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Heat is added to two identical samples of a monatomic ideal gas. In the first sample the heat is added while the volume of the gas is kept constant, and the heat causes the temperature to rise by \(75 \mathrm{~K}\). In the second sample, an identical amount of heat is added while the pressure (but not the volume) of the gas is kept constant. By how much does the temperature of this sample increase?

Short Answer

Expert verified
The temperature of the second sample increases by 45 K.

Step by step solution

01

Understand the Problem

We are given two scenarios with identical monatomic ideal gas samples. In the first scenario, heat is added at constant volume, causing a temperature increase of \(75 \mathrm{~K}\). In the second scenario, the same amount of heat is added but under constant pressure, and we need to find the temperature increase.
02

Describe the Behavior of Monatomic Ideal Gas

For a monatomic ideal gas, specific heat at constant volume \(C_v\) is \(\frac{3}{2} R\) and at constant pressure \(C_p\) is \(\frac{5}{2} R\), where \(R\) is the gas constant. The relationship \(C_p = C_v + R\) holds.
03

Apply Heat Capacity Formula at Constant Volume

For a process at constant volume, the heat added \(Q\) relates to temperature change \(\Delta T_v\) as \(Q = n C_v \Delta T_v\). Here, \(\Delta T_v = 75 \mathrm{~K}\). Thus, \(Q = n \frac{3}{2} R \times 75\).
04

Relate Heat Added to Increase in Temperature at Constant Pressure

For a process at constant pressure, the same amount of heat \(Q\) affects the temperature change \(\Delta T_p\) such that \(Q = n C_p \Delta T_p\). Substitute \(Q = n \frac{3}{2} R \times 75\) and \(C_p = \frac{5}{2} R\) into the equation, resulting in \(n \frac{3}{2} R \times 75 = n \frac{5}{2} R \Delta T_p\).
05

Calculate the Temperature Change at Constant Pressure

Cancel \(n\) and \(R\) from both sides since they appear in both terms, and solve for \(\Delta T_p\):\[\frac{3}{2} \times 75 = \frac{5}{2} \Delta T_p \]Solve for \(\Delta T_p\):\[\Delta T_p = \frac{3}{2} \times 75 \times \frac{2}{5} = 45 \mathrm{~K}\]
06

Conclusion

Hence, the temperature of the second sample increases by \(45 \mathrm{~K}\) when heat is added at constant pressure.

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Key Concepts

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

Ideal Gas Law
The Ideal Gas Law is a simple equation that describes the behavior of ideal gases. This law is useful in understanding how pressure, volume, and temperature of a gas are related. The equation is given by \[ PV = nRT \]where:
  • \( P \) is the pressure of the gas
  • \( V \) is the volume of the gas
  • \( n \) is the number of moles
  • \( R \) is the gas constant, approximately \( 8.314 \frac{J}{mol\cdot K} \)
  • \( T \) is the temperature in Kelvin
This relationship helps in predicting how a gas will behave when subjected to changes in temperature, volume, or pressure, assuming the gas behaves ideally. In our context, as heat is added at constant pressure, the volume may change according to this law, affecting the temperature of the gas.
Specific Heat Capacity
Specific heat capacity is a measure of how much heat energy is required to change the temperature of a substance. For gases, we must consider the conditions under which the heat is added—specifically, whether it's at constant volume or constant pressure. Specific heat capacity at constant volume is denoted as \( C_v \), and at constant pressure as \( C_p \). For monatomic ideal gases:
  • \( C_v = \frac{3}{2} R \)
  • \( C_p = \frac{5}{2} R \)
The relationship between these is known as Mayer's relation, which states:\[ C_p = C_v + R \]This tells us that it takes more heat to change the temperature of a gas at constant pressure than at constant volume, influencing how we calculate the temperature changes for different processes.
Monatomic Gas
A monatomic gas consists of single atoms as its basic particles. Noble gases like helium and argon are common examples. These gases have unique characteristics in thermodynamics, especially in processes involving heat exchange. Monatomic gases have simpler molecular structures compared to diatomic or polyatomic gases. This affects their specific heat capacities, as there are fewer degrees of freedom for energy storage. Therefore, their
  • Specific heat capacity at constant volume \(C_v\) is \(\frac{3}{2} R\)
  • Specific heat capacity at constant pressure \(C_p\) is \(\frac{5}{2} R\)
Understanding these values helps when applying concepts in problems involving temperature changes during different processes.
Constant Pressure Process
In a constant pressure process, the pressure of the gas remains unchanged even as it undergoes a temperature alteration due to heat addition or removal. Whereas the volume of the gas changes to compensate for this. Such processes in thermodynamics are used to analyze how heat addition affects gas behavior, chiefly temperature. In our case, though the same amount of heat is applied, the temperature rise is different than in constant volume. This is because the specific heat capacity at constant pressure, given by \(C_p = \frac{5}{2} R\), is higher than at constant volume. Therefore, the temperature increase is calculated using:\[ Q = n C_p \Delta T_p \]Knowing how these processes differ helps in understanding heat exchange accurately, especially in practical applications like engines and refrigerators.

