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) \(+300 \mathrm{~J}\)

Short answer

(c) +100 J

Step-by-step solution

Understanding the Problem

We need to calculate the change in internal energy of the gas. The first law of thermodynamics states that the change in internal energy \( \Delta U \) is given by \( \Delta U = Q - W \), where \( Q \) is the heat absorbed by the system and \( W \) is the work done by the system.

Define Given Values

We know \( Q = 200 \text{ J} \), the heat absorbed by the gas. The gas expands by \(500 \text{ cm}^3\) and the external pressure \( P = 2 \times 10^5 \text{ N/m}^2\).

Convert Volume Expansion Units

Convert the volume from \( \text{cm}^3 \) to \( \text{m}^3 \). Since \( 1 \text{ m}^3 = 10^6 \text{ cm}^3 \), so \( 500 \text{ cm}^3 = 500/10^6 \text{ m}^3 = 5 \times 10^{-4} \text{ m}^3 \).

Calculate Work Done

Work done \( W \) during a constant pressure process is given by \( W = P \Delta V \). Substituting the known values: \[ W = 2 \times 10^5 \times 5 \times 10^{-4} \] Therefore, \( W = 100 \text{ J} \).

Calculate Change in Internal Energy

Use the first law of thermodynamics: \( \Delta U = Q - W \). Substituting the known values: \[ \Delta U = 200 - 100 \] Therefore, \( \Delta U = 100 \text{ J} \).

Choose the Correct Answer

The calculated change in internal energy is \(+100 \text{ J}\), so the correct answer is (c) \(+100 \text{ J}\).

Key concepts

First Law of Thermodynamics

The first law of thermodynamics is a core principle that explains how energy is conserved within a system. It essentially states that energy cannot be created or destroyed, only transformed. This principle is often cited in the formula: \( \Delta U = Q - W \). This means that the change in internal energy \( \Delta U \) of a system is equal to the heat \( Q \) absorbed by the system minus the work \( W \) done by the system.
In simpler terms, if a gas within a system absorbs heat, that energy can either increase the system's internal energy or be used to do work on its surroundings.
This law is foundational because it helps us understand how different energy inputs and processes interact during thermodynamic processes.
This becomes highly relevant in predicting how various systems, from engines to natural processes, will behave when energy is added, removed, or transformed.

Change in Internal Energy

The change in internal energy \( \Delta U \) of a system is a measure of how the energy within the system has shifted. It represents the difference between the total energy entering the system and the energy leaving it.
In the context of the first law of thermodynamics, \( \Delta U \) is calculated by the equation \( \Delta U = Q - W \), where \( Q \) is the heat added to the system and \( W \) is the work done by the system.
If \( \Delta U \) is positive, the internal energy has increased, meaning the system has gained energy. Conversely, if it's negative, the system has lost energy. In our example, the system absorbs 200 J of heat and does 100 J of work, leading to a net increase in internal energy of 100 J. This calculation helps in understanding energy flow, especially in thermodynamic cycles.
The concept of internal energy change is vital for understanding processes like heating, cooling, compression, and expansion of gases.

Work Done by the Gas

When a gas does work, it typically expands against an external pressure. The work done by the gas \( W \) is often calculated using the formula \( W = P \Delta V \), where \( P \) is the pressure and \( \Delta V \) is the change in volume.
In this exercise, the gas expands by 500 \( \text{cm}^3 \) against a constant pressure of \( 2 \times 10^5 \text{ N/m}^2 \). We first convert the volume to cubic meters, resulting in \( 5 \times 10^{-4} \text{ m}^3 \). Using the formula, the calculation becomes
\[ W = 2 \times 10^5 \text{ N/m}^2 \times 5 \times 10^{-4} \text{ m}^3 = 100 \text{ J} \]
This tells us that 100 J of energy was used to do work, decreasing the energy available to increase the internal energy.
Understanding how gases perform work is crucial in mechanics and engineering, particularly when designing systems that utilize engines or pneumatically-powered machines.

Heat Absorbed by the System

Heat absorbed by the system \( Q \) pertains to the amount of thermal energy transferred into the system from its surroundings. In thermodynamics, this transfer usually results in a temperature change and potentially a phase change or work performed by the system.
In our problem, the gas absorbs 200 J of heat. This value is crucial because it represents the energy that becomes available for either increasing the internal energy of the system or doing work.
When calculating changes in energy, knowing the amount of heat absorbed helps in applying the first law of thermodynamics effectively. It sets the stage for determining the resultant internal energy, as we deduct the work done by the gas, represented by the equation \( \Delta U = Q - W \).
Understanding heat absorption is essential in fields that manage energy transformation, including power generation, heating systems, and cooling technologies.