Problem 15 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

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:\((\mathrm{R}=8.314 \mathrm{~J} / \mathrm{mol} \mathrm{K})(\ln 7.5=2.01)\) (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

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

Step 1: Identify A-B-C Relationships

In thermodynamics, for an ideal gas undergoing isothermal (constant temperature) reversible expansion, we use the formula \( q = \Delta U + w \). Since temperature is constant, the change in internal energy \( \Delta U = 0 \). Hence, \( q = w \).

Step 2: Given and Required

We are given that during the expansion the system absorbs heat: \( q = +208 \ \text{J} \). We need to find \( w \), the work done by the gas during expansion.

Step 3: Use the Isothermal Formula for Work

For isothermal expansion or compression of an ideal gas, the work done \( w \) by the gas can be determined by:\[ w = nRT \ln \left( \frac{V_f}{V_i} \right) \]where:- \( n = 0.04 \ \text{mol} \)- \( R = 8.314 \ \text{J/mol K} \)- \( T \) must be converted from Celsius to Kelvin: \( T = 37 + 273.15 = 310.15 \ \text{K} \)- \( V_i = 50.0 \ \text{mL} \) and \( V_f = 375 \ \text{mL} \)

Step 4: Calculate the Work Done

Plug the known values into the formula:\[ w = 0.04 \times 8.314 \times 310.15 \times \ln \left( \frac{375}{50} \right) \]\[ w = 0.04 \times 8.314 \times 310.15 \times \ln(7.5) \]Using \( \ln(7.5) = 2.01 \), compute:\[ w = 0.04 \times 8.314 \times 310.15 \times 2.01 \approx 208 \ \text{J} \]

Step 5: Analyze Sign of Work and Heat

Since the system is doing work on the surroundings during expansion, the work done \( w \) is negative, thus \( w = -208 \ \text{J} \). The heat absorbed \( q \) is positive, thus as given \( q = +208 \ \text{J} \).

Key Concepts

Isothermal Process

An isothermal process is characterized by a constant temperature throughout its duration. In this type of process, an ideal gas undergoes transformations while maintaining the same temperature. Because temperature remains constant, the internal energy of the gas does not change.
This means that any heat added to the system goes into doing work. The formula that dictates these processes is \( q = \Delta U + w \). Since the change in internal energy \( \Delta U = 0 \), the heat added \( q \) equals the work done \( w \).
In the original exercise, the gas remains at a constant temperature of \( 37^{\circ} \text{C} \), or \( 310.15 \text{K} \), throughout expansion. Therefore, understanding that \( q = w \) plays a fundamental role in solving for the unknown values in such thermodynamic processes.

Ideal Gas Law

The ideal gas law is a pivotal equation in thermodynamics describing the behavior of ideal gases. It is expressed as \( PV = nRT \), where:

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

It provides a simple relationship between these variables, which allows for the calculation of one when others are known.
In the isothermal and reversible expansion discussed in the exercise, the ideal gas law forms the backbone for further calculating the work done and understanding the gas behavior as its volume changes.

Work Done in Thermodynamics

In thermodynamic processes, work can be done by the system as it expands or compresses. For an isothermal and reversible expansion, the work done \( w \) by the gas is calculated using the formula:\[ w = nRT \ln \left( \frac{V_f}{V_i} \right) \] In this expression, \( V_f \) and \( V_i \) represent the final and initial volumes, respectively.
As demonstrated in the original solution, the calculation was shown as:\[ w = 0.04 \times 8.314 \times 310.15 \times \ln(7.5) \approx 208 \text{ J} \]It's essential to recognize the sign of work in thermodynamics. When a gas does work on the surroundings, as in an expansion, the work is traditionally considered negative.

Internal Energy Change

Internal energy refers to the total energy contained within a system from its molecular motions and interactions. In many thermodynamic processes, internal energy change \( \Delta U \) is a crucial component.
However, during isothermal processes for an ideal gas, there is no change in temperature, therefore \( \Delta U = 0 \). This is due to the direct relationship between internal energy and temperature in ideal gases.
Thus, for the isothermal process described in the exercise, \( \Delta U \) is zero, simplifying the energy relation to \( q = w \), where all heat absorbed is converted directly to work done by the gas.

Reversible Expansion

Reversible expansion refers to an idealized process in which the system expands in such a delicate manner that it can be reversed by an infinitesimal change in conditions.
In a reversible process, the system remains in near-equilibrium with its surroundings, allowing accurate calculation of work done, since each step is close to an equilibrium state.
Such expansions yield maximum work, denoted by the relation \( w = nRT \ln \left( \frac{V_f}{V_i} \right) \), used in the exercise, providing a crucial method for optimal energy use and calculation in thermodynamics.