Problem 65 - A Complete Resource Book in Chemistry for JEE Main

Question

2 moles of an ideal gas is expanded isothermally and reversibly from 1 litre of 10 litre at \(300 \mathrm{~K}\). The enthalpy change (in \(\mathrm{kJ}\) ) for the process is (a) \(11.4 \mathrm{~kJ}\) (b) \(-11.4 \mathrm{~kJ}\) (c) \(0 \mathrm{~kJ}\) (d) \(4.8 \mathrm{~kJ}\).

Step-by-Step Solution

Answer

The enthalpy change for the process is 0 kJ (option c).

Step 1: Understanding Isothermal Expansion

For an isothermal process, the temperature remains constant throughout the expansion. In our case, the temperature is given as \(300 \mathrm{~K}\). When an ideal gas undergoes an isothermal expansion or compression, the change in internal energy \(\Delta U\) is zero, because internal energy of an ideal gas is a function of temperature only.

Step 2: Relation Between Enthalpy and Internal Energy

The change in enthalpy \(\Delta H\) for an ideal gas process is given by \(\Delta H = \Delta U + \Delta (pV)\). Since the process is isothermal, \(\Delta U = 0\), and since it's a reversible expansion, the pressure and volume changes are interrelated.

Step 3: Calculate Change in Enthalpy

For isothermal processes, the formula \(\Delta H = nC_v \Delta T\) is used. Here, the change in temperature \(\Delta T = 0\) since the process is carried out at a constant temperature, hence \(\Delta H = 0\). This means there is no enthalpy change during the isothermal, reversible expansion of an ideal gas.

Key Concepts

Enthalpy Change

In thermodynamics, enthalpy change (,s an important concept that helps us understand the energy exchange in chemical processes and physical transformations. It is represented by the symbol , and it refers to the total heat content of a system. When a system undergoes a change, such as an expansion or compression, the enthalpy change ) is determined.

For an ideal gas, the change in enthalpy () is strongly linked to the system's temperature change, given by the formula:

• j cv 

where cv_7iC_vmtalpy chang is simpler to determine because for an isothermal process, the temperature does not vary. In such a scenario, since the temperature ( = 0) , we esumbe

Understanding this relationship between temperature and enthalpy can help us comprehend why there's no change in enthalpy during an isothermal process.

Ideal Gas

An ideal gas is a theoretical concept that simplifies how gases behave under different conditions. Essential to this concept are several assumptions:

• Molecules in an ideal gas are considered point particles, having no volume.
• There are no intermolecular forces between the gas molecules.
• The energy of the gas is entirely kinetic, meaning it moves in straight lines and only changes direction during elastic collisions.

The state of an ideal gas is described using well-known laws, such as the Ideal Gas Law:

\[ PV = nRT \]
where \( P \) is pressure, \( V \) is volume, \( n \) is the number of moles, \( R \) is the gas constant, and \( T \) is temperature in Kelvin.

For an isothermal expansion at constant temperature, according to the Ideal Gas Law, the product of pressure and volume remains constant (i.e., \( P_1V_1 = P_2V_2 \)). Hence, as volume increases, pressure decreases proportionately in a reversible manner.

These simplifications make the mathematics of thermodynamics manageable and are foundational for solving more complex problems.

Reversible Expansion

Reversible expansion is a key concept in thermodynamics, particularly in optimizing mechanical work done during gas expansion. It refers to a process that occurs infinitely slowly, such that the system remains in equilibrium at each stage. Though ideal and hypothetical in nature, reversible processes are significant because they maximize work done.

For an isothermal reversible expansion of an ideal gas, we consider the gas expanding at a consistent temperature while maintaining equilibrium pressures with its surroundings. The equations used relate pressure and volume at different stages, allowing for precise calculations of work done.

In practical terms, performing work reversibly leverages the gas to do maximum work output because it leads to efficient energy transfer. Each infinitesimal step influences subsequent steps, balancing the differences in internal pressure and external pressure precisely.

Understanding reversible expansions allows students to conceptualize the perfect exchange of energy within thermodynamic systems, functioning as a benchmark in efficiency.