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Chapter 20: Problem 8

For each of the following adiabatic processes, find the change in internal energy. ( \(a\) ) A gas does \(5 \mathrm{~J}\) of work while expanding adiabatically. (b) During an adiabatic compression, \(80 \mathrm{~J}\) of work is done on a gas. During an adiabatic process, no heat is transferred to or from the system. (a) \(\Delta U=\Delta Q-\Delta W=0-5 \mathrm{~J}=-5 \mathrm{~J}\) (b) \(\Delta U=\Delta Q-\Delta W=0-(-80 \mathrm{~J})=+80 \mathrm{~J}\)

Short Answer

Expert verified
(a) Internal energy decreases by 5 J; (b) Internal energy increases by 80 J.

Step by step solution

01

Understanding the Adiabatic Process

In an adiabatic process, the key characteristic is that there is no heat exchange with the surroundings, meaning that the change in heat, \( \Delta Q \), is zero. This simplifies the first law of thermodynamics equation down to \( \Delta U = - \Delta W \). Here, \( \Delta U \) is the change in internal energy, and \( \Delta W \) is the work done by or on the system.
02

Process (a): Expanding Gas

In this case, a gas does \(5\) J of work while expanding. Since the work is done by the system, \( \Delta W = 5 \) J. Using the equation \( \Delta U = - \Delta W \), we find that \( \Delta U = -5\) J, meaning the internal energy decreases by 5 J.
03

Process (b): Compressing Gas

For this process, \(80\) J of work is done on the gas, so the work is done on the system and \( \Delta W = -80 \) J. Plugging this into \( \Delta U = - \Delta W \), we get \( \Delta U = 80\) J, indicating that the internal energy increases by 80 J.

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Key Concepts

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

Internal Energy Change
When we talk about internal energy change, we are considering how the energy stored within a system alters due to various processes. In the context of an adiabatic process, where no heat is exchanged with the environment, the entire focus is on the work done by or on the gas.
In an adiabatic process:
  • The change in heat (\( \Delta Q \)) is zero.
  • Thus, any change in internal energy (\( \Delta U \)) is solely due to work done (\( \Delta W \)).
For a gas expanding adiabatically, like in process (a), work is done by the gas, resulting in a decrease in its internal energy. Conversely, during adiabatic compression, as shown in process (b), work is done on the gas, increasing its internal energy. Here鈥檚 the key takeaway:
  • If work is done by the system, internal energy decreases.
  • If work is done on the system, internal energy increases.
First Law of Thermodynamics
The First Law of Thermodynamics is a crucial principle in understanding energy transformations. It is expressed as:
\[\Delta U = \Delta Q - \Delta W\]Here,
  • \( \Delta U \) is the change in internal energy.
  • \( \Delta Q \) is the heat added to the system.
  • \( \Delta W \) is the work done by the system.
In an adiabatic process, \( \Delta Q = 0 \). This simplifies the first law to:
\[\Delta U = -\Delta W\]This means any work done by or on the system directly changes its internal energy.
In our examples:
  • For the expanding gas, \( \Delta U = -5 \mathrm{~J} \), because the work is done by the gas.
  • For the compressing gas, \( \Delta U = 80 \mathrm{~J} \), as work is applied onto the gas.
Work Done on/by Gas
Explaining the concept of work done in thermodynamics can be simplified by looking at how we measure energy changes. Work (\( W \)) is the energy transferred when a force is applied over a distance. In terms of gases, it relates to how the volume of the gas changes:
  • When a gas expands, it does work on the environment, consuming its own energy.
  • When a gas is compressed, work is done on it, increasing its internal energy.
For an adiabatic process:
- **Expansion (Process a):** - 5 J of work is done by the gas. - This energy use reduces the gas's internal energy by 5 J.- **Compression (Process b):** - 80 J of work is performed on the gas. - This input raises the gas's internal energy by 80 J.These scenarios help clarify that work affects energy states directly in the absence of heat exchange.

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Most popular questions from this chapter

A gas at a pressure of \(2.10 \times 10^{5}\) Pa occupies \(4.98 \times 10^{-3} \mathrm{~m}^{3}\) in a chamber that can change its volume. The gas is at an initial temperature of \(290 \mathrm{~K}\) when it is heated, so that it expands isobarically, thereupon doing \(200 \mathrm{~J}\) of work. Determine the new volume and the final temperature of the gas. [Hint: Use Eq. (20.1) and the Ideal Gas Law applied before and after the expansion.]

A \(2.0\) kg metal block \(\left(c=0.137 \mathrm{cal} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)\) is heated from \(15^{\circ} \mathrm{C}\) to \(90{ }^{\circ} \mathrm{C}\). By how much does its internal energy change?

A steam engine operating between a boiler temperature of 220 \({ }^{\circ} \mathrm{C}\) and a condenser temperature of \(35.0^{\circ} \mathrm{C}\) delivers \(8.00 \mathrm{hp}\). If its efficiency is \(30.0\) percent of that for a Carnot engine operating between these temperature limits, how many calories are absorbed each second by the boiler? How many calories are exhausted to the condenser each second? $$\text { Actual efficiency }=(0.30) \text { (Carnot efficiency) }=(0.300)\left(1-\frac{308 \mathrm{~K}}{493 \mathrm{~K}}\right)=0.113$$ We can determine the input heat from the relation for the efficiency Efficiency \(=\frac{\text { Output work }}{\text { Input heat }}\) and so every second $$\text { Input heat } / \mathrm{s}=\frac{\text { Output work } / \mathrm{s}}{\text { Efficiency }}=\frac{(8.00 \mathrm{hp})(746 \mathrm{~W} / \mathrm{hp})\left(\frac{1.00 \mathrm{cal} / \mathrm{s}}{4.184 \mathrm{~W}}\right)}{0.113}=12.7 \mathrm{kcal} / \mathrm{s}$$ To find the energy rejected to the condenser, we use the law of conservation of energy: Thus, $$ \begin{aligned} \text { Input energy } &=(\text { Output work })+(\text { Rejected energy }) \\\ \text { Rejected energy } / \mathrm{s} &=(\text { Input energy } / \mathrm{s})-(\text { Output work } / \mathrm{s}) \\ &=(\text { Input energy } / \mathrm{s})-(\text { Input energy } / \mathrm{s})(\text { Efficiency }) \\ &=(\text { Input energy } / \mathrm{s})[1-(\text { Efficiency })] \\ &=(12.7 \mathrm{kcal} / \mathrm{s})(1-0.113)=11.3 \mathrm{kcal} / \mathrm{s} \end{aligned} $$

Five moles of neon gas at \(2.00\) atm and \(27.0^{\circ} \mathrm{C}\) is adiabatically compressed to one-third its initial volume. Find the final pressure, final temperature, and external work done on the gas. For neon, \(\gamma\) \(=1.67, c_{v}=0.148 \mathrm{cal} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\), and \(M=20.18 \mathrm{~kg} / \mathrm{kmol} .\)

The specific heat of air at constant volume is \(0.175 \mathrm{cal} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C} .(a)\) By how much does the internal energy of \(5.0\) g of air change as it is heated from \(20^{\circ} \mathrm{C}\) to \(400{ }^{\circ} \mathrm{C} ?(b)\) Suppose that \(5.0\) g of air is adiabatically compressed so as to rise its temperature from \(20^{\circ} \mathrm{C}\) to \(400{ }^{\circ} \mathrm{C}\). How much work must be done on the air to compress it?
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