Question

The isotherms of an ideal gas at $ T_1, T_2, T_3 $ along with their slopes (m) (in the brackets) are shown here. If $ T_1 > T_2 > T_3 $, then the correct order of slopes of these isotherms is:

• A. \( m_2>m_1>m_3 \)
• B. \( m_3>m_2>m_1 \)
• C. \( m_2>m_3>m_1 \)
• D. \( m_1>m_2>m_3 \)

Hint:

The slope of an isotherm in an ideal gas increases with temperature. At higher temperatures, molecules have higher energy and exert more pressure at the same volume.

Answer 1

The Correct Option is D

Step 1: Understanding the relationship between isotherms and slopes.
For an ideal gas, the equation of state is given by the ideal gas law: \[ pV = nRT \] This implies that for a given temperature \(T\), the relationship between pressure \(p\) and volume \(V\) is inversely proportional.
The slope of the isotherm, denoted as \(m\), is given by: \[ m = \left( \frac{dP}{dV} \right)_T \] At higher temperatures, the gas molecules have higher kinetic energy, leading to a steeper slope of the isotherm. Therefore, for \( T_1>T_2>T_3 \), the isotherm at \( T_1 \) will have the steepest slope, followed by the isotherm at \( T_2 \), and the isotherm at \( T_3 \) will have the smallest slope.
Step 2: Analyzing the slopes.
Since \( T_1>T_2>T_3 \), the slope order will be: \[ m_1>m_2>m_3 \] Thus, the correct order of the slopes of these isotherms is \( m_1>m_2>m_3 \).