Chapter 19: Problem 7
Suppose \(1.80 \mathrm{~mol}\) of an ideal gas is taken from a volume of \(3.00 \mathrm{~m}^{3}\) to a volume of \(1.50 \mathrm{~m}^{3}\) via an isothermal compression at \(30^{\circ} \mathrm{C}\). (a) How much energy is transferred as heat during the compression, and (b) is the transfer to or from the gas?
Short Answer
Step by step solution
Understanding the Problem
Applying the Ideal Gas Law
Calculating Work Done
Calculating Specific Values
Evaluating Heat Transfer
Direction of Heat Transfer
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Isothermal Process
For an ideal gas in an isothermal process, the relationship is given by the equation \( PV = nRT \), where \( P \) is pressure, \( V \) is volume, \( n \) is the number of moles, \( R \) is the universal gas constant, and \( T \) is the absolute temperature. Even though pressure and volume may change, the product \( PV \) remains constant due to \( T \) remaining unchanged.
Isothermal compression, such as the scenario in the exercise, often involves a decrease in volume, which in turn means that pressure must increase to maintain the constant temperature provided that the amount of gas remains constant. Understanding this helps us analyze cycle engines and refrigerators, which heavily depend on isothermal processes.
Thermodynamics
One important principle in thermodynamics is the First Law: the conservation of energy, which states that energy cannot be created or destroyed in an isolated system. Instead, it only transforms from one type to another. In mathematical terms, 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.
In isothermal processes concerning ideal gases, the internal energy change \( \Delta U \) is zero since the temperature remains constant by definition. Therefore, the First Law simplifies to \( Q = W \). This means that all the heat transferred to the system is used to do work, or vice versa.
Work Done by Gas
During an isothermal compression, like the problem in the exercise, calculating the work done on the gas is straightforward using the formula: \[ W = nRT \ln \left( \frac{V_2}{V_1} \right) \]This formula directly links the amount of effort needed to compress or expand a gas under constant temperature conditions with the number of moles, the absolute temperature, and the initial and final volumes.
The sign of the work can tell us about the nature of the energy exchange. In the exercise, since the volume decreases, \( W \) is negative, indicating the work is done on the gas. Thus, energy is taken out as heat, which aligns with our understanding that compressing a gas involves doing work on it, thereby releasing energy out of the system as heat.