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Most popular questions from this chapter

The hot reservoir for a Carnot engine has a temperature of \(890 \mathrm{~K},\) while the cold reservoir has a temperature of \(670 \mathrm{~K}\). The heat input for this engine is \(4800 \mathrm{~J}\). The \(670-\mathrm{K}\) reservoir also serves as the hot reservoir for a second Carnot engine. This second engine uses the rejected heat of the first engine as input and extracts additional work from it. The rejected heat from the second engine goes into a reservoir that has a temperature of \(420 \mathrm{~K}\). Find the total work delivered by the two engines.

Heat is added isothermally to \(2.5\) mol of a monatomic ideal gas. The temperature of the gas is \(430 \mathrm{~K}\). How much heat must be added to make the volume of the gas double?

Concept Questions (a) A real (irreversible) engine operates between hot and cold reservoirs whose temperatures are \(T_{\mathrm{H}}\) and \(T_{\mathrm{C}}\). The engine absorbs heat of magnitude \(Q_{\mathrm{H}} \mid\) from the hot reservoir and performs work of magnitude \(|W| .\) What is the expression that gives the change in entropy \(\Delta S_{\text {universe }}\) of the universe associated with the operation of this engine? Express your answer in terms of \(T_{\mathrm{H}}, T_{\mathrm{C}},\left|Q_{\mathrm{H}}\right|,\) and \(\left|Q_{\mathrm{C}}\right|\) Would you expect the change in the entropy of the universe to be greater than, less than, or equal to zero? Provide a reason for your answer. (c) Suppose that a reversible engine (a Carnot engine) operates between the same hot and cold temperatures as the irreversible engine described above. For the same input heat \(\left|Q_{\mathrm{H}}\right|,\) would the reversible engine produce more, less, or the same work as the irreversible engine? Justify your answer. (d) In terms of \(\Delta\) S universe and \(T_{\mathrm{C}}\), how could one determine the difference, if any, between the work done by the reversible engine and that done by the irreversible engine? Problem (a) An irreversible engine operates between temperatures of 852 and \(314 \mathrm{~K}\). It absorbs \(1285 \mathrm{~J}\) of heat from the hot reservoir and does \(264 \mathrm{~J}\) of work. What is the change \(\Delta S_{\text {universe }}\) in the entropy of the universe associated with the operation of this engine? (b) If the engine were reversible, what is the magnitude \(|W|\) of the work it would have done, assuming it operated between the same temperatures and absorbed the same heat as the irreversible engine? (c) Using the results of parts (a) and (b), find the difference between the work produced by the reversible and irreversible engines.

When one gallon of gasoline is burned in a car engine, \(1.19 \times 10^{8} \mathrm{~J}\) of internal energy is released. Suppose that \(1.00 \times 10^{8} \mathrm{~J}\) of this energy flows directly into the surroundings (engine block and exhaust system) in the form of heat. If \(6.0 \times 10^{5} \mathrm{~J}\) of work is required to make the car go one mile, how many miles can the car travel on one gallon of gas?

From a hot reservoir at a temperature of \(\mathrm{T}_{1},\) Carnot engine A takes an input heat of 5550 \(\mathrm{J},\) delivers \(1750 \mathrm{~J}\) of work, and rejects heat to a cold reservoir that has a temperature of \(503 \mathrm{~K}\). This cold reservoir at \(503 \mathrm{~K}\) also serves as the hot reservoir for Carnot engine \(\mathrm{B}\) which uses the rejected heat of the first engine as input heat. Engine \(\mathrm{B}\) also delivers 1750 J of work, while rejecting heat to an even colder reservoir that has a temperature of \(T_{2}\). Find the temperatures (a) \(T_{1}\) and (b) \(T_{2}\).

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