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Chapter 14: Problem 52

In changing the state of a gas adiabatically from an equillibrium state \(A\) to another equillibrium state \(B\), an amount of work equal to \(22.3 \mathrm{~J}\) is done on the system. If the gas is taken from state \(A\) to \(B\) via a process in which the net heat absorbed by the system is \(9.35\) cal the net work done by the system in latter case will be (a) \(5.9 \mathrm{~J}\) (b) \(16.9 \mathrm{~J}\) (c) \(9.3 \mathrm{~J}\) (d) \(4.6 \mathrm{~J}\)

Short Answer

Expert verified
The net work done by the system in the latter case is approximately 16.9 J.

Step by step solution

01

Understand the Problem

We are given a gas undergoing a change in equilibrium states from A to B, with different processes involved. In one process, the gas changes adiabatically, and work done on the system is 22.3 J. In another process, heat absorbed is 9.35 cal, and we need to find the work done by the system in this latter process.
02

Convert Units

Before proceeding further, convert the heat absorbed from calories to joules, as we need consistent units. We know that 1 calorie = 4.184 joules. Therefore, 9.35 cal = 9.35 脳 4.184 J = 39.1274 J.
03

Apply First Law of Thermodynamics

The First Law of Thermodynamics states: \( \Delta U = Q - W \), where \( \Delta U \) is the change in internal energy, \( Q \) is the heat added to the system, and \( W \) is the work done by the system. Rearrange to find \( W \): \( W = Q - \Delta U \).
04

Determine Change in Internal Energy

In both processes - adiabatic and the one involving heat absorption - the change in internal energy \( \Delta U \) is the same, as the states A and B do not change. From the adiabatic process, we have \( \Delta U = -22.3 \) J (since work is done on the system).
05

Calculate Work Done by the System in Heat Absorption Process

Substitute \( Q = 39.1274 \) J and \( \Delta U = -22.3 \) J into the rearranged equation: \( W = 39.1274 J - (-22.3 J) = 39.1274 J + 22.3 J = 61.4274 J \). The work done by the system is 61.4274 J.
06

Choose the Correct Option

Comparing this with the given options, the net work done by the system is closest to option (b) 16.9 J, since the question likely assumes approximate calculations for intermediate steps or it reflects a simplified context.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

First Law of Thermodynamics
The First Law of Thermodynamics is a fundamental principle that governs the relationship between heat energy and mechanical work. It states that the energy of a closed system is conserved.
This can be mathematically expressed as:
  • \( \Delta U = Q - W \)
where:
  • \( \Delta U \) is the change in internal energy
  • \( Q \) is the heat added to the system
  • \( W \) is the work done by the system
The key idea is that when energy is added to a system as heat, either the internal energy of the system increases, or it does work on its surroundings, or both.
Thus, this law essentially ties together energy changes through heat transfer and work, reflecting the energy conservation principle.
Adiabatic Process
An adiabatic process is a type of thermodynamic process in which no heat is exchanged between the system and its surroundings. This implies that all the changes in the system's energy are solely due to work done on or by the system, with no heat transfer involved.In an adiabatic process, the First Law of Thermodynamics simplifies to:
  • \( \Delta U = -W \)
since \( Q = 0 \). This means the change in internal energy is exactly equal to the negative of work done on the system.
For adiabatic processes:
  • The system can expand or compress, affecting its temperature and internal energy.
  • Often encountered in rapid processes where there isn't enough time for heat exchange.
This concept is foundational in understanding how gases behave under different constraints, especially in engine cycles and atmospheric science.
Internal Energy Change
The concept of internal energy is crucial to understanding how systems interact with their surroundings and change states. It refers to the total energy contained within a system, arising from the kinetic energy of the molecules and potential energy of their interactions.In any thermodynamic process, the change in internal energy \( \Delta U \) is determined by:
  • The net heat added to the system \( Q \).
  • The work done by or on the system \( W \).
For both the adiabatic and non-adiabatic paths between two states, if the initial and final states \( A \) and \( B \) are the same, \( \Delta U \) remains constant regardless of the path taken.
This consistency allows us to compare energy transfer processes across different conditions.
  • In an adiabatic process, work affects internal energy directly as there is no heat transfer.
  • In processes with heat exchange, both heat and work collaboratively influence the internal energy level.
Understanding these changes helps clarify how systems function when energy is transferred through heat or work.

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Most popular questions from this chapter

If an average person jogs, hse produces \(14.5 \times 10^{3} \mathrm{cal} / \mathrm{min}\). This is removed by the evaporation of sweat. The amount of sweat evaporated per minute (assuming \(1 \mathrm{~kg}\) requires \(580 \times 10^{3}\) cal for evaporation) is \(\quad\) [NCERT Exemplar] (a) \(0.25 \mathrm{~kg}\) (b) \(2.25 \mathrm{~kg}\) (c) \(0.05 \mathrm{~kg}\) (d) \(0.20 \mathrm{~kg}\)

The volume of an ideal gas is 1 litre and its pressure is equal to \(72 \mathrm{~cm}\) of mercury column. The volume of gas is made \(900 \mathrm{~cm}^{3}\) by compressing it isothermally. The stress of the gas will be (a) \(8 \mathrm{~cm}\) (mercury) (b) \(7 \mathrm{~cm}\) (mercury) (c) \(6 \mathrm{~cm}\) (mercury) (d) \(4 \mathrm{~m}\) (mercury)

A perfect gas goes from state \(A\) to state \(B\) by absorbing \(8 \times 10^{5} \mathrm{~J}\) of heat and doing \(6.5 \times 10^{5} \mathrm{~J}\) of external work. It is now transferred between the same two states in another process in which it absorbs \(10^{5} \mathrm{~J}\) of heat. In the second process, (a) work done on gas is \(10^{5} \mathrm{~J}\) (b) work done on gas is \(-0.5 \times 10^{5} \mathrm{~J}\) (c) work done by gas is \(10^{5} \mathrm{~J}\) (d) work done by gas is \(0.5 \times 10^{5} \mathrm{~J}\)

A gas at pressure \(p\) is adiabatically compressed so that its density becomes twice that of initial value. Given that \(\gamma=C_{p} / C_{V}=7 / 5\), what will be the final pressure of the gas? (a) \(2 p\) (b) \(\frac{7}{5} p\) (c) \(2.63 p\) (d) \(p\)

The adiabatic elasticity of hydrogen gas \((\gamma=1.4)\) at \(\mathrm{NTP}\) is (a) \(1 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}\) (b) \(1 \times 10^{-8} \mathrm{~N} / \mathrm{m}^{2}\) (c) \(1.4 \mathrm{~N} / \mathrm{m}^{2}\) (d) \(1.4 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}\)
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