/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 29 Starting with \(H=U+P V\), show ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Starting with \(H=U+P V\), show that $$ \left(\frac{\partial U}{\partial T}\right)_{P}=C_{P}-P\left(\frac{\partial V}{\partial T}\right)_{P} $$ Interpret this result physically.

Short Answer

Expert verified
The result shows the internal energy change at constant pressure, accounting for work done by volume expansion against external pressure.

Step by step solution

01

Write down the expression for enthalpy

We start with the given formula for enthalpy, which is:\[ H = U + PV \] where \( H \) is the enthalpy, \( U \) is the internal energy, \( P \) is the pressure, and \( V \) is the volume.
02

Differentiate the enthalpy expression with respect to temperature at constant pressure

Take the partial derivative of \( H \) with respect to temperature \( T \) while keeping pressure constant. Thus:\[ \left( \frac{\partial H}{\partial T} \right)_{P} = \left( \frac{\partial U}{\partial T} \right)_{P} + P \left( \frac{\partial V}{\partial T} \right)_{P} \]
03

Recognize the definition of heat capacity at constant pressure

The heat capacity at constant pressure \( C_{P} \) is defined as the partial derivative of enthalpy with respect to temperature at constant pressure:\[ C_{P} = \left( \frac{\partial H}{\partial T} \right)_{P} \]
04

Substitute the expression for heat capacity

Substitute the value of \( C_{P} \) from Step 3 into the equation obtained in Step 2:\[ C_{P} = \left( \frac{\partial U}{\partial T} \right)_{P} + P \left( \frac{\partial V}{\partial T} \right)_{P} \]
05

Rearrange the equation to isolate the desired derivative

Rearrange the expression obtained in Step 4 to solve for \( \left( \frac{\partial U}{\partial T} \right)_{P} \):\[ \left( \frac{\partial U}{\partial T} \right)_{P} = C_{P} - P \left( \frac{\partial V}{\partial T} \right)_{P} \]

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 91Ó°ÊÓ!

Key Concepts

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

Enthalpy
Enthalpy, denoted by the symbol \(H\), is a crucial concept in thermodynamics. It represents the total heat content of a system and is defined as:
  • \(H = U + PV\)
Here, \(U\) is the internal energy, \(P\) stands for pressure, and \(V\) signifies volume. Enthalpy accounts for both the internal energy of a system and the energy needed to make room for it in its environment by displacing its surroundings.
This makes understanding enthalpy essential when studying processes at work in chemical reactions, phase changes, and other thermodynamic phenomena. Consider it as the energy a system has to "spend" in work and heat exchange, under constant pressure conditions. A higher enthalpy indicates a system is capable of delivering or absorbing more energy, which influences how we manipulate these processes in practical applications.
Heat Capacity
Heat capacity is an important concept when dealing with changes in temperature. At constant pressure, it is often defined using the partial derivative of enthalpy with respect to temperature:
  • \(C_P = \left(\frac{\partial H}{\partial T}\right)_P\)
Heat capacity, \(C_P\), tells us how much heat is required to change a system's temperature by a single degree under constant pressure.
This quantifies the "thermal inertia" of a system, indicating its ability to resist temperature changes.Heat capacity can often vary widely depending on the material or substance in question and is crucial when considering how different substances react to heating and cooling. It's a critical parameter in designing heating processes, chemical reactions, and even cooling systems.
Partial Derivatives
Partial derivatives are at the heart of understanding thermodynamic systems. They allow us to examine how a particular variable, such as volume or internal energy, changes regarding another, such as temperature, while keeping other parameters constant.In our context:
  • \(\left(\frac{\partial U}{\partial T}\right)_P\) measures the change in internal energy with temperature at constant pressure.
  • \(P\left(\frac{\partial V}{\partial T}\right)_P\) gives insight into how volume changes with temperature under constant pressure.
By using partial derivatives, we can break down complex relationships in thermodynamics into more manageable components.
This approach helps us understand precisely how changes in one variable, like temperature, can affect others, such as internal energy or volume. It's a powerful tool in both theoretical studies and practical applications.
Thermodynamic Equations
Thermodynamic equations, like \(H = U + PV\), are the backbone of thermodynamics. They describe relationships between different state variables, providing a mathematical framework to understand energy exchanges and transformations.
  • Effects of temperature, volume, pressure, and heat can be examined using equations derived from basic thermodynamic laws.
  • The equation \(\left(\frac{\partial U}{\partial T}\right)_P = C_P - P \left(\frac{\partial V}{\partial T}\right)_P\) is an example of how these foundations elucidate the roles of variables like pressure-volume work and heat capacity.
Using thermodynamic equations allows us to predict how a system behaves under various conditions, offering insight into optimally designing experiments or processes.
Mastering these equations is key to harnessing the principles of thermodynamics to solve real-world problems and innovate in fields such as chemical engineering, material science, and energy management.

