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(I) A gas is enclosed in a cylinder fitted with a light frictionless piston and maintained at atmospheric pressure. When \(1250 \mathrm{kcal}\) of heat is added to the gas, the volume is observed to increase slowly from \(12.0 \mathrm{~m}^{3}\) to \(18.2 \mathrm{~m}^{3}\). Calculate \((a)\) the work done by the gas and \((b)\) the change in internal energy of the gas.

Short Answer

Expert verified
Work done by the gas: \(6.2786 \times 10^5 \, \text{J}\); Change in internal energy: \(4.60214 \times 10^6 \, \text{J}\).

Step by step solution

01

Understand the Given Information

We have a gas in a cylinder with an initial volume of \(V_i = 12.0 \, \text{m}^3\) and a final volume of \(V_f = 18.2 \, \text{m}^3\). It is maintained at atmospheric pressure, which we will take as \(P = 1.013 \, \text{bar} = 1.013 \times 10^5 \, \text{Pa}\). When \(1250 \, \text{kcal}\) of heat is added, it converts to \(Q = 1250 \, \text{kcal} = 1250 \times 4.184 \times 10^3 \, \text{J}\) (since 1 kcal = 4184 J).
02

Calculate the Work Done by the Gas

The work done by the gas, \(W\), is given by the formula \(W = P \Delta V\), where \(\Delta V = V_f - V_i\). First, calculate the change in volume: \(\Delta V = 18.2 \, \text{m}^3 - 12.0 \, \text{m}^3 = 6.2 \, \text{m}^3\). Now, calculate the work: \[ W = 1.013 \times 10^5 \, \text{Pa} \times 6.2 \, \text{m}^3 = 6.2786 \times 10^5 \, \text{J} \].
03

Determine the Change in Internal Energy

Use the First Law of Thermodynamics, which states that \(\Delta U = Q - W\), where \(\Delta U\) is the change in internal energy, \(Q\) is the heat added, and \(W\) is the work done. Substitute the known values: \( Q = 1250 \times 4.184 \times 10^3 \, \text{J} = 5.23 \times 10^6 \, \text{J}\), and \(W = 6.2786 \times 10^5 \, \text{J}\). Calculate the change in internal energy: \[ \Delta U = 5.23 \times 10^6 \, \text{J} - 6.2786 \times 10^5 \, \text{J} = 4.60214 \times 10^6 \, \text{J} \].

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Key Concepts

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

First Law of Thermodynamics
The first law of thermodynamics is a core principle that explains how energy operates within a system. It's essentially a statement of energy conservation, meaning energy cannot be created or destroyed. This law can be expressed in the simple formula \( \Delta U = Q - W \), where:
  • \( \Delta U \) represents the change in the internal energy of the system.
  • \( Q \) is the heat added to the system.
  • \( W \) is the work done by the system.

In an intuitive sense, the law tells us that any heat energy put into a system will either increase its internal energy or be used to do work. If you add heat to a gas, it can make the particles move faster (increasing internal energy), or it can push the walls of its container, doing work.
This principle is what makes engines, refrigerators, and all sorts of thermodynamic devices work. It's versatile and applies to a wide range of physical systems, from the air in a balloon to the stars in the sky.
Work Done by Gas
When a gas expands against an external pressure, it performs work. The concept of work done by gas is central in thermodynamics because it represents one way energy is transferred. In this context, work done by gas is calculated using the formula \( W = P \Delta V \), where:
  • \( W \) is the work done by the gas.
  • \( P \) is the constant external pressure against which the gas is expanding.
  • \( \Delta V \) is the change in volume of the gas.

In our exercise, the initial and final volumes of gas were given, which allowed us to find the change in volume \( \Delta V \). By multiplying this change with the atmospheric pressure, we calculated the work done during the expansion. This work is a sign that energy has moved from the gas to something else, like a piston being pushed up, which is useful in machines like engines.
It's important to remember that when a gas does work on its surroundings, its internal energy decreases if no extra heat is provided to make up for the energy expended in doing work.
Change in Internal Energy
The change in internal energy of a system is a concept that describes how the internal energy of a system changes due to heat transfer and work done. In our problem, the First Law of Thermodynamics helps in determining this change.
To find the change in internal energy, \( \Delta U \), we use the expression \( \Delta U = Q - W \). This formula tells us that the change in internal energy is the difference between the heat added to the system and the work done by the system.
By calculating \( Q \) (heat added) and \( W \) (work done), we can plug these values into our formula to find \( \Delta U \). A positive \( \Delta U \) indicates an increase in the gas's internal energy, meaning the particles within have gained kinetic energy. A negative value would mean a loss in internal energy.
  • More heat added than work done results in increased internal energy.
  • More work done than heat added results in decreased internal energy.

In real-world terms, knowing how the internal energy changes helps us understand states of matter and energy efficiency in systems like engines, where fuel's potential energy is converted into useful work.

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Most popular questions from this chapter

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