/*! 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 42 As an ideal gas is compressed is... [FREE SOLUTION] | 91Ó°ÊÓ

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As an ideal gas is compressed isothermally, the compressing agent does \(36 \mathrm{~J}\) of work on the gas. How much heat flows from the gas during the compression process?

Short Answer

Expert verified
-36 J of heat flows from the gas.

Step by step solution

01

Identify the Process

This is an isothermal process, which means the temperature of the gas remains constant throughout the compression.
02

Apply the First Law of Thermodynamics

The First Law of Thermodynamics states that the change in internal energy (\(\Delta U\)) of a system is equal to the heat added to the system (\(Q\)) minus the work done by the system (\(W\)). For isothermal processes of ideal gases, the change in internal energy is zero, \(\Delta U = 0\). The equation simplifies to \(0 = Q - W\).
03

Solve for Heat Flow

Since \(\Delta U = 0\), we have the relation \(Q = W\), where \(W\) is the work done by the gas. Given the problem states that \(36 \,\mathrm{J}\) is the work done on the gas, it implies the same amount of heat, \(-36 \,\mathrm{J}\), flows out of the gas (where \(Q = -W\)).
04

Translate Work Done on Gas to Work Done by Gas

In the context of work done 'on the gas' versus 'by the gas', when \(36 \,\mathrm{J}\) is done on the gas, it means \(-36 \,\mathrm{J}\) is the work done by it in an isothermal process, thereby releasing equal heat.

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Key Concepts

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

Ideal Gas
An ideal gas is a theoretical concept used in physics and chemistry to simplify the behavior of gases. This model assumes that the gas particles are small, randomly moving points with no volume and that they do not exert any forces on each other except during elastic collisions. These assumptions simplify calculations and provide a useful approximation of the behavior of real gases under many conditions.

The important properties of ideal gases include:
  • The pressure, volume, and temperature are related by the equation of state known as the ideal gas law: \( PV = nRT \), where \( P \) is pressure, \( V \) is volume, \( n \) is the number of moles, \( R \) is the ideal gas constant, and \( T \) is temperature in Kelvin.
  • Ideal gases follow the laws of thermodynamics perfectly, which makes them important in theoretical models such as isothermal and adiabatic processes.
The simplicity of the ideal gas model allows us to study complex thermodynamic processes with ease, serving as a foundational tool in understanding real-world gas behavior.
First Law of Thermodynamics
The First Law of Thermodynamics, also known as the Law of Energy Conservation, is crucial for understanding energy changes in a system. It states that the total energy contained within a closed system is constant. Energy can neither be created nor destroyed, only transformed from one form to another.

Mathematically, the First Law 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.
This principle serves as the bedrock for many thermodynamic processes, including isothermal processes, where the temperature is kept constant.

In such processes for an ideal gas, since the internal energy depends solely on temperature, any heat flow into or out of the system is exactly balanced by work done by or on the system, ensuring \( \Delta U = 0 \). This deeply relates the concepts of work and heat in a simple yet significant way.
Internal Energy
Internal energy refers to the total energy stored within a system, encompassing both potential and kinetic energy of its molecules. In the case of an ideal gas, internal energy is primarily a function of temperature because there are no intermolecular forces other than during collisions.

Key aspects include:
  • For an ideal gas, internal energy is directly proportional to the temperature. This denotes that a temperature change is synonymous with a change in internal energy.
  • During an isothermal process (constant temperature), the internal energy of an ideal gas remains unchanged, meaning \( \Delta U = 0 \).
  • Any heat exchange with the surroundings during an isothermal process results in a corresponding amount of work being done.
Thus, throughout isothermal compression of an ideal gas, even though work is done, the internal energy remains constant as all energy changes are balanced by heat transfer.
Heat Flow
Heat flow is the transfer of thermal energy from a hotter object to a cooler one. In thermodynamic processes, it's essential to understand how heat flow interacts with work and internal energy.

