Q29 / 150 · MHT CET 2019 (Shift 1) · Chemistry

A gas performs 0.320 kJ work on surrounding and absorbs 120 J of heat from the surrounding. Hence, change in internal energy is

A
200 J
B
120.32 J
Correct answer
C
-200 J
D
440 J

Solution

According to first law of thermodynamics,

Δ U = q + W

Since, work is does on the surrounding.
So,

W = - 0.320 K J

= - 320 J

q = 120 J

∴ Δ U = - 320 + 120

= - 200 J

Hence, the correct answer is option: 'C'.

Relevant Questions

If mole of an ideal gas expand isothermally and reversibly at C from 1 to calculate work done?20 Apr, 2026 (M)First Law and Basic Fundamentals of Thermodynamics
Which of the following processes exhibits the decrease in internal energy of a system?20 Apr, 2026 (E)First Law and Basic Fundamentals of Thermodynamics
Calculate the change in internal energy if an ideal gas expands from to at constant pressure 2 bar by absorbing 1800 J heat energy.20 Apr, 2026 (E)First Law and Basic Fundamentals of Thermodynamics
An ideal gas expands from initial volume to against a constant external pressure of atm. What is the work done by the gas ?19 Apr, 2026 (M)First Law and Basic Fundamentals of Thermodynamics
A gas absorbs J of heat and expands by against a constant external pressure of . What will be the change in internal energy of the system ?19 Apr, 2026 (M)First Law and Basic Fundamentals of Thermodynamics
Calculate the work done for the following reaction at ( )19 Apr, 2026 (E)First Law and Basic Fundamentals of Thermodynamics