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Chapter 3: Problem 93

A monoatomic gas 'A' and a diatomic gas ' \(\mathrm{B}\) ', both initially at the same temperature and pressure are compressed adiabatically from a volume \(\mathrm{V}\) to \(\mathrm{V} / 2\). The gas which has higher temperature is (a) \(\mathrm{B}\) (b) \(\mathrm{A}\) (c) both have same temperature (d) can not be said

Short Answer

Expert verified
Gas A has a higher final temperature than gas B after adiabatic compression.

Step by step solution

01

Identify Adiabatic Process Formula

In an adiabatic process, the formula that relates pressure \( P \), volume \( V \), and temperature \( T \) with the ratio of specific heats \( \gamma \) is given by:\[ TV^{\gamma-1} = \text{constant} \]We will use this formula to compare the final temperatures of monoatomic gas \( A \) and diatomic gas \( B \) after adiabatic compression from \( V \) to \( V/2 \).
02

Determine the Specific Heat Ratios

For a monoatomic gas \( A \), the ratio of specific heats \( \gamma_A \) is 5/3. For a diatomic gas \( B \), the ratio of specific heats \( \gamma_B \) is 7/5. Knowing these values will help us calculate the final temperatures.
03

Use Adiabatic Relation for Gas A

Apply the adiabatic process formula for monoatomic gas \( A \):\[ T_{A2} V_{A2}^{\gamma_A - 1} = T_{A1} V_{A1}^{\gamma_A - 1} \]Plug in \( V_{A2} = V/2 \) and \( V_{A1} = V \):\[ T_{A2} (V/2)^{5/3 - 1} = T_{A1} V^{5/3 - 1} \]Rearranging will give an expression for the final temperature \( T_{A2} \).
04

Use Adiabatic Relation for Gas B

Similarly, apply the adiabatic process formula for diatomic gas \( B \):\[ T_{B2} V_{B2}^{\gamma_B - 1} = T_{B1} V_{B1}^{\gamma_B - 1} \]With \( V_{B2} = V/2 \) and \( V_{B1} = V \):\[ T_{B2} (V/2)^{7/5 - 1} = T_{B1} V^{7/5 - 1} \]Rearranging this will give an expression for the final temperature \( T_{B2} \).
05

Compare Final Temperatures of A and B

Compare the expressions obtained for \( T_{A2} \) and \( T_{B2} \). Since \( \gamma_A - 1 = 2/3 \) and \( \gamma_B - 1 = 2/5 \), this makes the exponent for gas \( A \) smaller, leading to a larger temperature increase because \( V/2 < V \). Thus, \( T_{A2} > T_{B2} \).

