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First law of thermodynamics is not adequate in predicting the direction of a process.

Short Answer

Expert verified
The first law of thermodynamics does not predict process direction; the second law is required for that.

Step by step solution

01

Understanding the First Law of Thermodynamics

The first law of thermodynamics states that energy cannot be created or destroyed, only transferred or changed in form. Mathematically, it can be expressed as \( \Delta U = Q - W \), where \( \Delta U \) is the change in internal energy, \( Q \) is the heat added to the system, and \( W \) is the work done by the system.
02

Analyzing Process Direction

To determine the direction of a process, one must consider factors beyond energy conservation. The first law does not specify which processes are spontaneous or the direction in which a process will naturally occur, as it only focuses on energy balance.
03

Limitations of First Law for Process Direction

While the first law accounts for the quantitative energy relationship, it does not account for entropy or the quality of energy, which are crucial for predicting direction. The second law of thermodynamics, which involves entropy, is necessary for determining process direction.
04

Applying the Second Law of Thermodynamics

To predict the direction of a process, the second law, which involves the concept of entropy, should be applied. The second law states that the total entropy of an isolated system can never decrease over time, indicating the preferred direction of natural processes.

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Key Concepts

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

Energy Conservation
In the realm of thermodynamics, energy conservation is a fundamental principle. It is encapsulated by the First Law of Thermodynamics, which tells us that energy in a closed system is constant. This means energy cannot vanish into thin air or appear from nowhere.
This law is expressed mathematically as \( \Delta U = Q - W \), where:
  • \( \Delta U \) is the change in internal energy of the system.
  • \( Q \) represents the heat added to the system.
  • \( W \) is the work done by the system.
It is important to note that different forms of energy, such as thermal, mechanical, and chemical, can transform into one another, but the total energy remains the same. However, while the First Law is excellent for tracking energy conversions, it does not tell us everything about how these processes affect the system or surroundings.
Process Direction
Energy conservation through the First Law of Thermodynamics gives no clue about the natural direction a process should take. It's similar to having a map without a compass. To determine process direction, factors like entropy and energy quality must be considered.
The First Law ensures the energy equation balances but doesn't indicate whether a process occurs spontaneously. For instance:
  • Heat might transfer from a hot object to a cold one naturally, but never the other way around.
  • Processes like melting ice will happen in a particular direction unless constrained by external conditions.
Thus, to predict or understand how and why processes proceed in certain directions, we must introduce concepts beyond mere energy balance. This is where understanding entropy and the more detailed Second Law of Thermodynamics comes into play.
Second Law of Thermodynamics
The Second Law of Thermodynamics provides the missing piece to predict process direction by introducing entropy, a measure of disorder or randomness in a system. This law explains that the total entropy of an isolated system can never decrease. It tells us that natural processes favor states of higher entropy.
This law is essential for understanding phenomena such as:
  • Why heat travels from hot to cold spontaneously.
  • Why gases diffuse from areas of high concentration to low concentration.
The Second Law also highlights the quality of energy, not just its quantity. While the First Law tells us energy is conserved, the Second Law teaches us about the degradation of energy as it transforms, where usable energy tends to diminish.
In essence, it connects the concepts of energy with the inherent directionality of natural processes, showing us that without this law, we cannot fully grasp the behavior of systems as they evolve over time.

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Most popular questions from this chapter

