Problem 116 - A Complete Resource Book in Chemistry for JEE Main

Question

Calculate \(\mathrm{Q}\) and \(\mathrm{W}\) for the isothermal reversible expansion of one mole of an ideal gas from an initial pressure of \(1.0\) bar to a final pressure of \(0.1\) bar at a constant temperature of \(273 \mathrm{~K}\). (a) \(5.22 \mathrm{~kJ},-5.22 \mathrm{~kJ}\) (b) \(-27.3 \mathrm{~kJ}, 27.3 \mathrm{~kJ}\) (c) \(27.3 \mathrm{~kJ},-27.3 \mathrm{~kJ}\) (d) \(-5.22 \mathrm{~kJ}, 5.22 \mathrm{~kJ}\)

Step-by-Step Solution

Answer

(d) -5.22 kJ, 5.22 kJ

Step 1: Understand the process

The problem involves isothermal reversible expansion of one mole of an ideal gas. The temperature remains constant at 273 K throughout this process.

Step 2: Apply the Ideal Gas Law

Since it is an isothermal process, we use the ideal gas equation in the form: \[ P_1 V_1 = P_2 V_2 \] Where \( P_1 = 1.0 \text{ bar} \), \( P_2 = 0.1 \text{ bar} \). This allows us to express the volumes in terms of pressures assuming constant temperature.

Step 3: Calculate the Work Done

For an isothermal reversible process, the work done \( W \) on or by the gas can be described by:\[ W = nRT \ln \frac{V_2}{V_1} \]Substitute \( V_1 \) and \( V_2 \) in terms of pressure:\[ W = nRT \ln \frac{P_1}{P_2} \]where \( n = 1 \), \( R = 8.314 \text{ J/mol·K} \), and \( T = 273 \text{ K} \).Calculate the work:\[ W = (1)(8.314)(273) \ln \frac{1.0}{0.1} \]\[ W = 2270.622 \ln 10 \approx 2270.622 \times 2.302 \approx 5230 \text{ J} \approx 5.23 \text{ kJ} \]

Step 4: Determine the Relationship Between Q and W

In an isothermal process, the internal energy change \( \Delta U \) is zero for an ideal gas. Thus, the heat absorbed \( Q \) is equal to the work done \( W \):\[ Q = -W \]Therefore, \( Q = -5.23 \text{ kJ} \) when \( W = 5.23 \text{ kJ} \).

Step 5: Match to Answer Choices

The calculated values are \( Q = -5.23 \text{ kJ} \) and \( W = 5.23 \text{ kJ} \). Compare this with the given options. The closest match is option (d): \(-5.22 \text{ kJ}, 5.22 \text{ kJ}\).

Key Concepts

Ideal Gas Law

The ideal gas law is a fundamental principle in chemistry and physics used to relate the pressure, volume, and temperature of an ideal gas. In this exercise, the situation is the isothermal expansion of an ideal gas, meaning the temperature remains constant. The formula that embodies the ideal gas law is:

• \[ PV = nRT \]

Where:

• \(P\) is the pressure of the gas,
• \(V\) is the volume,
• \(n\) is the number of moles,
• \(R\) is the universal gas constant (8.314 J/mol·K), and
• \(T\) is the temperature in Kelvin.

In this problem, since the temperature is constant, we assume that the product of the initial pressure and volume (\(P_1V_1\)) must equal the product of the final pressure and volume (\(P_2V_2\)). This concept allows us to solve problems involving changes in gas states by linking pressure and volume changes directly with temperature constant.

Work Done

In thermodynamics, the concept of work done is vital to understanding how energy is transferred during processes. When dealing with gas in a cylinder with a movable piston, work is done by the gas on the surroundings when it expands. Specifically for isothermal reversible processes involving ideal gases, the work done \( W \) can be calculated by the integral of pressure with respect to volume. Ultimately, this can be simplified to:

• \[ W = nRT \ln \frac{V_2}{V_1} \]

Substituting the volumes with pressures (to avoid calculating volume directly), the equation becomes:

• \[ W = nRT \ln \frac{P_1}{P_2} \]

Here, the natural logarithm (\(\ln\)) function indicates that the work done is proportional to the logarithm of the pressure ratio. For this problem, once the constant and known values are substituted in, the result is given in kilojoules \((kJ)\), showing that expanding the gas does work on the surroundings.

Temperature

Temperature plays a crucial role in gas behavior and thermodynamic processes. In this specific exercise, the process described is isothermal, meaning the temperature remains constant throughout. The significance of constant temperature in an isothermal process is that it simplifies calculations and makes energy exchange straightforward, particularly for ideal gases. For an ideal gas at constant temperature:

• The internal energy \( \Delta U \) of the system remains unchanged.
• An increase in volume leads to a corresponding decrease in pressure, without altering the internal energy.

Therefore, any work done by the system on its surroundings is matched by an equal amount of heat entering the system, maintaining thermal equilibrium. Understanding the concept of temperature as constant unravels how variables like work and heat are interrelated.

Heat Absorbed

Heat absorbed by a gas during expansion is a critical part of thermodynamic studies. In the case of an isothermal process for an ideal gas, the internal energy change \( \Delta U \) is zero, because internal energy depends only on temperature, which is constant. This leads to a straightforward relationship between heat \( Q \) and work \( W \).

• The first law of thermodynamics states that: \[ \Delta U = Q - W \]
• For isothermal processes, since \( \Delta U = 0 \), it implies that:\[ Q = W \]

This reversal of roles implies the quantity of heat absorbed aligns directly with the work done by the gas. For this exercise, where the work done \( W \) by the system is 5.23 kJ, an equal amount but opposite in sign of heat is absorbed, hence \( Q = -W = -5.23 \) kJ. This negative sign signifies heat flow into the system, indicating an endothermic process, sustaining the gas expansion at constant temperature.