Problem 49 - A Complete Resource Book in Chemistry for JEE Main

Question

When 1 mole gas is heated at constant volume tem perature is raised from 298 to \(308 \mathrm{~K}\). Heat supplied to the gas is \(500 \mathrm{~J}\). Then which of the following state ments is correct? (a) \(\mathrm{Q}=\mathrm{W}=500 \mathrm{~J}, \Delta \mathrm{U}=0\) (b) \(\mathrm{Q}=\Delta \mathrm{U}=500 \mathrm{~J}, \mathrm{~W}=0\) (c) \(\mathrm{Q}=\mathrm{W}=500 \mathrm{~J}, \Delta \mathrm{U}=0\) (d) \(\Delta U=0, Q=W=-500 \mathrm{~J}\)

Step-by-Step Solution

Answer

The correct statement is (b) \(Q = \Delta U = 500 \mathrm{~J}\), \(W = 0\).

Step 1: Understand the Process

The problem involves heating a gas at constant volume. When volume is constant, no work is done by or on the gas, meaning work (W) is zero. We need to find the relationship between heat (Q), change in internal energy (94U), and work (W) during this process.

Step 2: Use the First Law of Thermodynamics

According to the first law of thermodynamics, the change in internal energy is given by \(\Delta U = Q - W\). For a constant volume process, \(W = 0\), so \(\Delta U = Q\).

Step 3: Apply Known Values

Given that heat supplied (Q) to the gas is 500 J, and since no work is done (W = 0), it follows that the change in internal energy \(\Delta U = 500 \mathrm{~J}\).

Step 4: Choose the Correct Option

From Step 3, the correct statements are \(Q = \Delta U = 500 \mathrm{~J}\) and \(W = 0\). Comparing with the options given, option (b) is correct.

Key Concepts

First Law of Thermodynamics

The First Law of Thermodynamics is a foundational principle in physics that describes the conservation of energy in thermodynamic processes. It states that energy cannot be created or destroyed, only transferred or transformed. This can be expressed mathematically as: \[\Delta U = Q - W\] where:

• \(\Delta U\) is 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 essence, the first law ensures that any change in a system's internal energy is equal to the heat added to the system minus the work done by the system on its surroundings. Therefore, if no work is done, the heat supplied to or removed from the system results in a change in internal energy. When approaching problems using this law, it's essential to carefully identify these energy changes under different conditions.
Understanding this law helps us navigate questions involving thermal processes, like heating gases at constant volume, allowing these discussions to be rooted in energy conservation principles.

Constant Volume Process

A constant volume process, often referred to as an isochoric process, occurs when a gas is heated or cooled in a way that its volume remains unchanged. In such processes, work done by or on the system is zero because work is a function of volume change. Work (\(W\)) is defined as:

• \(W = P\Delta V\)

where:

• \(P\) is pressure
• \(\Delta V\) is the change in volume

Hence, in a constant volume process, \(\Delta V = 0\), meaning \(W = 0\). As a result, the entire heat energy added to the system at constant volume translates into a change in internal energy.
This makes analyzing these processes simpler since the first law of thermodynamics (\(\Delta U = Q - W\)) reduces to \(\Delta U = Q\). The constant volume process is an essential concept, especially when dealing with closed systems like those encountered in thermodynamic cycles and other engineered systems where volume is constrained.

Internal Energy Change

Internal energy, a key component of thermodynamics, is the total energy contained within a system. It includes both kinetic energy due to the motion of molecules and potential energy arising from molecular interactions. The change in internal energy (\(\Delta U\)) during a process reveals how energy is distributed among the particles in the system.
In any process, including heating or cooling, the internal energy changes as a function of heat added or removed and work done. For a constant volume process, since no work is done (\(W = 0\)), all the heat supplied translates into a change in internal energy: \[\Delta U = Q\]
This is important in practical applications, as knowing the heat addition can directly determine how much energy the system's particles gain or lose, impacting temperature and phase of the material. Understanding internal energy change enables one to predict system responses under various thermal conditions, aiding in energy management and optimization in systems like engines or refrigeration units.