Problem 119 - A Complete Resource Book in Chemistry for JEE Main

Question

Which of the following is /are true about the isothermal expansion of an ideal gas? (a) \(\Delta \mathrm{U}=0\) (b) \(\Delta \mathrm{T}=0\) (c) \(\mathrm{q}=2.303 \mathrm{nRT} \log _{10}\left(\frac{\mathrm{v}_{1}}{\mathrm{v}_{2}}\right)\) (d) \(\mathrm{q}=0\)

Step-by-Step Solution

Answer

Statements (a), (b), and (c) are true; statement (d) is false.

Step 1: Understand the process

In an isothermal process, the temperature of the system remains constant. Therefore, for an ideal gas, the internal energy change, \(\Delta U\), depends only on temperature variation.

Step 2: Evaluate \(\Delta U = 0\) for isothermal process

Since the temperature remains constant in an isothermal process, the change in internal energy, \(\Delta U\), is zero for an ideal gas. Thus, statement (a) is true.

Step 3: Verify \(\Delta T = 0\) for isothermal process

In an isothermal process, by definition, the temperature \(\Delta T\) does not change. Thus, statement (b) is true.

Step 4: Evaluate the heat exchange expressions

For an isothermal expansion, the work done \(W\) by the gas is expressed as \(W = q = nRT \ln\left(\frac{v_2}{v_1}\right)\). The expression \(q = 2.303 nRT \log_{10}\left(\frac{v_1}{v_2}\right)\) can be derived from this by converting natural log to log base 10.

Step 5: Validate \(q = 2.303 nRT \log_{10}\left(\frac{v_1}{v_2}\right)\)

The expression \(q = 2.303 nRT \log_{10}\left(\frac{v_1}{v_2}\right)\) represents the heat added to the system during isothermal expansion, if the reference is made to log base 10. Hence, this statement (c) is true.

Step 6: Evaluate \(q = 0\)

An isothermal process involves heat exchange to maintain constant temperature while doing work. As such, \(q = 0\) cannot be correct for an isothermal expansion, so statement (d) is false.

Key Concepts

Ideal Gas

An ideal gas is a theoretical gas composed of a set of randomly moving point particles that interact only via elastic collisions. It's an important concept and simplifies the study of gaseous behavior under various conditions since real gases approximate this ideal behavior under many conditions.

Internal Energy

For an ideal gas, internal energy is the energy contained within the gas molecules as a result of their motion. An important aspect of the internal energy of an ideal gas is that it depends solely on its temperature, not on the volume or pressure. During an isothermal expansion, the temperature remains constant, and therefore, there is no change in internal energy (\(\Delta U = 0\)). This implies that for an isothermal expansion of an ideal gas, the internal energy remains steady.

Heat Exchange

Heat exchange during an isothermal process can be a bit of a puzzle. In an isothermal expansion, the gas performs work on its surroundings. To keep the temperature constant, the gas must absorb heat equal to this work from the surroundings. The formula \[ q = 2.303 nRT \log_{10}\left(\frac{v_1}{v_2}\right) \] expresses the amount of heat that must be absorbed, using log base 10. This formula reflects the quantitative aspect of how the gas exchanges heat to maintain a stable temperature while performing work.

Work Done

The work done by the gas during an isothermal expansion is intricately linked with the heat exchanged. In this specific process, the work done is calculated as the product of the number of moles, the gas constant, the temperature, and the logarithm of the volume change.The natural logarithm base is normally used, leading to:\[ W = nRT \ln\left(\frac{v_2}{v_1}\right) \]During an isothermal expansion, as the gas expands and does work on the surroundings, it maintains temperature by absorbing an equivalent amount of heat from the environment. This relationship is crucial as it demonstrates how energy conservation is always maintained during these thermodynamic processes.