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Key Concepts

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

Monoatomic Gas
A monoatomic gas is made up of atoms that do not bond with others, existing individually. These gases are simpler in structure than their diatomic counterparts, which translates to certain distinct properties.
For example, when such gases undergo an adiabatic process鈥攚here no heat is exchanged with the environment鈥攖heir behavior can be predicted using thermodynamic principles.
  • In the given problem, the monoatomic gas A undergoes adiabatic compression.
  • Its specific heat ratio, represented by gamma (\( \gamma \)), is crucial for determining how temperature changes.
The specific heat ratio for a monoatomic gas is typically 5/3, compared to diatomic gases where it's lower.
Diatomic Gas
Diatomic gases consist of molecules made up of two atoms. Common examples include oxygen \( O_2 \) and nitrogen \( N_2 \), and they generally have more complex interactions due to their molecular structure.
This complexity affects their responses in thermodynamic processes such as compression or expansion.
  • As diatomic gases compress adiabatically, they exhibit different temperature changes compared to monoatomic gases.
  • They have a different specific heat ratio (\( \gamma \)), which is often 7/5.
The presence of two atoms allows diatomic gases to share energy in more ways than monoatomic gases, influencing how their temperature varies under various conditions.
Specific Heat Ratio
The specific heat ratio (\( \gamma \)) is a key parameter in understanding how gases behave under adiabatic processes. It is defined as the ratio of specific heat at constant pressure (\( C_p \)) to specific heat at constant volume (\( C_v \)).
  • For a monoatomic gas, \( \gamma \) is 5/3.
  • For a diatomic gas, \( \gamma \) is 7/5.
This ratio affects how the temperature changes when the gas expands or compresses without heat exchange.
The smaller the value of \( \gamma - 1 \), the larger the temperature increase, given the same volume change. This is why, in the exercise, the monoatomic gas increases more in temperature compared to the diatomic gas when both are compressed to a smaller volume.
Temperature Comparison
Comparing the temperatures of compressed gases helps in practical applications such as engines and refrigerators. When both gases A and B are compressed to half their initial volumes, the differences in their temperature changes become apparent.
  • The change is more pronounced in monoatomic gas than in diatomic gas, thanks to its higher specific heat ratio.
  • The formula \( TV^{\gamma-1} \) = constant provides insights into the temperature comparison.
Through this comparison, one can conclude that the final temperature \( T_{A2} \) for the monoatomic gas A is higher than \( T_{B2} \) for the diatomic gas B after the same adiabatic compression.
Compression in Thermodynamics
Compression is a key process in thermodynamics, transforming the state variables of a gas. Adiabatic compression is particularly significant because it involves reducing volume while keeping the system insulated so no heat is lost or gained externally.
Due to the conservation of energy, the work done on the gas increases its internal energy, manifesting as a temperature rise.
  • In both gases A and B, initial states are the same, but due to different \( \gamma \), their final temperatures differ after compression.
  • The specific formula used in such cases is \( TV^{\gamma-1} = \text{constant} \).
Understanding how these principles affect temperature changes during compression helps in designing efficient systems like engines where control over temperature and pressure conditions is crucial.

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Most popular questions from this chapter

At \(25^{\circ} \mathrm{C}\) and \(730 \mathrm{~mm}\) pressure, \(380 \mathrm{~mL}\) of dry oxygen was collected. If the temperature is constant, what volume will the oxygen occupy at \(760 \mathrm{~mm}\) pressure? (a) \(365 \mathrm{~mL}\) (b) \(2 \mathrm{~mL}\) (c) \(10 \mathrm{~mL}\) (d) \(20 \mathrm{~mL}\)

\(10 \mathrm{~mL}\) of propane are mixed with \(70 \mathrm{~mL}\) of oxygen and exploded. What would be the volume of residual gases after explosion and after exposure of the residual gases to alkali? All volume measurements are made at the same temperature and pressure. (a) \(55 \mathrm{~mL}, 25 \mathrm{~mL}\) (b) \(40 \mathrm{~mL}, 15 \mathrm{~mL}\) (c) \(50 \mathrm{~mL}, 20 \mathrm{~mL}\) (d) \(35 \mathrm{~mL}, 40 \mathrm{~mL}\)

Mark the correct statements (a) At low pressure, the van der Waal's equation is written as $$ \left(\mathrm{P}+\frac{\mathrm{a}}{\mathrm{V}^{2}}\right) \times \mathrm{V}=\mathrm{RT} $$ (b) When \(\mathrm{Z}>1\), at \(\mathrm{STP}, \mathrm{V}_{\text {real }}>\mathrm{V}_{\text {ideal }}\) (c) Mean free path of \(\mathrm{O}_{2}\) is greater than that of \(\mathrm{H}_{2}\). (d) At \(273 \mathrm{~K}\), the total kinetic energy of \(\mathrm{O}_{2}\) will be eight times that of one mole of He.

A closed vessel contains equal number of molecules of \(\mathrm{O}_{2}\) and \(\mathrm{H}_{2} .\) Consider the following statements: 1\. The average speed of hydrogen molecules will be greater. 2\. Weight of hydrogen is \(1 / 8\) th of the weight of oxygen. 3\. Hydrogen molecules strike the walls more often. 4\. The two gases have different average energies. The statements are wrong in (a) 2 and 4 (b) 2 and 3 (c) 1 and 4 (d) 1 and 3

An ideal gas obeying kinetic gas equation can be liquefied if (a) it cannot be liquefi ed at any value of \(\mathrm{P}\) and \(\mathrm{T}\) (b) its temperature is more than Boyle's temperature (c) its temperature is more than critical temperature (d) its pressure is more than critical pressure
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