Problem 93 - A Complete Resource Book in Chemistry for JEE Main

Question

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

Step-by-Step Solution

Answer

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

Step 1: 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 \( V/2 \).

Step 2: 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.

Step 3: 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 \( T_{A2} \).

Step 4: 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 \( T_{B2} \).

Step 5: 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, \( T_{A2} > T_{B2} \).

Key Concepts

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—where no heat is exchanged with the environment—their 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.