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Chapter 6: Problem 83

Show that isobars and isochores have positive slopes in the single-phase regions of a \(T S\) diagram. Suppose that \(C_{P}=a+b T,\) where \(a\) and \(b\) are positive constants. Show that the curvature of an isobar is also positive. For specified \(\mathrm{T}\) and \(S\), which is steeper: an isobar or an isochore? Why? Note that \(C_{P}>C_{V}\).

Short Answer

Expert verified
Isobars and isochores have positive slopes; the curvature of an isobar is positive. Isochores are steeper than isobars.

Step by step solution

01

Recall Definitions

In a T-S (Temperature-Entropy) diagram, an isobar is a line of constant pressure, and an isochore is a line of constant volume. Both describe thermodynamic processes or states.
02

Understand Positive Slopes

For a process to have a positive slope in a T-S diagram, the derivative \( \frac{dT}{dS} \) must be positive. We need to explore these for both isobars and isochores.
03

Analyze the Isobar Slope

Using the Maxwell relation and isobaric conditions (constant pressure), we know \( dP=0 \). The slope can be analyzed using \( C_P = T \left( \frac{\partial S}{\partial T} \right)_P \). Hence, \( \left( \frac{dT}{dS} \right)_P = \frac{T}{C_P} \), which is positive as long as \(C_P > 0\).
04

Analyze the Isochore Slope

Under isochoric conditions (constant volume), using similar relations, \( C_V = T \left( \frac{\partial S}{\partial T} \right)_V \). Thus, \( \left( \frac{dT}{dS} \right)_V = \frac{T}{C_V} \). As \(C_V > 0\), this slope is also positive.
05

Examine Isobar Curvature

Using \( C_P = a + bT \), the curvature of an isobar, \( \frac{d^2T}{dS^2} \), depends on \( \frac{d}{dS} \left( \frac{dT}{dS} \right)_P \). Since \(b > 0\), and \( \frac{dT}{dS} \) increases with \(T\), \( \frac{d^2T}{dS^2} > 0\), indicating a positive curvature.
06

Compare Steepness of Isobars and Isochores

The slope for an isobar \( \left( \frac{dT}{dS} \right)_P = \frac{T}{C_P} \) vs the slope for an isochore \( \left( \frac{dT}{dS} \right)_V = \frac{T}{C_V} \). Since \( C_P > C_V \), \( \frac{dT}{dS} \) for isobars is less than \( \frac{dT}{dS} \) for isochores. Thus, isochores are steeper than isobars.

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Key Concepts

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

Isobar
In thermodynamics, an isobar represents a process where the pressure remains constant. On the Temperature-Entropy (T-S) diagram, an isobar is depicted as a line with a uniform pressure level throughout its length. Since the relationship between temperature and entropy involves heat exchange, along an isobar, this relationship is dictated by the heat capacity at constant pressure, denoted as \(C_P\).
  • For positive slopes on a T-S diagram, the derivative \( \frac{dT}{dS} \) must be positive.
  • Under constant pressure, the slope can be calculated as \( \left( \frac{dT}{dS} \right)_P = \frac{T}{C_P} \).
  • As long as \(C_P > 0\), the slope remains positive, denoting an upward incline as entropy increases.
Understanding the isobar curvature involves exploring the nature of \(C_P\). If \(C_P = a + bT\) where \(b > 0\), it implies the slope steepens with increasing temperature, showcasing positive curvature on the T-S diagram.
Isochore
An isochore pertains to a thermodynamic process conducted at a constant volume. On the T-S diagram, isochores appear as vertical paths because the volume remains unchanged. A line of constant volume provides insight into how changes in temperature affect entropy when no volume displacement occurs.
  • Similar to isobars, the positive slope condition \( \left( \frac{dT}{dS} \right)_V = \frac{T}{C_V} \) ensures that temperature increases with entropy.
  • Here, \(C_V\) is the heat capacity at constant volume, and since \(C_V > 0\), the resulting slope remains positive.
Since processes involving an isochore deal with a fixed volume, no work is performed by the system during the process, allowing a clear focus on how thermal energy impacts the system's entropy.
Temperature-Entropy Diagram
The Temperature-Entropy (T-S) diagram plays a vital role in visualizing thermodynamic processes by plotting temperature against entropy. This tool aids in understanding how temperature and entropy evolve, offering crucial insights into work and heat exchanges across different states.
  • Isobars and isochores plotted on a T-S diagram reflect constant pressure and volume processes, respectively.
  • The T-S diagram enables the analysis of positive slopes, contributing crucial background to identifying thermal properties and behaviors of substances.
Such diagrams provide guidance on efficiency and energy transformations, making it essential for learning thermodynamic principles and performance assessments.
Heat Capacity at Constant Pressure
The heat capacity at constant pressure, \(C_P\), is a parameter defining how much heat energy a substance requires to raise its temperature by one unit while maintaining constant pressure. This property directly affects the slope of isobars on a T-S diagram, as \(C_P\) plays a substantial role in thermodynamic calculations.
  • If \(C_P = a + bT\) with \(a, b > 0\), it indicates that the heat required increases with temperature.
  • The relation to entropy change manifests through the derivative \( \left( \frac{dT}{dS} \right)_P = \frac{T}{C_P} \), showing that higher \(C_P\) leads to shallower slopes compared to isochores, where \(C_P > C_V\).
This property signifies how a substance reacts to thermal energy input, influencing practical applications such as heating, cooling, and energy efficiency considerations.
Heat Capacity at Constant Volume
Heat capacity at constant volume, \(C_V\), is an analogous measure of the thermal energy required to increase a substance's temperature by one unit while maintaining safe constant volume conditions. In thermodynamics, understanding \(C_V\) is crucial for examining processes that occur without volume alteration.
  • \(C_V\) determines the slope of isochores on a T-S diagram, reflected as \( \left( \frac{dT}{dS} \right)_V = \frac{T}{C_V} \).
  • Typically, \(C_V\) is less than \(C_P\) because no expansion work is included at constant volume, allowing isochores to be steeper.
Appreciating \(C_V\) helps perceive how substances respond to heat when expansion is not an option, offering perspectives into efficiency around restrained volume systems like enclosed gases and liquids.

