Verified Solution • 2025–26 Session moderate

Q11.5: NCERT Physics - Part 2 THERMODYNAMICS 11.5 Solution: In changing the state of a gas adiabatically from an…

In changing the state of a gas adiabatically from an equilibrium state A to another equilibrium state B, an amount of work equal to 22.3 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, how much is the net work done by the system in the latter case? (Take 1 cal = 4.19 J)

Step-by-Step Verified Solution

**Answer:** Work done = 16.9 J **Step-by-Step Explanation:** For the adiabatic process (A to B):
Q=0,Wad=−22.3 J (work done on the system)Q = 0, \quad W_{ad} = -22.3 \text{ J (work done on the system)}
ΔU=Q−W=0−(−22.3)=22.3 J\Delta U = Q - W = 0 - (-22.3) = 22.3 \text{ J}
For the second process:
Q=9.35 cal=9.35×4.19=39.18 JQ = 9.35 \text{ cal} = 9.35 \times 4.19 = 39.18 \text{ J}
Since internal energy change is path-independent:
ΔU=22.3 J\Delta U = 22.3 \text{ J}
W=Q−ΔU=39.18−22.3=16.88 J≈16.9 JW = Q - \Delta U = 39.18 - 22.3 = 16.88 \text{ J} \approx 16.9 \text{ J}

All Questions in this Chapter

Q11.1 Q11.2 Q11.3 Q11.4 Q11.5 Q11.6 Q11.7 Q11.8

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