Question

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)

Answer 1

The work done (W) on the system while the gas changes from state A to state B is 22.3 J.

This is an adiabatic process. Hence, change in heat is zero.

∴ ΔQ = 0

ΔW = –22.3 J (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

∴ ΔU = ΔQ – ΔW = – (– 22.3 J)

ΔU = + 22.3 J

When the gas goes from state A to state B via a process, the net heat absorbed by the system is:

ΔQ = 9.35 cal = 9.35 × 4.19 = 39.1765 J

Heat absorbed, ΔQ = ΔU + ΔQ

∴ΔW = ΔQ – ΔU

= 39.1765 – 22.3

= 16.8765 J

Therefore, 16.88 J of work is done by the system

Answer 2

Here, when the change is adiabatic, `triangleQ = 0, triangle W = - 22.3 J`

if `triangle uu` is change in internal energy of the system then

as `triangle Q = triangle uu + triangle W`

`0 = triangleuu - 22.3 or triangle uu = 22.3 J`

In second case, `triangle Q = 9.35 cal = 9.35 xx 4.2 J = 39.3 J`

`triangle W = ? `

As` triangle uu + triangle W =  triangle Q`

`:.triangle W =  triangle Q - triangle uu = 39.3 - 22.3 = 17.0 J`