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

Consider an ideal gas that occupies \(2.25 \mathrm{L}\) at 1.33 bar. Calculate the work required to compress the gas isothermally to a volume of \(1.50 \mathrm{L}\) at a constant pressure of 2.00 bar followed by another isothermal compression to \(0.800 \mathrm{L}\) at a constant pressure of 2.50 bar (Figure 19.4 ). Compare the result with the work of compressing the gas isothermally and reversibly from 2.25 L to 0.800 L.

Suppose that a 10 -kg mass of iron at \(20^{\circ} \mathrm{C}\) is dropped from a height of 100 meters. What is the kinetic energy of the mass -just before it hits the ground? What is its speed? What would be the final temperature of the mass if all its kinetic energy at impact is transformed into internal energy? Take the molar heat capacity of iron to be \(\bar{C}_{P}=25.1 \mathrm{J} \cdot \mathrm{mol}^{-1} \cdot \mathrm{K}^{-1}\) and the gravitational acceleration constant to be \(9.80 \mathrm{m} \cdot \mathrm{s}^{-2}\)

In this problem, we will discuss a famous experiment called the Joule-Thomson experiment. In the first half of the 19 th century, Joule tried to measure the temperature change when a gas is expanded into a vacuum. The experimental setup was not sensitive enough, however, and he found that there was no temperature change, within the limits of his error. Soon afterward, Joule and Thomson devised a much more sensitive method for measuring the temperature change upon expansion. In their experiments (see Figure 19.11 ), a constant applied pressure \(P_{1}\) causes a quantity of gas to flow slowly from one chamber to another through a porous plug of silk or cotton. If a volume, \(V_{1},\) of gas is pushed through the porous plug, the work done on the gas is \(P_{1} V_{1}\). The pressure on the other side of the plug is maintained at \(P_{2},\) so if a volume \(V_{2}\) enters the right-side chamber, then the net work is given by \\[ w=P_{1} V_{1}-P_{2} V_{2} \\] The apparatus is constructed so that the entire process is adiabatic, so \(q=0 .\) Use the First Law of Thermodynamics to show that \\[ U_{2}+P_{2} V_{2}=U_{1}+P_{1} V_{1} \\] or that \(\Delta H=0\) for a Joule-Thomson expansion. Starting with \\[ d H=\left(\frac{\partial H}{\partial P}\right)_{T} d P+\left(\frac{\partial H}{\partial T}\right)_{P} d T \\] show that \\[ \left(\frac{\partial T}{\partial P}\right)_{H}=-\frac{1}{C_{P}}\left(\frac{\partial H}{\partial P}\right)_{T} \\] Interpret physically the derivative on the left side of this equation. This quantity is called the Joule-Thomson coefficient and is denoted by \(\mu_{\mathrm{JT}} .\) In Problem 19-54 you will show that it equals zero for an ideal gas. Nonzero values of \((\partial T / \partial P)_{H}\) directly reflect intermolecular interactions. Most gases cool upon expansion [a positive value of \(\left.(\partial T / \partial P)_{H}\right]\) and a JouleThomson expansion is used to liquefy gases.

Suppose that a \(10-\mathrm{kg}\) mass of iron at \(20^{\circ} \mathrm{C}\) is dropped from a height of 100 meters. What is the kinetic energy of the mass just before it hits the ground? What is its speed? What would be the final temperature of the mass if all its kinetic energy at impact is transformed into internal energy? Take the molar heat capacity of iron to be \(\bar{C}_{P}=25.1 \mathrm{~J} \cdot \mathrm{mol}^{-1} \cdot \mathrm{K}^{-1}\) and the gravitational acceleration constant to be \(9.80 \mathrm{~m} \cdot \mathrm{s}^{-2}\).

Calculate the work involved when one mole of a monatomic ideal gas at \(298 \mathrm{K}\) expands reversibly and adiabatically from a pressure of 10.00 bar to a pressure of 5.00 bar.

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.