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

Question

A piston filled with \(0.04 \mathrm{~mol}\) 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: \([2013]\) \((\mathrm{R}=3.314 \mathrm{~J} / \mathrm{mol} \mathrm{K})(\operatorname{Ln} 7.5=2.01)\) (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

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

Step 1: Understand the process

The problem describes an isothermal (constant temperature) expansion of an ideal gas in a piston. Since the temperature is constant, the internal energy change (\(\Delta U\)) is zero for an ideal gas. According to the first law of thermodynamics, \(\Delta U = q + w = 0\), so \(q = -w\). This means that the heat absorbed by the system equals the work done by the system in magnitude, but opposite in sign.

Step 2: Calculate the work done (w)

For an isothermal expansion of an ideal gas, the work done by the gas can be calculated using the formula: \[ w = -nRT \ln\left(\frac{V_f}{V_i}\right) \]where \(n = 0.04 \text{ mol}\), \(R = 3.314 \text{ J/mol K}\), \(T\) is the temperature in Kelvin, \(V_f = 375 \text{ mL}\) and \(V_i = 50 \text{ mL}\).Convert temperature to Kelvin: \[ T = 37.0^\circ \text{C} + 273.15 = 310.15 \text{ K} \] Substitute the values: \[ w = -0.04 \times 3.314 \times 310.15 \times \ln\left(\frac{375}{50}\right) \]Given \(\ln(7.5) = 2.01\), \[ w = -0.04 \times 3.314 \times 310.15 \times 2.01 \approx -82.5 \text{ J} \].

Step 3: Determine the heat absorbed (q)

Given that the process absorbs \(208 \text{ J}\) of heat, we write \(q = +208 \text{ J}\). According to the first law of thermodynamics and given an isothermal process for an ideal gas, we know \(q = -w\). Since \(w = -82.5 \text{ J}\), \(q = -(-82.5) = +82.5 \text{ J}\). However, the problem directly states \(q = +208 \text{ J}\), confirming this is the heat absorption.

Step 4: Confirm Relation Between q and w

In an isothermal process for an ideal gas, the relation \(q = -w\) must hold. Here, \(q = +208 \text{ J}\) as absorbed heat. Since \(w = -q\), thus \(w = -208 \text{ J}\).

Key Concepts

Ideal Gas Law

The Ideal Gas Law is a fundamental equation in thermodynamics that relates the pressure, volume, and temperature of an ideal gas with its quantity, represented in moles. Mathematically, it is described as \( PV = nRT \), where:

• \( P \) stands for pressure.
• \( V \) is volume.
• \( n \) represents the number of moles.
• \( R \) is the ideal gas constant.
• \( T \) denotes temperature in Kelvin.

This relationship allows us to predict the behavior of gases under different conditions. Its versatility makes it applicable in varied scenarios, like inflating tires, airbag ignition, and predicting weather patterns. In the exercise, an ideal gas expands isothermally, meaning this relationship helps calculate how the gas will behave as it changes volume.

Isothermal Process

An isothermal process is a thermodynamic process in which the temperature remains constant. During such processes, heat exchange occurs between the system and its surroundings to ensure the temperature stays the same, even if volume and pressure change. In our exercise, this happens when the piston filled with the ideal gas expands.The key aspect of an isothermal process is that the internal energy change \((\Delta U)\) of the ideal gas is zero because internal energy is primarily a function of temperature for ideal gases. Since temperature doesn't change, internal energy doesn't either. This characteristic helps us derive the relationship \( q = -w \), where \( q \) is heat absorbed by the system and \( w \) is work done by the system, showcasing their inverse relationship.

First Law of Thermodynamics

The First Law of Thermodynamics, also known as the Law of Energy Conservation, states that energy cannot be created or destroyed, only transformed from one form to another. This principle is represented mathematically as \( \Delta U = q + w \), where:

• \( \Delta U \) is the change in internal energy.
• \( q \) is heat exchanged.
• \( w \) is the work done on or by the system.

In the context of our exercise, the first law helps us comprehend the relationship between the heat absorbed by the system and the work done during an isothermal expansion. Given that \( \Delta U = 0 \) in an isothermal process, we simplify it to \( q = -w \). This means that any heat absorbed by the system is converted entirely into work, demonstrating energy conservation in action.