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\((a)\) A monatomic ideal gas initially at \(19.0^{\circ} \mathrm{C}\) is suddenly compressed to one-tenth its original volume. What is its temperature after compression? (b) Make the same calculation for a diatomic gas.

Short Answer

Expert verified
The final temperature for a monatomic gas is approximately 493 K. For a diatomic gas, it's approximately 553 K.

Step by step solution

01

Convert Temperature to Kelvin

The initial temperature is given in Celsius. It needs to be converted to Kelvin, as this is the unit of temperature in the International System of Units (SI). The conversion can be done using the formula \(T(K) = T(C) + 273.15\). Thus, \(T_0 = 19.0C + 273.15 = 292.15 K\).
02

Calculate the Final Temperature for a Monatomic Gas

The adiabatic equation \(T_{0}V_{0}^{\gamma - 1} = T_{1}V_{1}^{\gamma - 1}\) should be used to find the final temperature for a monatomic gas. Given the initial volume \(V_0\) is ten times the final volume \(V_1\), we can write the equation as \(T_{0}10^{\gamma - 1} = T_{1}\). Solving this equation for \(T_1\) yields \(T_1 = T_{0}10^{\gamma - 1}\). For a monatomic gas, \(\gamma = 5/3\). Substituting \(\gamma\) and \(T_0 = 292.15K\) into the equation yields the final temperature \(T_{1} = 292.15 K \times 10^{5/3 - 1} \approx 493 K\).
03

Calculate the Final Temperature for a Diatomic Gas

Use the same process for a diatomic gas. Only the value of \(\gamma\) changes to \(7/5\) in case of a diatomic gas. Substituting \(\gamma = 7/5\) and \(T_0 = 292.15K\) into the equation yields \(T_{1} = 292.15 K \times 10^{7/5 - 1}\), or approximately \(T_{1} \approx 553 K\).

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Key Concepts

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

Monatomic Ideal Gas
A monatomic ideal gas is a simple conceptual model used in thermodynamics. It consists of gas particles that are single atoms, like helium or neon. These atoms have no molecular bonds and move freely.
Because the gas is monatomic, it has a smaller heat capacity compared to complex gases.
  • Fluctuations in this type of gas are predictable when undergoing processes like adiabatic compression.
  • The adiabatic index, represented as \( \gamma \), is calculated as \( \frac{5}{3} \) for a monatomic gas. This index plays a crucial role in determining how temperature changes during volume changes without heat exchange.
The concept helps us understand basic thermodynamic properties and behaviors of gases through theoretical explorations.
Diatomic Ideal Gas
Diatomic ideal gases consist of molecules that have two atoms. Common examples include oxygen (\(O_2\)) and nitrogen (\(N_2\)). These gases are more complex than monatomic gases due to their molecular structure.
In thermodynamic calculations, diatomic gases have different properties:
  • They have higher specific heat capacities because the molecules have more degrees of freedom than monatomic gases.
  • The adiabatic index \( \gamma \) for diatomic gases is usually \( \frac{7}{5} \), reflecting these additional freedoms and complexity.
Understanding diatomic gases is essential for working with real-world gases, where such molecular structures are common.
Temperature Conversion
Converting temperatures from Celsius to Kelvin is essential for many scientific calculations. Kelvin is the SI unit of temperature and is often used in equations dealing with temperature, since it begins from absolute zero.
The formula for conversion is straightforward:
  • \( T(K) = T(°C) + 273.15 \)
  • For example, an initial temperature of 19°C converts to 292.15K.
This conversion ensures consistency and accuracy in thermodynamic calculations, enabling precise evaluations in processes such as adiabatic compression.
Adiabatic Compression
In adiabatic processes, there is no exchange of heat with the surroundings. Adiabatic compression refers to increasing a gas's pressure by reducing its volume, leading to a rise in temperature without heat transfer.
This can be described using the equation:
  • \( T_{0}V_{0}^{\gamma - 1} = T_{1}V_{1}^{\gamma - 1} \), where \( T \) is temperature, \( V \) is volume, and \( \gamma \) is the adiabatic index.
For both monatomic and diatomic gases, this equation allows us to calculate the new temperature after compression, based on their specific \( \gamma \) values. This principle is fundamental in understanding thermodynamic cycles and applications like engines and refrigeration.

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Most popular questions from this chapter

Gas occupies a volume of \(4.33 \mathrm{~L}\) at a pressure of \(1.17 \mathrm{~atm}\) and a temperature of \(310 \mathrm{~K}\). It is compressed adiabatically to a volume of \(1.06 \mathrm{~L}\). Determine \((a)\) the final pressure and \((b)\) the final temperature, assuming the gas to be an ideal gas for which \(\gamma=1.40 .(c)\) How much work was done on the gas?

The average rate at which heat flows out through the surface of the Earth in North America is \(54 \mathrm{~mW} / \mathrm{m}^{2}\) and the average thermal conductivity of the near surface rocks is \(2.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Assuming a surface temperature of \(10^{\circ} \mathrm{C}\), what should be the temperature at a depth of \(33 \mathrm{~km}\) (near the base of the crust)? Ignore the heat generated by radioactive elements; the curvature of the Earth can also be ignored.

(a) Two \(50-\mathrm{g}\) ice cubes are dropped into \(200 \mathrm{~g}\) of water in a glass. If the water were initially at a temperature of \(25^{\circ} \mathrm{C}\), and if the ice came directly from a freezer at \(-15^{\circ} \mathrm{C}\), what is the final temperature of the drink? ( \(b\) ) If only one ice cube had been used in (a), what would be the final temperature of the drink? Neglect the heat capacity of the glass.

A certain substance has a molar mass of \(51.4 \mathrm{~g} / \mathrm{mol}\). When \(320 \mathrm{~J}\) of heat are added to a 37.1-g sample of this material, its temperature rises from \(26.1\) to \(42.0^{\circ} \mathrm{C}\). \((a)\) Find the specific heat of the substance. (b) How many moles of the substance are present? (c) Calculate the molar heat capacity of the substance.

a) Calculate the rate at which body heat flows out through the clothing of a skier, given the following data: the body surface area is \(1.8 \mathrm{~m}^{2}\) and the clothing is \(1.2 \mathrm{~cm}\) thick; skin surface temperature is \(33^{\circ} \mathrm{C}\), whereas the outer surface of the clothing is at \(1.0^{\circ} \mathrm{C} ;\) the thermal conductivity of the clothing is \(0.040 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K} .(b)\) How would the answer change if, after a fall, the skier's clothes become soaked with water? Assume that the thermal conductivity of water is \(0.60 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\)

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