Question

In a cyclic process ABCA for an ideal gas . In AB , BC and CA processes $50J$ , $20J$ and $5J$ heat is supplied to an ideal gas . In process AB internal energy of gas increases by $60J$ and in process BC work done by gas is $30J$ . The increases in internal energy of gas in process CA is :
A.$50J$
B.$ - 50J$
C.$75J$
D.$55J$

Answer 1

Hint: A cyclic process in simple terms is a process that returns to the same thermodynamic state at which it started . In a cyclic process the sum of the internal energy of all the processes involved is always zero .

Complete step by step answer:
Let us first write down all the values which are given to us
Process AB - $q = 50J$ , $\Delta U = 60J$
Process BC - $q = 20J$ , $w = - 30J$ (since work is done by the gas so we take it as negative )
Process CA - $q = 5J$
Now , we know according to first law of thermodynamics
$\Delta U = q + w$
$\Delta U$ = Change in internal energy of the system
$q$ = Total Heat transferred into the system
$w$ = Net work done on the system
On substituting the values of process AB in above equation we get ,
$60J = 50J + w$
$ \Rightarrow w = 10J$
On substituting the values of process BC we get ,
$\Delta U = 20J - 30J$
$ \Rightarrow \Delta U = - 10J$
Now , for process CA we have to find the increase in internal energy
As we know that the sum of internal energy in a cyclic process is zero , therefore we can find the internal energy of process CA with the help of internal energy of processes AB and BC .
So , $\Delta U = \Delta {U_{AB}} + \Delta {U_{BC}} + \Delta {U_{CA}} = 0$
So on substituting the values we get
$0 = 60J - 10J + \Delta {U_{CA}}$
$ \Rightarrow \Delta {U_{CA}} = - 50J$
Hence the increase in internal energy in process CA is $ - 50J$ .
Hence the option B is correct .

Additional Information:
Alternatively this question can be solved by first taking out the work done by the gas in process CA and then taking out its internal energy .
As internal energy of the total process is zero therefore the work done is equal to the heat supplied , that is , $(q = - w)$ and the total heat supplied is given as $75J$ .
So work done in process CA is
${w_{AB}} + {w_{BC}} + {w_{CA}} = - 75$
$10J - 30J + {w_{CA}} = 75J$
$ \Rightarrow {w_{CA}} = - 55J$
So , internal energy is
$\Delta U = 5J - 55J = - 50J$

Note:
According to the first law of thermodynamics, change in internal energy of the system is equal to the sum of the total heat transferred into the system and the total work done on the system. it also varies at different conditions.