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

The answer is (a) 5.22 kJ, -5.22 kJ.

Step 1: Objective Clarification

We need to find the quantities of heat \(Q\) and work \(W\) for an isothermal reversible expansion of an ideal gas.

Step 2: Determine the Formula for Work

For an isothermal reversible expansion of an ideal gas, the work \(W\) done by the gas can be calculated using the formula: \[ W = -nRT \ln \left( \frac{V_f}{V_i} \right) \]where \(n\) is the number of moles, \(R\) is the ideal gas constant \(8.314 \text{ J/(mol K)}\), \(T\) is the temperature in Kelvin, and \(V_f/V_i\) is the ratio of final to initial volume.

Step 3: Relate Volume to Pressure

Since this is an ideal gas undergoing isothermal processes, \[ P_1 V_1 = P_2 V_2 \]Hence, the ratio \(\frac{V_f}{V_i} = \frac{P_i}{P_f}\). Substitute \(P_i = 1.0\) bar and \(P_f = 0.1\) bar to find \(\frac{V_f}{V_i}\).

Step 4: Calculate Work (W)

Substitute \(n = 1\) mole of gas, \(R = 8.314\) J/(mol·K), \(T = 273\) K, \(P_i = 1.0\) bar, and \(P_f = 0.1\) bar into the formula: \[ W = -1 \times 8.314 \times 273 \times \ln \left( \frac{1.0}{0.1} \right) \]\[ W = -1 \times 8.314 \times 273 \times \ln (10) \]Calculate \(W\) using \(\ln (10) = 2.302\). \[ W = -1 \times 8.314 \times 273 \times 2.302 \approx -5220 \text{ J} = -5.22 \text{ kJ} \]

Step 5: Identify that Heat (Q) Equals Work in Isothermal Expansion

For an isothermal process in an ideal gas, the change in internal energy is zero. Hence, by the first law of thermodynamics, \(\Delta U = 0\), we have \(Q = -W\). Thus, \(Q = 5.22 \text{ kJ} \).

Step 6: Final Answer Selection

Looking at the options, we identify \((a)\) as the correct answer: \(Q = 5.22\ \text{kJ}, W = -5.22\ \text{kJ}\).

Key Concepts

Understanding the Ideal Gas Law

The ideal gas law is a cornerstone of thermodynamics in chemistry. It relates the pressure (P), volume (V), temperature (T), and number of moles (n) of an ideal gas. The formula is expressed as:

• \[ PV = nRT \]

where \( R \) is the ideal gas constant, valued at 8.314 J/(mol·K). In this particular exercise involving isothermal reversible expansion, one essential aspect is the temperature stays constant. Therefore, the product of pressure and volume must also remain constant due to the law. When calculating changes where volume is involved, knowing that \[ P_1 V_1 = P_2 V_2 \] helps us determine the relationships and changes in properties during such expansions.

Calculating Work in Isothermal Processes

In thermodynamics, the work done by or on a system indicates energy transfer. Specifically for isothermal processes for an ideal gas, the formula for work \( W \) is vital. Here, since the temperature remains constant, we use:

• \[ W = -nRT \ln \left( \frac{V_f}{V_i} \right) \]

In these calculations, \( V_f \) and \( V_i \) can be derived from the pressures assuming constant temperature and using the ideal gas law. As we calculate work, it represents energy flowing out of the system, marked by the negative sign, which is standard for expansion. For example, when pressure decreases during expansion from 1.0 bar to 0.1 bar, the work done here was \(-5.22\, \text{kJ}\).

How Heat Calculation Complements Work

Calculating heat in an isothermal expansion of an ideal gas aligns with the work calculation. By thermodynamics' first law, the heat added to the system \( Q \) balances out the work done if there's no change in internal energy (\( \Delta U = 0 \)). Therefore, we equate:

• \( Q = -W \)

This relation shows that for a reversible process, whatever energy the system spends performing work is precisely gained in heat input. In our scenario, with an imposed work of \(-5.22\, \text{kJ}\), the heat added to keep temperature constant amounts to \(5.22\, \text{kJ}\). This perfect balance is typical in theoretical calculations of isothermal reversible processes.

Exploring Thermodynamics in Chemistry

Thermodynamics in chemistry governs energy transformations in chemical processes. It emphasizes studies on energy changes and system equilibrium. In an isothermal reversible expansion of an ideal gas, the concepts of work, heat, and energy conservation are underlined.

• The first law of thermodynamics establishes that energy cannot be created or destroyed, only transformed.
• This exercise exemplifies that while the ideal gas expands, energy used as work equals the heat added to maintain constant temperature.

Thus, understanding these principles provides insights into energy efficiency and energy transfer processes in various chemical reactions and systems. It prepares students for mastering more complex topics involving every kind of thermodynamic analysis.