Problem 16 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

Which of the following statements/relationships is not correct in thermodynamic changes?(a) \(\Delta \mathrm{U}=0\) (isothermal reversible expansion of a gas) (b) \(\mathrm{w}=-\mathrm{nRT} \ln \frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}}\) (isothermal reversible expansion of an ideal gas) (c) \(\mathrm{w}=\mathrm{nRT} \ln \frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}}\) (isothermal reversible expansion of an ideal gas) (d) For a system of constant volume heat involved directly changes to internal energy.

Step-by-Step Solution

Answer

(c) is the incorrect statement.

Step 1: Understand Thermodynamic Terms

In thermodynamics, \(\Delta U\) represents the change in internal energy, \( w \) is the work done by or on the system, \( n \) is the number of moles, \( R \) is the gas constant, \( T \) is the temperature, and \( V_1 \) and \( V_2 \) are initial and final volumes respectively.

Step 2: Analyze Option (a)

In an isothermal process (constant temperature), the change in internal energy \( \Delta U \) is zero for ideal gases since internal energy depends only on temperature. So, statement (a) is correct.

Step 3: Examine Option (b)

For an isothermal reversible expansion of an ideal gas, the work done by the gas can be calculated using the formula \( w = -nRT \ln \frac{V_2}{V_1} \). Hence, statement (b) is correct.

Step 4: Assess Option (c)

The expression given is \( w = nRT \ln \frac{V_2}{V_1} \), which does not have the negative sign as expected for work done by the system. Work is considered negative when the gas expands (loses energy). Hence, statement (c) is not correct.

Step 5: Consider Option (d)

In a process at constant volume, no work is done by or on the system (\( w = 0 \)). Therefore, any heat added or removed from the system changes the internal energy directly as \( q = \Delta U \). Thus, statement (d) is correct.

Key Concepts

Isothermal Process

An isothermal process occurs when the temperature of a system remains constant throughout a thermodynamic change.
This means that for an ideal gas undergoing an isothermal process, the internal energy remains unchanged because, for ideal gases, internal energy is solely a function of temperature.

• In an isothermal expansion, a gas does work on its surroundings as it expands.
• Since the system is at constant temperature, any energy input as heat will go into doing work rather than changing internal energy.
• The term \(\Delta U = 0\) appropriately describes this, as the internal energy change is zero in isothermal processes.

Understanding an isothermal process helps clarify why options like statement (a) in the original exercise show that internal energy change is indeed zero, reaffirming how a consistent temperature ensures conservation of energy between heat absorbed and work done.

Internal Energy

Internal energy, denoted by \( U \), is the total energy contained within a system.
It encompasses both the kinetic and potential energy of the molecules.

• In thermodynamics, the change in internal energy \( \Delta U \) is crucial as it helps determine how energy is transferred or transformed in a process.
• For ideal gases, internal energy change only occurs with a change in temperature. Hence, \( \Delta U = 0 \) during isothermal processes, as these processes are at a constant temperature.
• At constant volume, any heat transfer directly alters the internal energy because there is no work being done, based on the principle \( q = \Delta U \).

This is why in statement (d) from the exercise, the concept of constant volume indicates that heat change directly corresponds to internal energy change.

Work Done

Work done by or on a system during a thermodynamic process is a key concept.
It represents energy transfer due to volume change under external pressure.

• In an isothermal reversible expansion, the formula \( w = -nRT \ln \frac{V_2}{V_1} \) calculates the work done.
• This formula indicates that during expansion, work is done by the gas and reflects a decrease in the system's energy, hence the negative sign.
• It’s important to note that for expansions, we expect work to be negative, as energy is used to perform the expansion.

The distinction between work sign conventions is crucial, as seen when comparing statements (b) and (c) from the exercise. Understanding these sign conventions provides clarity on energy flow during thermodynamic processes, especially when interpreting reversibility and direction of work.