CHEM 101 Lecture Notes - Lecture 5: Calorimetry, Thermodynamic System, Calorie

32 views2 pages

michellele8outlook-com

2 Mar 2020

School

Department

Course

Professor

For unlimited access to Class Notes, a Class+ subscription is required.

Document Summary

Energy-the ability to do work or transfer heat, two forms: kinetic and potential. Work-energy used to move objects, w = f x d w = -p delta v. Heat-energy used to raise temperature, flows from warmer objects to cooler objects- Kinetic energy-energy an object possess by virtue of its motion. Potential energy-energy an object possess by virtue of its position or chemical composition. System-the thing we want to study, i. e. hydrogen and oxygen molecules. First law of thermodynamics-energy is neither created nor destroyed, but it can be transferred. Internal energy-the sum of all kinetic and potential energies of all components of the system, e. Change in internal energy-delta e = e final - e initial, exchanged as either heat (q) or work (w), delta e = q + w. Endergonic-e final > e initial, the system absorbed energy from the surroundings. Exergonic-e final < e initial, the system released energy to the surroundings.

Related Questions

Energy cannot be created nor destroyed, but it can be transferred between a reaction and its surroundings. The change in energy, Delta E, is positive if the reaction absorbs energy, and it is negative if the reaction releases energy. You may also see this expressed in terms of change in internal energy, Delta U. For the purposes of this question, Delta U and Delta E are equal. One way a reaction can transfer energy to or from the surroundings is by releasing or absorbing heat. A reaction can also transfer energy, in the form of work, through a change in volume. The total change in energy is the sum of the heat and work: Delta E = q + w At constant pressure, q = Delta H and w = -P Delta V and so Delta E = Delta H - P Delta V Assuming constant pressure, rank these reactions from most energy released by the system to most energy absorbed by the system, based on the following descriptions: Surroundings get colder and the system decreases in volume. Surroundings get hotter and the system expands in volume. Surroundings get hotter and the system decreases in volume. Surroundings get hotter and the system does not change in volume. Also assume that the magnitude of the volume and temperature changes are similar among the reactions. Rank from most energy released to most energy absorbed. To rank items as equivalent, overlap them. Energy cannot be created nor destroyed, but it can be transferred between a reaction and its surroundings. The change in energy, Delta E, is positive if the reaction absorbs energy, and it is negative if the reaction releases energy. You may also see this expressed in terms of change in internal energy, Delta U. For the purposes of this question, Delta U and Delta E are equal. One way a reaction can transfer energy to or from the surroundings is by releasing or absorbing heat. A reaction can also transfer energy, in the form of work, through a change in volume. The total change in energy is the sum of the heat and work: Delta E = q + w At constant pressure, q = Delta H and w = -P Delta V and so Delta E = Delta H - P Delta V A mole of X reacts at a constant pressure of 43.0 atm via the reaction X(g) + 4Y(g) - 2Z(g), Delta H degree = -75.0 kJ Before the reaction, the volume of the gaseous mixture was 5.00 L. After the reaction, the volume was 2.00 L Calculate the value of the total energy change. AE, in kilojoules. Express your answer with the appropriate units.

You are testing some novel fuel mixtures for potential use in automobiles. Your testing apparatus is a 2.54-cm inner diameter cylinder fitted with a piston, designed to mimic a cylinder found in a car engine. A fuel/air mixture is ignited while a force of 286.0 N is applied to the piston. If there are 0.100 mol of gas after combustion at 240.0Â°C, and the atmospheric pressure is 1.01325 Ã 105 Pa, what is the pressure of gas inside the cylinder at these initial conditions? Assume the combustion products behave ideally.

What is the volume of gas in the cylinder?

The force applied to the piston is then reduced to 108.0 N and the gas responds in a two step process. The first, quick step is adiabatic. During the second, slower step, the gas returns to 240.0Â°C from heat provided by the cylinder walls. Assume that the change in kinetic and potential energies in both steps is negligible. Consider the energy balance equation for a closed system

where ÎU is the change in internal energy, ÎEk is the change in kinetic energy, ÎEp is the change in potential energy, Q is heat and W is work. For the first step of the two step process, sort the terms in the energy balance equation based on if they are zero or non-zero.

For the second step of the process, sort the terms in the energy balance equation based on if they are zero or non-zero.

For the third step of the process, sort the terms in the energy balance equations.

Then find the final pressure and volume of the gas?

How much heat was added to the gas in the process?

â.

CHEM 101 Lecture Notes - Lecture 5: Calorimetry, Thermodynamic System, Calorie

Document Summary

Get access

Related textbook solutions

Chemistry: Structure and Properties

Basic Chemistry

Principles of Chemistry Molecular Approach

Principles of Chemistry Molecular Approach

Chemistry: Structure and Properties

Chemistry: A Molecular Approach

Chemistry: A Molecular Approach

Principles of Chemistry: A Molecular Approach

Chemistry: The Central Science

Related Documents

01:160:161 Chapter Notes - Chapter 9.6-9.8: Exothermic Reaction, Endothermic Process, Enthalpy

Chemistry 1027A/B Chapter Notes - Chapter 4: Heat Capacity, Enthalpy, Jmol

CHM 2046L Study Guide - Final Guide: Dynamic Equilibrium, Equilibrium Constant, Reversible Reaction

Related Questions