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Home » 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)
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)
CBSE | Class 11 | NCERT | Thermodynamics
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)

Given

22.3 joules of work (W) are performed on the system while the gas transitions between states A and B.

This is an example of an adiabatic process.
As a result, the change in heat is zero.

So, ΔQ = 0

ΔW = – 22.3 (Since the work is done on the system)

From the first law of thermodynamics, we have:

ΔQ = ΔU + ΔW

Where, ΔU = Change in the internal energy of the gas

Hence,

ΔU = ΔQ – ΔW

= 0 – (-22.3 J)

We get,

ΔU = + 22.3 J

When a gas transitions from state A to state B as a result of a process, the net heat absorbed by the system is as follows.

ΔQ = 9.35 cal = 9.35 x 4.19 J

On calculation, we get,

= 39.1765 J

Heat absorbed, ΔQ = ΔU + ΔW

Thus, ΔW = ΔQ – ΔU

ΔW= 39.1765 – 22.3 = 16.8765 J

Hence, 16.88 J of work is done by the system

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