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Chapter 6: Problem 18

The work done to compress a gas is \(74 \mathrm{~J}\). As a result, \(26 \mathrm{~J}\) of heat is given off to the surroundings. Calculate the change in energy of the gas.

Short Answer

Expert verified
The change in energy of the gas is 48 J.

Step by step solution

01

Understand the First Law of Thermodynamics

The First Law of Thermodynamics is 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

Identify Known Values and Sign Conventions

In this problem, 74 J of work is done on the gas, which means \( W = -74 \) J since the work is done on the system (work done by the system is negative). The heat given off to the surroundings is \( 26 \) J, so \( Q = -26 \) J because heat is leaving the system.
03

Substitute Values into the First Law of Thermodynamics Equation

Using the equation \( \Delta U = Q - W \), substitute the identified values: \( \Delta U = -26 - (-74) \).
04

Simplify the Equation

Simplify the equation: \( \Delta U = -26 + 74 = 48 \).
05

Interpret the Result

The calculation shows that the change in energy of the gas \( \Delta U \) is 48 J. This means the internal energy of the gas increased by 48 J as a result of the work done and heat exchanged in this process.

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Key Concepts

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

Internal Energy
Internal energy refers to the total energy contained within a system. It includes all forms of energy associated with the random motion and interactions of the molecules within the system.
A change in internal energy (\( \Delta U \)) can be caused by:
  • Heat transfer
  • Work done on or by the system
According to the First Law of Thermodynamics, which states that energy cannot be created or destroyed, only transformed, the relationship involving internal energy is expressed as: \[ \Delta U = Q - W \] where:
  • \( \Delta U \) is the change in internal energy of the system
  • \( Q \) is the net heat added to or removed from the system
  • \( W \) is the work done by the system (positive if done by the system and negative if done on the system)
In simpler terms, if heat is added, or work is done on the system, the internal energy usually increases, as the molecules have more energy to move.
Heat Transfer
Heat transfer refers to the process of heat energy moving from a hotter to a colder region. This is a natural phenomenon that always occurs when there is a temperature difference.
When considering heat transfer in thermodynamics, it is essential to define the direction of heat flow:
  • When heat is absorbed by a system, the heat quantity \( Q \) is positive.
  • When heat is released or lost to the surroundings, \( Q \) is negative.
In our exercise:
  • The system gave off 26 J of heat.
  • Therefore, \( Q = -26 \) J, indicating heat left the system.
The sign convention is crucial because it helps describe how the energy in the system is changing and determines the next steps in energy calculations.
Work Done on Gas
Work done on a gas system is another way energy changes within the system. Generally, work done on a gas results in the compression or expansion of the gas.
In thermodynamics, the work done by or on a gas can be tricky due to the choice of sign conventions:
  • Work done by the system (e.g., gas expanding) is considered positive.
  • Work done on the system (e.g., gas compressing) is typically negative.
For our problem, the work done on the gas was 74 J.
  • This results in \( W = -74 \) J because the work was done on the system, compressing the gas.
Understanding how to apply these conventions helps ensure the First Law of Thermodynamics is accurately used to calculate changes, such as determining that internal energy increases in this scenario.

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Most popular questions from this chapter

The combustion of \(0.4196 \mathrm{~g}\) of a hydrocarbon releases \(17.55 \mathrm{~kJ}\) of heat. The masses of the products are \(\mathrm{CO}_{2}=1.419 \mathrm{~g}\) and \(\mathrm{H}_{2} \mathrm{O}=0.290 \mathrm{~g} .\) (a) What is the empirical formula of the compound? (b) If the approximate molar mass of the compound is \(76 \mathrm{~g}\), calculate its standard enthalpy of formation.

Predict the value of \(\Delta H_{\mathrm{f}}^{\circ}\) (greater than, less than, or equal to zero) for these elements at \(25^{\circ} \mathrm{C}:\) (a) \(\mathrm{Br}_{2}(g)\) and \(\operatorname{Br}_{2}(l),\) (b) \(\mathrm{I}_{2}(g)\) and \(\mathrm{I}_{2}(s)\)

The first step in the industrial recovery of copper from the copper sulfide ore is roasting, that is, the conversion of \(\mathrm{CuS}\) to \(\mathrm{CuO}\) by heating: $$ \begin{aligned} 2 \mathrm{CuS}(s)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CuO}(s) &+2 \mathrm{SO}_{2}(g) \\ \Delta H=&-805.6 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $$ Calculate the heat evolved (in \(\mathrm{kJ}\) ) per gram of \(\mathrm{CuS}\) roasted.

Describe how chemists use Hess's law to determine the \(\Delta H_{\mathrm{f}}^{\circ}\) of a compound by measuring its heat (enthalpy) of combustion.

The enthalpy of combustion of benzoic acid \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\right)\) is commonly used as the standard for calibrating constant-volume bomb calorimeters; its value has been accurately determined to be \(-3226.7 \mathrm{~kJ} / \mathrm{mol}\). When \(1.9862 \mathrm{~g}\) of benzoic acid are burned in a calorimeter, the temperature rises from \(21.84^{\circ} \mathrm{C}\) to \(25.67^{\circ} \mathrm{C}\). What is the heat capacity of the bomb? (Assume that the quantity of water surrounding the bomb is exactly \(2000 \mathrm{~g} .)\)
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