
When 1 mole of gas is heated at constant volume, the temperature is raised from 298 to 308 K and heat supplied to gas is 500 J. Which of the following statements is true?
A. q = -w = 500J, $\Delta $U=0
B. q = $\Delta $U = 500J, w=0
C. q=w=500J, $\Delta $U=0
D. $\Delta $U=0, q=w= -500J
Answer
233.1k+ views
Hint: The first law of thermodynamics is the version of the law of conservation of energy, adapted from thermodynamics processes, distinguishing two kinds of transfer of energy as heat and as thermodynamic work and relating them to a function of the body's state called internal energy.
The law of conservation of energy states that the total energy of an isolated system is constant, energy can be transferred from one form to another, but can neither be created nor destroyed.
Complete step by step answer:
First law of thermodynamics states that the total energy of an isolated system is constant, energy can be transformed from one form to another but can neither be created nor destroyed.
$\Delta U = q + W$
Where, $\Delta U$represents the change in internal energy of a closed system.
Q denotes the quantity of energy supplied to the system as heat.
W denotes the amount of thermodynamic work done by the system on its surrounding.
An equivalent statement is that a perpetual motion machine of first kind is impossible.
So, from first law of thermodynamics:
$\Delta U = q + W$
and in the given problem, $q = 500J$
$W = - p\Delta V$
where, $\Delta V = 0$
so, $W = 0$
So, $\Delta U = q = 500J$
So, the correct answer is “Option B”.
Note:
In a non-cyclic process, the change in internal energy of the system is equal to net energy added as heat to the system minus the net work done by the system, both being measured in mechanical units.
Taking $\Delta U$ as a change in internal energy one can write,
$\Delta U = q - W$
However, by using IUPAC convention the first law is formulated with work done on the system by its surrounding having a positive sign. With this now often use sign convention for work, the first law for close system can be written as:
$\Delta U = q + W$.
The law of conservation of energy states that the total energy of an isolated system is constant, energy can be transferred from one form to another, but can neither be created nor destroyed.
Complete step by step answer:
First law of thermodynamics states that the total energy of an isolated system is constant, energy can be transformed from one form to another but can neither be created nor destroyed.
$\Delta U = q + W$
Where, $\Delta U$represents the change in internal energy of a closed system.
Q denotes the quantity of energy supplied to the system as heat.
W denotes the amount of thermodynamic work done by the system on its surrounding.
An equivalent statement is that a perpetual motion machine of first kind is impossible.
So, from first law of thermodynamics:
$\Delta U = q + W$
and in the given problem, $q = 500J$
$W = - p\Delta V$
where, $\Delta V = 0$
so, $W = 0$
So, $\Delta U = q = 500J$
So, the correct answer is “Option B”.
Note:
In a non-cyclic process, the change in internal energy of the system is equal to net energy added as heat to the system minus the net work done by the system, both being measured in mechanical units.
Taking $\Delta U$ as a change in internal energy one can write,
$\Delta U = q - W$
However, by using IUPAC convention the first law is formulated with work done on the system by its surrounding having a positive sign. With this now often use sign convention for work, the first law for close system can be written as:
$\Delta U = q + W$.
Recently Updated Pages
JEE Main 2023 April 6 Shift 1 Question Paper with Answer Key

JEE Main 2023 April 6 Shift 2 Question Paper with Answer Key

JEE Main 2023 (January 31 Evening Shift) Question Paper with Solutions [PDF]

JEE Main 2023 January 30 Shift 2 Question Paper with Answer Key

JEE Main 2023 January 25 Shift 1 Question Paper with Answer Key

JEE Main 2023 January 24 Shift 2 Question Paper with Answer Key

Trending doubts
JEE Main 2026: Session 2 Registration Open, City Intimation Slip, Exam Dates, Syllabus & Eligibility

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

Understanding the Angle of Deviation in a Prism

Hybridisation in Chemistry – Concept, Types & Applications

How to Convert a Galvanometer into an Ammeter or Voltmeter

Understanding the Electric Field of a Uniformly Charged Ring

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Hydrocarbons Class 11 Chemistry Chapter 9 CBSE Notes - 2025-26

Thermodynamics Class 11 Chemistry Chapter 5 CBSE Notes - 2025-26

Equilibrium Class 11 Chemistry Chapter 6 CBSE Notes - 2025-26

Organic Chemistry Some Basic Principles And Techniques Class 11 Chemistry Chapter 8 CBSE Notes - 2025-26

NCERT Solutions For Class 11 Chemistry Chapter 7 Redox Reactions (2025-26)

