Problem 10 - A Complete Resource Book in Chemistry for JEE Main

Question

If a gas absorbs \(200 \mathrm{~J}\) of heat and expands by 500 \(\mathrm{cm}^{3}\) against a constant pressure of \(2 \times 10^{\mathrm{s}} \mathrm{Nm}^{-2}\), then change in internal energy is (a) \(-200 \mathrm{~J}\) (b) \(-100 \mathrm{~J}\) (c) \(+100 \mathrm{~J}\) (d) \(+300 \mathrm{~J}\)

Step-by-Step Solution

Answer

The change in internal energy is +100 J (option c).

Step 1: Understand the Given Values

We are given that the gas absorbs 200 J of heat (
Q = 200 J). The gas expands by 500 cm³, which we need to convert to m³. Since 1 m³ = 1,000,000 cm³, we have a volume change (
V) = 500 cm³ = 500 × 10^{-6} m³. The constant pressure is given as 2 × 10^{5} Nm^{-2}.

Step 2: Find Work Done by the Gas

To find the work done by the gas (
W), we use the formula: \[ \Delta W = P \Delta V \] Substituting the given values: \[ \Delta W = (2 \times 10^{5} \text{ Nm}^{-2})(500 \times 10^{-6} \text{ m}^3) \] \[ \Delta W = 100 \text{ J} \] Thus, the work done by the gas is 100 J.

Step 3: Apply the First Law of Thermodynamics

The first law of thermodynamics states that: \[ \Delta U = \Delta Q - \Delta W \] where \( \Delta U \) is the change in internal energy, \( \Delta Q \) is the heat absorbed, and \( \Delta W \) is the work done by the system. Substituting the known values: \[ \Delta U = 200 \text{ J} - 100 \text{ J} \] \[ \Delta U = 100 \text{ J} \] Therefore, the change in internal energy is 100 J.

Key Concepts

Internal Energy Change

In thermodynamics, internal energy change is a fascinating concept. Imagine it as the total energy stored within a system of particles—in this case, a gas. This energy can change when the gas absorbs heat or performs work. When heat is absorbed, it can increase the internal energy. The gas particles move faster or can even change their arrangement, both of which increase their energy. But, when the gas does work, like expanding and pushing against something, it uses some of this internal energy.The change in internal energy, denoted as \( \Delta U \), can be calculated using the First Law of Thermodynamics. This law is expressed as:\[ \Delta U = \Delta Q - \Delta W \]where \( \Delta Q \) is the heat absorbed and \( \Delta W \) is the work done by the system. It essentially tells us how much energy remains in the system after accounting for these exchanges.

Work Done by Gas

When a gas expands, it can perform work on its surroundings. This involves the gas pushing against an external pressure. The amount of work done by the gas is directly proportional to the volume change and the constant external pressure it holds.To find this work, we use the formula:\[ \Delta W = P \Delta V \]Where:

• \( \Delta W \) is the work done by the gas,
• \( P \) is the constant external pressure, and
• \( \Delta V \) is the volume change.

For instance, if a gas expands its volume by \( 500 \text{ cm}^3 \) (convert this to \( 500 \times 10^{-6} \text{ m}^3 \)) at a constant pressure of \( 2 \times 10^5 \text{ Nm}^{-2} \), then the calculation for work done would be:\[ \Delta W = 2 \times 10^5 \times 500 \times 10^{-6} = 100 \text{ J} \]This tells us that the gas expended 100 J of energy to do the work.

Heat Absorption

Heat absorption in gases happens when they take in energy from their surroundings. This absorbed heat can cause the particles in the gas to move more vigorously. The First Law of Thermodynamics relates this absorbed heat \( \Delta Q \) to changes in internal energy and work done. For a gas absorbing 200 J of heat, it shows us that not all the absorbed energy results in a temperature increase or a change in internal energy because some of it may also be used to do work. So we calculate the change in internal energy due to heat absorption using:\[ \Delta U = \Delta Q - \Delta W \]If the gas absorbs 200 J of heat and does 100 J of work, the remaining energy that increases the internal energy is:\[ \Delta U = 200 \text{ J} - 100 \text{ J} = 100 \text{ J} \]This calculation shows that after doing work, the gas has 100 J of energy left, increasing its internal energy.