mathrm{kPa}\) (d) \(437.4 \mathrm{kPa}\)

If one mole of a monoatomic gas \((\gamma=5 / 3)\) is mixed with one mole of a diatomic gas \((\gamma=7 / 5)\), the value of \(\gamma\) for the mixture is (a) 1 (b) \(1.5\) (c) 2 (d) \(3.0\)

An ideal gas \((\gamma=1.5)\) undergoes a change in state such that the magnitude of heat absorbed by the gas is equal to the magnitude of work done by the gas. The molar heat capacity of the gas in this process is (a) \(2 R\) (b) \(R\) (c) \(3 R\) (d) \(1.5 R\)

One mole of a certain ideal gas is contained under a weightless piston of a vertical cylinder at a temperature \(T\). The space over the piston opens into the atmosphere. What work has to be performed in order to increase isothermally the gas volume under the piston \(n\) times by slowly raising the piston? The friction of the piston against the cylinder walls is negligibly small. (a) \(R T(n-1-\ln n)\) (b) \(R T(1-n+\ln n)\) (c) \(R T \ln n\) (d) \(-R T \ln n\)

For which of the following process, \(\Delta S\) is negative? (a) \(\mathrm{H}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{H}(\mathrm{g})\) (b) \(\mathrm{N}_{2}(\mathrm{~g}, 1 \mathrm{~atm}) \rightarrow \mathrm{N}_{2}(\mathrm{~g}, 8 \mathrm{~atm})\) (c) \(2 \mathrm{SO}_{3}(\mathrm{~g}) \rightarrow 2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g})\) (d) \(\mathrm{C}\) (graphite) \(\rightarrow \mathrm{C}\) (diamond)

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free