Answer :

We know in an adiabatic change there is no transfer of heat either from system to surrounding or from surrounding to system, so there is no change in total heat Energy of system i.e.


ΔQ = 0


Now amount of work equal to 22.3 J is done on the system so we have


ΔW = -22.3 J


(Negative sign states that work is done on the system)


We know The first law of thermodynamics is the general law of conservation of energy applied to any system in which energy transfer from or to the surroundings (through heat and work) and is stated as


ΔQ = ΔU + ΔW


where ΔQ is the heat supplied to the system, ΔW is the work done by the system and ΔU is the change in internal energy of the system.


So to find the change in internal energy of system we will apply the first law of thermodynamics to find change in internal Energy


ΔQ = 0 and ΔW = -22.3 J , ΔU = ?


0 = ΔU + (-22.3 J)


i.e. ΔU = 22.3 J


so change in internal energy of gas in adiabatically going from an equilibrium state A to another equilibrium state B is 22.3 J


now gas is again taken from equilibrium state A to equilibrium state B, since internal energy of a system is a state variable, this means it depends only upon initial and final state of the system irrespective of the type of process so here also change in internal energy would have same value


i.e. ΔU = 22.3 J


this time heat absorbed by the system is 9.35 cal


1 cal = 4.19 J


So ΔQ = 9.35 × 4.19 = 39.17 J


(ΔQ is positive because heat is absorbed by the system)


Now again using first law of thermodynamics to find the work done by the gas


ΔQ = ΔU + ΔW


where ΔQ is the heat supplied to the system, ΔW is the work done by the system and ΔU is the change in internal energy of the system.


Here


ΔQ = 39.17 J , ΔU = 22.3 J , ΔW = ?


So we have


39.17 J = 22.3 J + ΔW


i.e. ΔW = 39.17 – 22.3 J = 16.87 J


so the work done by the gas in this process is 16.87 J


Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
caricature
view all courses
RELATED QUESTIONS :

Is it possible toPhysics - Exemplar

A system goes froPhysics - Exemplar

Air pressure in aPhysics - Exemplar

Figure 12.5 showsPhysics - Exemplar

Explain whyNCERT - Physics Part-II

In changingNCERT - Physics Part-II

Can a system be hPhysics - Exemplar

What amountNCERT - Physics Part-II

Explain whyNCERT - Physics Part-II

Explain whyNCERT - Physics Part-II