Q. 115.0( 1 Vote )

A gas is taken through a cyclic process ABCA as shown in the figure. If 2.4 cal of heat is given in the process, what is the value of J?

Answer :

‘J’ is mechanical equivalent of heat a conversion factor between two different units of energy: calorie to joule

From the graph we can write



PA=PC=100kPa=100×103 Pa


We know that work done by the gas is given as


Work done in path AB=0 as VA=VB.

Work done in path CA ΔW1=PA(VA-VC)

=100×103× (500-700)×10-6


Work done in path BC ΔW2= Pavg(VC-VB)



Total work done in process ABCA=ΔW=ΔW1+ΔW2


We know that in the cyclic process the system returns to its initial state. So, change internal energy in the cyclic process will be zero as internal energy is a state function.

From first law of thermodynamics, we know that,


Where ΔQ=heat supplied to the system

ΔU=change in internal energy

ΔW=work done by the system

Since ΔU=0, first law becomes


But it is given in question that ΔQ=2.4cal

So, 2.4×J=10Joule

value of ‘J’ is 4.17J/cal.

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.