In a thermodynamic process pressure of a fixed mass of a gas is changed in such a manner that gas releases $20J$ of heat and $8J$ of work is done on the gas. If the initial internal energy of the gas was $30J$, what will be the fixed internal energy?
(A) 9J
(B) 27J
(C) 18J
(D) 12J
Answer
299.7k+ views
Hint First of all let us find out the change in internal energy. This change in internal energy is equal to the difference between the total heat energy and the work done by the system. The work done on the system is mentioned as negative and energy released is mentioned as negative. These all may help you to solve this question.
Complete Step-by-step answer
The change in internal energy is expressed as,
$\Delta U = {U_{final}} - {U_{initial}}$
Following the first law of thermodynamics, change in internal energy is equal to the difference of the total heat energy and the work done by the system. This can be expressed mathematically as,
$\Delta U = Q - W$
Therefore, we can equate both the equations together,
${U_{final}} - {U_{initial}} = Q - W$
As stated in the question, the heat energy has been released from the system.
According to the conventions, the heat energy released is specified as negative and vice-versa.
Therefore we can write that,
$Q = - Q$
And the work done is mentioned as work done on the system. As per the conventions, the work done on the system is mentioned as negative. Therefore we can write that,
$W = - W$
Hence we can substitute this in the equation,
That is,
${U_{final}} - {U_{initial}} = - Q + W$
The value of work done of the gas is given as,
$W = 8J$
The heat energy released is given as,
$Q = 20J$
The initial internal energy is given as,
${U_{initial}} = 30J$
Substituting these values in the equation will give,
${U_{final}} - 30 = - 20 + 8$
${U_{final}} = 18J$
So, the correct answer is (C) $18J$.
Note The first law of thermodynamics is defined as a version of the law of conservation of energy. When it is taken for the thermodynamic processes, we can distinguish this into two. One is the transfer of energy as heat and the other one is the transfer of energy in the form of work.
Complete Step-by-step answer
The change in internal energy is expressed as,
$\Delta U = {U_{final}} - {U_{initial}}$
Following the first law of thermodynamics, change in internal energy is equal to the difference of the total heat energy and the work done by the system. This can be expressed mathematically as,
$\Delta U = Q - W$
Therefore, we can equate both the equations together,
${U_{final}} - {U_{initial}} = Q - W$
As stated in the question, the heat energy has been released from the system.
According to the conventions, the heat energy released is specified as negative and vice-versa.
Therefore we can write that,
$Q = - Q$
And the work done is mentioned as work done on the system. As per the conventions, the work done on the system is mentioned as negative. Therefore we can write that,
$W = - W$
Hence we can substitute this in the equation,
That is,
${U_{final}} - {U_{initial}} = - Q + W$
The value of work done of the gas is given as,
$W = 8J$
The heat energy released is given as,
$Q = 20J$
The initial internal energy is given as,
${U_{initial}} = 30J$
Substituting these values in the equation will give,
${U_{final}} - 30 = - 20 + 8$
${U_{final}} = 18J$
So, the correct answer is (C) $18J$.
Note The first law of thermodynamics is defined as a version of the law of conservation of energy. When it is taken for the thermodynamic processes, we can distinguish this into two. One is the transfer of energy as heat and the other one is the transfer of energy in the form of work.
Recently Updated Pages
Four persons A B C and D initially at the corners of class 11 physics JEE_Main

What is the difference between Conduction and conv class 11 physics JEE_Main

Moment of inertia of solid sphere about its diameter class 11 physics JEE_Main

If a piece of ice floating on the surface of water class 11 physics JEE_Main

At what temperature speed of sound in air will be doubled class 11 physics JEE_Main

A closed organ pipe and an open organ pipe are tuned class 11 physics JEE_Main

Trending doubts
JEE Main 2026: Exam Dates, Session 2 Updates, City Slip, Admit Card & Latest News

Understanding the Electric Field of a Uniformly Charged Ring

Electron Gain Enthalpy and Electron Affinity Explained

Derivation of Equation of Trajectory Explained for Students

Understanding Atomic Structure for Beginners

How to Convert a Galvanometer into an Ammeter or Voltmeter

Other Pages
CBSE Notes Class 11 Physics Chapter 1 - Units And Measurements - 2026-27 PDF Download (Login Required)

NCERT Solutions For Class 11 Physics Chapter 1 Units And Measurements - 2026-27 Free PDF Download (Login Required)

NCERT Solutions For Class 11 Physics Chapter 2 Motion In A Straight Line - 2026-27 Free PDF Download (Login Required)

Important Questions For Class 11 Physics Chapter 1 Units and Measurement - 2026-27 Free PDF Download (Sign-in Required)

CBSE Notes Class 11 Physics Chapter 2 - Motion in a Straight Line - 2026-27 PDF Download (Login Required)

NCERT Solutions For Class 11 Physics Chapter 3 Motion In A Plane - 2026-27 Free PDF Download (Sign-in Required)