The standard state Gibbs free energies of formation of C(graphite) and \(\mathrm{C}\) (diamond) at \(T=298 \mathrm{~K}\) are \(\Delta_{f} G^{0}[\mathrm{C}\) (graphite) \(]=0 \mathrm{~kJ} \mathrm{~mol}^{-1}\) \(\Delta_{f} G^{0}[\mathrm{C}\) (diamond) \(]=2.9 \mathrm{~kJ} \mathrm{~mol}^{-1}\) The standard state means that the pressure should be 1 bar, and substance should be pure at a given temperature. The conversion of graphite \([\mathrm{C}\) (graphite) \(]\) to diamond \(\left[C\right.\) (diamond)] reduces its volume by \(2 \times 10^{-}\) \({ }^{6} \mathrm{~m}^{3} \mathrm{~mol}^{-1} .\) If \(C\) (graphite) is converted to \(\mathrm{C}\) (diamond) isothermally at \(T\) \(=298 \mathrm{~K}\), the pressure at which \(\mathrm{C}\) (graphite) is in equilibrium with \(\mathrm{C}\) (diamond), is [Useful information : \(\left.1 \mathrm{~J}=1 \mathrm{~kg} \mathrm{~m}^{2} \mathrm{~s}^{-2} ; 1 \mathrm{~Pa}=1 \mathrm{~kg} \mathrm{~m}^{-1} \mathrm{~s}^{-2} ; 1 \mathrm{bar}=10^{5} \mathrm{~Pa}\right]\) [Adv. 2017] (a) 14501 bar (b) 58001 bar (c) \(1450 \mathrm{bar}\) (d) 29001 bar

For the process \(\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) at \(T=100^{\circ} \mathrm{C}\) and 1 atmosphere pressure, the correct choice is (a) \(\Delta S_{\text {system }}>0\) and \(\Delta S_{\text {surroundings }}>0\) (b) \(\Delta S_{\text {system }}>0\) and \(\Delta S_{\text {surroundings }}<0\) (c) \(\Delta S_{\text {system }}<0\) and \(\Delta S_{\text {surroundings }}>0\) (d) \(\Delta S_{\text {system }}<0\) and \(\Delta S_{\text {surroundings }}<0\)

The surface of copper gets tarnished by the formation of copper oxide. \(\mathrm{N}_{2}\) gas was passed to prevent the oxide formation during heating of copper at \(1250 \mathrm{~K}\). However, the \(\mathrm{N}_{2}\) gas contains 1 mole \(\%\) of water vapour as impurity. The water vapour oxidises copper as per the reaction given below: \(2 \mathrm{Cu}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightarrow \mathrm{Cu}_{2} \mathrm{O}(\mathrm{s})+\mathrm{H}_{2}(\mathrm{~g})\) \(p_{\mathrm{H} 2}\) is the minimum partial pressure of \(\mathrm{H}_{2}\) (in bar) needed to prevent the oxidation at \(1250 \mathrm{~K}\). The value of \(\ln \left(p_{\mathrm{H} 2}\right)\) is (Given: total pressure \(=1\) bar, \(R\) (universal gas constant \()=8 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}, \ln\) \((10)=2.3 . \mathrm{Cu}(\mathrm{s})\) and \(\mathrm{Cu}_{2} \mathrm{O}(\mathrm{s})\) are mutually immiscible. At \(1250 \mathrm{~K}: 2 \mathrm{Cu}(\mathrm{s})+1 / 2 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{Cu}_{2} \mathrm{O}(\mathrm{s}) ;\) $$ \Delta G^{\circ}=-78,000 \mathrm{~J} \mathrm{~mol}^{-1} $$ $$ \begin{aligned} &\mathrm{H}_{2}(\mathrm{~g})+1 / 2 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) ; \Delta G^{\circ}=-1,78,000 \mathrm{~J} \mathrm{~mol}^{-1}\\\ &\text { ( } G \text { is the Gibbs energy) } \end{aligned} $$

The standard enthalpy of combustion at \(25^{\circ} \mathrm{C}\) of hydrogen, cyclohexene \(\left(\mathrm{C}_{6} \mathrm{H}_{10}\right)\) and cyclohexane \(\left(\mathrm{C}_{6} \mathrm{H}_{12}\right)\) are \(-241,-3800\) and \(-\) \(3920 \mathrm{~kJ} /\) mole respectively. Calculate the heat of hydrogenation of cyclohexene.

Among the following, the intensive property is (properties are) (a) molar conductivity (b) electromotive force (c) resistance (d) heat capacity

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