During an isothermal process involving an ideal gas, heat flow is tightly linked to work done:
  • Since temperature remains constant, any work done on the gas results in an equivalent amount of heat flowing out to maintain energy balance.
  • The equation \( Q = -W \) is used to express this relationship during isothermal processes. This indicates that the heat lost by the system equals the work done on the system.
  • In the given problem, when 36 J of work is done on the gas, an equal amount of heat flows out, maintaining the energy equilibrium.
Heat flow ensures that the gas maintains its initial temperature even during compression or expansion under an isothermal condition, illustrating the interplay between heat, work, and internal energy in thermodynamic systems.

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Most popular questions from this chapter

Molecular hydrogen gas having a mass of \(6.44 \mathrm{~g}\) at \(26.0^{\circ} \mathrm{C}\) is heated until its volume doubles while it is held at a constant pressure. How much work was done by the gas? [Hint: Take it to be an ideal gas.]

A gas does \(100.0 \mathrm{~J}\) of work while receiving \(110.0 \mathrm{~J}\) heat. What is the resulting change in the gas's internal energy?

For nitrogen gas, \(c_{v}=740 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\). Assuming it to behave like an ideal gas, find its specific heat at constant pressure. (The molecular mass of nitrogen gas is \(28.0 \mathrm{~kg} / \mathrm{kmol} .\) ) Method 1 $$c_{p}=c_{v}+\frac{R}{M}=\frac{740 \mathrm{~J}}{\mathrm{~kg} \cdot \mathrm{K}}+\frac{8314 \mathrm{~J} / \mathrm{kmol} \cdot \mathrm{K}}{28.0 \mathrm{~kg} / \mathrm{kmol}}=1.04 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}$$ Method 2 Since \(\mathrm{N}_{2}\) is a diatomic gas, and since \(\gamma=c_{p} / c_{v}\) for such a gas, $$ c_{p}=1.40 c_{v}=1.40(740 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K})=1.04 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K} $$

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?

Find \(\Delta W\) and \(\Delta U\) for a \(6.0\) -cm cube of iron as it is heated from 20 \({ }^{\circ} \mathrm{C}\) to \(300{ }^{\circ} \mathrm{C}\) at atmospheric pressure. For iron, \(c=0.11 \mathrm{cal} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\) and the volume coefficient of thermal expansion is \(3.6 \times 10^{-5}{ }^{\circ} \mathrm{C}\) \- 1 . The mass of the cube is 1700 g. Given that \(\Delta T=300^{\circ} \mathrm{C}-20^{\circ} \mathrm{C}=280^{\circ} \mathrm{C}\), \(\Delta Q=c m \Delta T=\left(0.11 \mathrm{cal} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)(1700 \mathrm{~g})\left(280^{\circ} \mathrm{C}\right)=52 \mathrm{kcal}\) To find that the work done by the expansion of the cube, we need to determine \(\Delta V\). The volume of the cube is \(V=(6.0 \mathrm{~cm})^{3}=216 \mathrm{~cm}^{3} .\) Using \((\Delta V) / V\) \(=\beta \Delta T\), $$\begin{array}{l} \Delta V=V \beta \Delta T=\left(216 \times 10^{-6} \mathrm{~m}^{3}\right)\left(3.6 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1}\right)\left(280^{\circ} \mathrm{C}\right)=2.18 \times 10^{-6} \mathrm{~m}^{3} \end{array}$$ But the First Law tells us that $$\begin{aligned} \Delta U &=\Delta Q-\Delta W=(52000 \mathrm{cal})(4.184 \mathrm{~J} / \mathrm{cal})-0.22 \mathrm{~J} \\ &=218000 \mathrm{~J}-0.22 \mathrm{~J} \approx 2.2 \times 10^{5} \mathrm{~J} \end{aligned}$$ Notice how very small the work of expansion against the atmosphere is in comparison to \(\Delta U\) and \(\Delta Q .\) Often \(\Delta W\) can be neglected when dealing with liquids and solids.

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