/*! 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 30 A system undergoes a process con... [FREE SOLUTION] | 91影视

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A system undergoes a process consisting of the following two steps: Step 1: The system absorbs \(72 \mathrm{~J}\) of heat while \(35 \mathrm{~J}\) of work is done on it. Step 2: The system absorbs \(35 \mathrm{~J}\) of heat while performing \(72 \mathrm{~J}\) of work. Calculate \(\Delta E\) for the overall process.

Short Answer

Expert verified
The change in internal energy (鈭咵) for the overall process is 144 J.

Step by step solution

01

Calculate 鈭咵 for the first step.

Using the first law of thermodynamics: 鈭咵鈧 = Q鈧 - W鈧 Here, Q鈧 = 72 J (heat absorbed) and W鈧 = 35 J (work done on the system). 鈭咵鈧 = 72 J - 35 J = 37 J The change in internal energy for the first step is 37 J.
02

Calculate 鈭咵 for the second step.

Using the first law of thermodynamics again: 鈭咵鈧 = Q鈧 - W鈧 Here, Q鈧= 35 J (heat absorbed) and W鈧 = -72 J (since the system is performing work). 鈭咵鈧 = 35 J - (-72 J) = 35 J + 72 J = 107 J The change in internal energy for the second step is 107 J.
03

Calculate the total 鈭咵 for the overall process.

To find the total change in internal energy for the overall process, we can sum up the calculated 鈭咵 values for each step: 鈭咵_total = 鈭咵鈧 + 鈭咵鈧 = 37 J + 107 J = 144 J The change in internal energy for the overall process is 144 J.

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Key Concepts

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

Internal Energy Change
The concept of internal energy change is based on the first law of thermodynamics. This law states that the total energy of an isolated system remains constant. Essentially, energy cannot be created or destroyed, only transformed or transferred.
To understand internal energy change, we consider the change in a system's energy due to heat transfer and work done. When a system absorbs or releases energy in the form of heat, or when work is done on or by the system, its internal energy changes:
\[\Delta E = Q - W\]
Here, \(\Delta E\) represents the change in internal energy, \(Q\) stands for heat absorbed by the system, and \(W\) represents the work done by the system.

It is important to note the sign conventions used:
  • Positive \(Q\): Heat is absorbed by the system.
  • Negative \(W\): Work is done on the system.
  • Positive \(W\): Work is done by the system.
In the problem provided, the internal energy change for each step was calculated using these conventions and formulas, yielding a total \(\Delta E = 144 \text{ J}\). This illustrates how energy changes can be tracked in a multi-step process.
Heat Absorption
Heat absorption refers to the process by which a system takes in energy. This can change the internal energy of a system, causing it to rise. When energy is transferred into a system as heat, it can increase the temperature or cause other changes such as a phase transformation.
In thermodynamics, we pay close attention to the amount of heat exchanged, because it directly influences the system's energy state. In our exercise, the system absorbed heat in two steps鈥攆irst \(72 \text{ J}\) and then \(35 \text{ J}\).

It's useful to remember that heat, denoted by \(Q\), is a path function. This means the amount of heat absorbed or released can vary depending on how the process occurs. In the context of the first law of thermodynamics, heat absorption affects the calculation of internal energy as it directly adds to or deducts from the system's total energy."
  • Step 1: Absorbs \(72 \text{ J}\).
  • Step 2: Absorbs \(35 \text{ J}\).
Overall, the way heat is absorbed can reveal a lot about system processes and their total energy impact.
Work Done on System
Work done on a system is another vital component when considering energy changes. Like heat, work can alter the internal energy, affecting how a system evolves during a process.
In our example, during the first step, \(35 \text{ J}\) of work was done on the system, meaning that energy was added to the system, contributing to an increase in internal energy. During the second step, the system performed \(72 \text{ J}\) of work, which is energy leaving the system. For calculation purposes, this was treated as negative work done on the system.

Here鈥檚 how work affects energy change:
  • If work is done on the system, it increases internal energy, and is considered a negative \(W\) when calculating \(\Delta E\).
  • If the system does work, it results in a decrease of internal energy, and is considered a positive \(W\) when calculating \(\Delta E\).
This highlights the role of external interactions on the system's energy, showing that work鈥攍ike heat鈥攄irectly ties into how internal energy is calculated and equilibrated during any thermodynamic process.

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Most popular questions from this chapter

In a bomb calorimeter, the reaction vessel is surrounded by water that must be added for each experiment. Since the amount of water is not constant from experiment to experiment, the mass of water must be measured in each case. The heat capacity of the calorimeter is broken down into two parts: the water and the calorimeter components. If a calorimeter contains \(1.00 \mathrm{~kg}\) water and has a total heat capacity of \(10.84 \mathrm{~kJ} /{ }^{\circ} \mathrm{C}\), what is the heat capacity of the calorimeter components?

The overall reaction in a commercial heat pack can be represented as $$ 4 \mathrm{Fe}(s)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{Fe}_{2} \mathrm{O}_{3}(s) \quad \Delta H=-1652 \mathrm{~kJ} $$ a. How much heat is released when \(4.00 \mathrm{~mol}\) iron is reacted with excess \(\mathrm{O}_{2}\) ? b. How much heat is released when \(1.00 \mathrm{~mol} \mathrm{Fe}_{2} \mathrm{O}_{3}\) is produced? c. How much heat is released when \(1.00 \mathrm{~g}\) iron is reacted with excess \(\mathrm{O}_{2} ?\) d. How much heat is released when \(10.0 \mathrm{~g} \mathrm{Fe}\) and \(2.00 \mathrm{~g} \mathrm{O}_{2}\) are reacted?

Calculate the internal energy change for each of the following. a. One hundred (100.) joules of work is required to compress a gas. At the same time, the gas releases \(23 \mathrm{~J}\) of heat. b. A piston is compressed from a volume of \(8.30 \mathrm{~L}\) to \(2.80 \mathrm{~L}\) against a constant pressure of \(1.90 \mathrm{~atm} .\) In the process, there is a heat gain by the system of \(350 . \mathrm{J}\). c. A piston expands against \(1.00 \mathrm{~atm}\) of pressure from \(11.2 \mathrm{~L}\) to \(29.1 \mathrm{~L}\). In the process, \(1037 \mathrm{~J}\) of heat is absorbed.

For the following reactions at constant pressure, predict if \(\Delta H>\) \(\Delta E, \Delta H<\Delta E\), or \(\Delta H=\Delta E\) a. \(2 \mathrm{HF}(g) \longrightarrow \mathrm{H}_{2}(g)+\mathrm{F}_{2}(g)\) b. \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)\) c. \(4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)\)

Consider the following changes: a. \(\mathrm{N}_{2}(g) \longrightarrow \mathrm{N}_{2}(l)\) b. \(\mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g) \longrightarrow \mathrm{H}_{2}(g)+\mathrm{CO}_{2}(g)\) c. \(\mathrm{Ca}_{3} \mathrm{P}_{2}(s)+6 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 3 \mathrm{Ca}(\mathrm{OH})_{2}(s)+2 \mathrm{PH}_{3}(g)\) d. \(2 \mathrm{CH}_{3} \mathrm{OH}(l)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(l)\) e. \(\mathrm{I}_{2}(s) \longrightarrow \mathrm{I}_{2}(g)\) At constant temperature and pressure, in which of these changes is work done by the system on the surroundings? By the surroundings on the system? In which of them is no work done?

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