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

Option (a) 5.22 kJ, -5.22 kJ.

Step 1: Understand the Concept

For an isothermal, reversible expansion of an ideal gas, the heat absorbed (\(Q\)) is equal to the work done (\(W\)) on the system. This is derived from the first law of thermodynamics and the properties of isothermal processes: \( \Delta U = 0 \), hence \( Q = -W \).

Step 2: Write the Ideal Gas Equation

For an isothermal process, the ideal gas equation \( PV = nRT \) can be used where \(P\) is pressure, \(V\) is volume, \(n\) is the amount of substance in moles, \(R\) is the universal gas constant \(8.31 \text{ J}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}\), and \(T\) is temperature in Kelvin.

Step 3: Work Done in Isothermal Expansion

The work done (\(W\)) in an isothermal expansion from \(P_1\) to \(P_2\) is given by:\[ W = -nRT \ln \(\frac{V_2}{V_1}\) \]Where \(V_2/V_1\) can be replaced by \(P_1/P_2\) for an ideal gas, thus:\[ W = -nRT \ln \frac{P_1}{P_2} \]

Step 4: Substituting Values into the Work Equation

Insert the values for \(n = 1\) mol, \(R = 8.31\), \(T = 273\) K, \(P_1 = 1.0\) bar, and \(P_2 = 0.1\) bar:\[ W = -(1)(8.31)(273) \ln \frac{1.0}{0.1} \]\[ W = -2270.23 \cdot \ln 10 \]\[ W = -2270.23 \times 2.302 \approx -5230.92 \text{ J} \approx -5.23 \text{ kJ} \]

Step 5: Calculate Heat Absorbed

Since \(Q = -W\) for an isothermal process, substitute the calculated value of \(W\):\[ Q = 5.23 \text{ kJ} \]

Step 6: Match with Options

The calculated heat \(Q = 5.23 \text{ kJ}\) and work \(W = -5.23 \text{ kJ}\) best match option (a) \(5.22 \text{ kJ}, -5.22 \text{ kJ}\) when rounding to two decimal places.

Key Concepts

Ideal Gas Law

The ideal gas law is a foundation of thermodynamics that describes the behavior of an ideal gas. It shows the relationship between pressure (P), volume (V), and temperature (T) for a given amount of gas. The equation is expressed as:
\[PV = nRT\]Here:

• \(P\) is the pressure of the gas.
• \(V\) is the volume.
• \(n\) is the amount of substance in moles.
• \(R\) is the universal gas constant, which is approximately \(8.31 \ ext{ J}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}\).
• \(T\) is the temperature in Kelvin.

This formula helps you predict how changing one property, such as pressure, will affect another, like volume, at constant temperature. It's essential in isothermal processes because temperature remains constant, and this constancy simplifies the calculations by maintaining the equation's balance.

First Law of Thermodynamics

The first law of thermodynamics is all about the conservation of energy. It states that energy cannot be created or destroyed, only transformed. This is reflected in the equation:
\[\Delta U = Q - W\]- \(\Delta U\) stands for 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 an isothermal process, there's no change in temperature, and thus, no change in internal energy (\Delta U = 0). With this simplification, the first law simplifies further:\[ Q = W \]This is especially useful in isothermal, reversible expansions, where the heat absorbed by the system is equal to the work done on it, thus making calculation straightforward.

Work Done in Isothermal Process

During an isothermal process, the work done by or on the gas is tied to the pressure and volume changes. To calculate it, you can use the equation:
\[W = -nRT \ln \frac{P_1}{P_2}\]This equation gives the work done (W) when a gas expands under constant temperature conditions. Here's how it works:

• \(n\) is the number of moles of gas.
• \(R\) is the universal gas constant.
• \(T\) is the absolute temperature.
• \(P_1\) and \(P_2\) are the initial and final pressures, respectively.

The negative sign indicates that the system's internal energy decreases as the gas does work on its surroundings. Understanding this concept is vital for mastering thermodynamic processes involving ideal gases.

Heat Absorbed During Expansion

In isothermal expansions of ideal gases, the heat absorbed by the gas is equal to the work done but with opposite sign.
Since \(Q = -W\), if the gas does work on the surroundings (energy leaves the gas), heat of the same magnitude enters the gas. This ensures that the internal energy doesn't change because the process is isothermal.
During an expansion, as the gas pushes against the external pressure, it absorbs energy in the form of heat to maintain a constant temperature. Calculating this involves:

• Using the same calculation for \(W\), we find \(Q\) using the relationship \(Q = -W\).
• This results in the concept being both intuitive and mathematically coherent due to the laws of thermodynamics.

This feature of isothermal processes helps us better understand how energy interactions work in closed systems.