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Most popular questions from this chapter

A vessel contains \(1 \mathrm{kg}\) of \(\mathrm{H}_{2} \mathrm{O}\) as liquid and vapor in equilibrium at \(1000 \mathrm{kPa}\). If the vapor occupies \(70 \%\) of the volume of the vessel, determine \(\mathrm{H}\) and \(S\) for the \(1 \mathrm{kg}\) of \(\mathrm{H}_{2} \mathrm{O}\).

Wet steam at \(1100 \mathrm{kPa}\) expand-sat constant enthalpy (as in a throttling process) to 101.325 \(\mathrm{kPa},\) where its temperature is \(378.15 \mathrm{K}\left(105^{\circ} \mathrm{C}\right) .\) What is the quality of the steam in its initial state?

Hot nitrogen gas at \(673.15 \mathrm{K}\left(400^{\circ} \mathrm{C}\right)\) and atmospheric pressure flows into a waste-heat boiler at the rate of \(20 \mathrm{kg} \mathrm{s}^{-1}\), and transfers heat to water boiling at \(101.33 \mathrm{kPa}\). The water feed to the boiler is saturated liquid at \(101.33 \mathrm{kPa}\), and it leaves the boiler as superheated steam at \(101.33 \mathrm{kPa}\) and \(423.15 \mathrm{K}\left(150^{\circ} \mathrm{C}\right)\). If the nitrogen is cooled to \(443.15 \mathrm{K}\left(170^{\circ} \mathrm{C}\right)\) and if heat is lost to the surroundings at a rate of \(80 \mathrm{kJ}\) for each kilogram of steam generated, what is the steam-generation rate? If the surroundings are at \(298.15 \mathrm{K}\left(25^{\circ} \mathrm{C}\right)\) what is \(\hat{S}_{G}\) for the process? Assume nitrogen to be an ideal gas for which \(C_{P}=(7 / 2) R\).

A \(2-m^{3}\) tank initially contains a mixture of saturated-vapor steam and saturated-liquid water at \(3000 \mathrm{kPa}\). Of the total mass, \(10 \%\) is vapor. Saturated-liquid water is bled from the tank through a valve until the total mass in the tank is \(40 \%\) of the initial total mass. If during the process the temperature of the contents of the tank is kept constant, how much heat is transferred?

Determine the maximum amount of work obtainable in a flow process from 1 kg of steam at \(3000 \mathrm{kPa}\) and \(723.15 \mathrm{K}\left(450^{\circ} \mathrm{C}\right)\) for surrounding conditions of \(300 \mathrm{K}\) and \(101.33 \mathrm{kPa}\).
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