When one mole of an ideal gas is compressed to half of its initial volume and simultaneously heated to twice its initial temperature, the change in entropy of gas \((\Delta S)\) is : (a) \(C_{p, m} \ln 2\) (b) \(C_{v, m} \ln 2\) (c) \(R \ln 2\) (d) \(\left(C_{v, m}-R\right) \ln 2\)

Short Answer

Expert verified
\(\left(R - C_{v, m}\right) \ln 2\)

Step by step solution

01

Understand the process

The exercise describes a thermodynamic process in which one mole of an ideal gas is compressed to half its initial volume while its temperature is doubled. This implies an isentropic process and uses the ideal gas law, combined with the thermodynamic relation for entropy change.
02

Calculate entropy change

To find the change in entropy \(\Delta S\), use the formula for a reversible process: \[\Delta S = nC_{v, m} \ln\frac{V_2}{V_1} + nR\ln\frac{T_2}{T_1}\] Since the volume is halved, \(V_2/V_1 = 1/2\), and the temperature is doubled, \(T_2/T_1 = 2\), the formula simplifies to: \[\Delta S = nC_{v, m} \ln(1/2) + nR\ln(2)\]
03

Simplify the entropy change formula

Using the properties of logarithms, we can simplify the formula as: \[\Delta S = nC_{v, m} (\ln 1 - \ln 2) + nR(\ln 2 - \ln 1)\] Since \ln 1 = 0, the formula further simplifies to: \[\Delta S = -nC_{v, m} \ln 2 + nR \ln 2\] Given that \(n = 1\) mole, we combine terms to get the final result: \[\Delta S = (R - C_{v, m}) \ln 2\]

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.

Thermodynamic Processes
Thermodynamic processes describe the dynamics of systems in which energy in the form of heat or work is transferred to or from the surroundings. These include processes like heating, cooling, compressing, or expanding a substance.

Understanding thermodynamic processes is crucial for calculating changes in properties such as temperature, pressure, volume, and entropy. Entropy, a central concept in thermodynamics, quantifies the amount of disorder within a system. It's associated with the number of microscopic configurations that correspond to a thermodynamic system's macroscopic state.

An important aspect of thermodynamics is the laws that govern these processes. The first law relates to the conservation of energy, indicating that energy cannot be created or destroyed. The second law introduces the concept of entropy and suggests that for any spontaneous process, the total entropy of a system and its environment always increases.

Understanding Reversible and Irreversible Processes

Reversible processes are idealized scenarios where the system remains in equilibrium as it changes states, meaning the process can be reversed without any entropy change in the universe. In contrast, an irreversible process is one that increases the total entropy of the universe.

The exercise example highlights a thermodynamic process involving entropy change and requires applying these principles to reach a solution.
Ideal Gas Law
The Ideal Gas Law is a fundamental equation that relates the pressure (P), volume (V), temperature (T), and amount of substance (n) in moles of an ideal gas. Expressed mathematically as:
\[ PV = nRT \]
The constant R is known as the ideal or universal gas constant, and the law serves as a good approximation for the behavior of real gases under many conditions, although it assumes that gas particles have no volume and do not interact.

Understanding the ideal gas law is critical when studying thermodynamic processes since it allows us to predict how a gas will respond to changes in conditions such as pressure, volume, or temperature. For instance, in the given exercise, knowing that the volume is compressed to half and the temperature is doubled provides essential information for calculating the change in entropy of the gas using this law.
Isentropic Process
An isentropic process is a thermodynamic process that occurs at a constant entropy, which means there is no change in the entropy of the system as the process unfolds. Isentropic processes are reversible adiabatic processes; they involve no heat transfer with the surroundings (hence adiabatic), and the total entropy remains unchanged (hence isentropic).

In an idealized isentropic process, entropy is conserved due to the absence of any external entropy exchange, and it's manifested in perfect energy conservation within the system. This implies that all changes are internally reversible and there are no frictional losses or non-conservative forces at play.
  • Significance in Real-World Applications: Processes that approximate isentropic behavior are important in engineering, for example in the functioning of steam turbines or compressors, where they are used to maximize efficiency.
  • Practical Example: The exercise provided is similar to an isentropic process, but since there is a change in entropy due to heating, it strictly doesn't qualify as one. Understanding where real-life deviates from the ideal helps in better understanding system behavior.

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

Which of the following expressions is known as Clausius inequality? (a) \(\oint \frac{d q}{T} \leq 0\) (b) \(\oint \frac{d s}{T}=0\) (c) \(\oint \frac{T}{d q} \leq 0\) (d) \(\oint \frac{d q}{T} \geq 0\)

One mole of an ideal gas at \(25^{\circ} \mathrm{C}\) expands in volume from \(1.0 \mathrm{~L}\) to \(4.0 \mathrm{~L}\) at constant . temperature. What work (in \(J\) ) is done if the gas expands against vacuum \(\left(P_{\text {cxtemal }}=0\right)\) ? (a) \(-4.0 \times 10^{2}\) (b) \(-3.0 \times 10^{2}\) (c) \(-1.0 \times 10^{2}\) (d) Zero

Calculate the final temperature of a monoatomic ideal gas that is compressed reversible and adiabatically from \(16 \mathrm{~L}\) to \(2 \mathrm{~L}\) at \(300 \mathrm{~K}\) : (a) \(600 \mathrm{~K}\) (b) \(1044.6 \mathrm{~K}\) (c) \(1200 \mathrm{~K}\) (d) \(2400 \mathrm{~K}\)

Calculate the standard enthalpy of reaction for the following reaction using the listed enthalpies of reaction : $$ \begin{aligned} 3 \mathrm{Co}(s)+2 \mathrm{O}_{2}(g) & \longrightarrow \mathrm{Co}_{3} \mathrm{O}_{4}(s) \\ 2 \mathrm{Co}(s)+\mathrm{O}_{2}(g) & \longrightarrow 2 \mathrm{CoO}(s) ; \quad \Delta H_{1}^{\circ}=-475.8 \mathrm{~kJ} \\ 6 \mathrm{CoO}(s)+\mathrm{O}_{2}(g) & \longrightarrow 2 \mathrm{Co}_{3} \mathrm{O}_{4}(s) ; \Delta H_{2}^{\circ}=-355.0 \mathrm{~kJ} \end{aligned} $$ (a) \(-891.2 \mathrm{~kJ}\) (b) \(-120.8 \mathrm{~kJ}\) (c) \(+891.2 \mathrm{~kJ}\) (d) \(-830.8 \mathrm{~kJ}\)

Which of the following conditions will always lead to a non-spontaneous, change? (a) \(\Delta H\) and \(\Delta S\) both \(+\) ve (b) \(\Delta H\) is -ve and \(\Delta S\) is \(+\) ve (c) \(\Delta H\) and \(\Delta S\) both \(-v e\) (d) \(\Delta H\) is +ve and \(\Delta S\) is -ve

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