Problem 175 - A Complete Resource Book in Chemistry for JEE Main

Question

A piston filled with \(0.04\) mol of an ideal gas expands reversibly from \(50.0 \mathrm{~mL}\) to \(375 \mathrm{~mL}\) at a constant temperature of \(37.0^{\circ} \mathrm{C}\). As it does so, it absorbs \(208 \mathrm{~J}\) of heat. The values of \(\mathrm{q}\) and \(\mathrm{w}\) for the process will be: \((\mathrm{R}=3.314 \mathrm{~J} / \mathrm{mol} \mathrm{K})(\operatorname{Ln} 7.5=2.01)\) (a) \(\mathrm{q}=-208 \mathrm{~J}, \mathrm{w}=+208 \mathrm{~J}\) (b) \(\mathrm{q}=+208 \mathrm{~J}, \mathrm{w}=+208 \mathrm{~J}\) (c) \(\mathrm{q}=+208 \mathrm{~J}, \mathrm{w}=-208 \mathrm{~J}\) (d) \(\mathrm{q}=-208 \mathrm{~J}, \mathrm{w}=-208 \mathrm{~J}\)

Step-by-Step Solution

Answer

The correct answer is (c) \( q = +208 \text{ J}, \ w = -208 \text{ J} \).

Step 1: Understanding the System

The exercise provides the initial and final volumes of an ideal gas, the number of moles, temperature, and the heat absorbed during the expansion. It asks to find the heat and work done during this process. Since the system expands at a constant temperature (isothermal process), the internal energy change \( \Delta U = 0 \). The first law of thermodynamics tells us \( \Delta U = q + w \), thus \( q = -w \).

Step 2: Analyze and Apply Formulas

From Step 1, we have \( q = -w \). Given that \( q = +208 \text{ J} \), it means heat is absorbed. Since \( \Delta U = 0 \), the work \( w = -q = -208 \text{ J} \). Therefore, the gas does 208 J of work on the surroundings because the work done by the system is negative.

Step 3: Check Options and Conclusion

Now, compare the values of \( q = +208 \text{ J} \) and \( w = -208 \text{ J} \) to the given options. Option (c) is \( q = +208 \text{ J}, \ w = -208 \text{ J} \), which matches the computed values.

Key Concepts

First Law of Thermodynamics

The First Law of Thermodynamics is one of the foundational principles in understanding energy interactions within a system. It essentially states that energy cannot be created or destroyed, but only transferred or converted from one form to another.
In mathematical terms, the First Law is expressed as: \[ \Delta U = q + w \] where:

• \( \Delta U \) is the change in internal energy of the system.
• \( q \) is the heat added to the system.
• \( w \) is the work done by the system.

In this exercise, the piston undergoes a process where it absorbs heat, with \( q \) given as +208 J. Since the system is an isothermal process (the temperature remains constant), the change in internal energy \( \Delta U \) is zero, leading to the relation \( q = -w \). This tells us the work done by the system (in this case, the ideal gas in the piston) is equal in magnitude but opposite in sign to the heat added.

Isothermal Process

An isothermal process is a type of thermodynamic process where the system's temperature remains constant throughout the entire process. This can occur only if the system is perfectly insulated and the process is conducted very slowly, allowing heat exchange to maintain a constant temperature.
For an ideal gas undergoing an isothermal process, this implies \( \Delta T = 0 \) and, thereby, \( \Delta U = 0 \). This principle plays a crucial role in determining the values of heat \( q \) and work \( w \) in an isothermal expansion or compression of an ideal gas.
Since \( \Delta U = 0 \) in an isothermal process for an ideal gas, the First Law reduces to \( q = -w \). This means that any heat added to the system is entirely converted into work done by the system, with no change in internal energy.

Ideal Gas

The term "ideal gas" refers to a theoretical gas composed of many randomly moving point particles that interact only through elastic collisions. Ideal gas behavior is described by the Ideal Gas Law, which is expressed as: \[ PV = nRT \] where:

• \( P \) is the pressure of the gas.
• \( V \) is the volume.
• \( n \) is the number of moles.
• \( R \) is the ideal gas constant.
• \( T \) is the absolute temperature.

In the given exercise, the ideal gas law helps to understand how volume changes relate to temperature and pressure during the expansion. The assumption that the gas behaves ideally allows us to employ simple calculations to predict the behavior of the gas, such as determining that \( \Delta U = 0 \) for an isothermal process, given the nature of ideal gases. This assumption is key to simplifying real-world gas behaviors for fundamental study and creating a baseline for deeper thermodynamics exploration.