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Chemistry Question 54 – JEE-MAIN 2026

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Gas 'A' undergoes change from state 'X' to state 'Y'. In this process, the heat absorbed and work done by the gas is 10 J and 18 J respectively. Now gas is brought back to state 'X' by another process during which 6 J of heat is evolved. In the reverse process of 'Y' to 'X',

The First Law of Thermodynamics relates the change in internal energy of a system to the heat absorbed by the system and the work done by the system.

Step 1: Calculate Change in Internal Energy for X to Y✦ Active

For the process from state 'X' to state 'Y':

Heat absorbed by the gas, QXY=+10 J.

Work done by the gas, WXY=+18 J.

Using the First Law of Thermodynamics, ΔU=Q−W:

ΔUXY=QXY−WXY=10 J−18 J=−8 J
Step 2: Determine Change in Internal Energy for Y to X○ Expand

Internal energy is a state function, meaning its change depends only on the initial and final states. Therefore, for the reverse process from 'Y' to 'X':

ΔUYX=−ΔUXY=−(−8 J)=+8 J
💡 Teacher's Secret Hint

Ensure correct sign convention for internal energy change in the reverse process.

Step 3: Calculate Work Done in Y to X Process○ Expand

For the process from state 'Y' to state 'X':

Heat evolved by the gas, QYX=−6 J.

Using the First Law of Thermodynamics, ΔUYX=QYX−WYX:

8 J=−6 J−WYX

Solving for WYX (work done by the gas):

WYX=−6 J−8 J=−14 J

A negative value for WYX means work is done *on* the gas. Therefore, 14 J of work is done on the gas 'A' by the surrounding.

💡 Teacher's Secret Hint

Pay close attention to the sign of work. Negative work done by the gas means work is done on the gas by the surroundings.

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