One mole of a monatomic gas is taken  through a cycle ABCDA as shown in the P-V diagram.  Column II give the characteristics involved in the cycle. Match them with each of the processes given in Column I. 

Column I		Column II
A)	Process  $A\to B$	p)	Internal energy decreases
B)	Process  $B\to C$	q)	Internal energy increases
C)	Process  $C\to D$	r)	Heat is lost
D)	Process  $D\to A$	s)	Heat is gained
		t)	Work is done on the gas

A) Process  $A\to B$
This is an  isobaric process, P = constant and volume (V) 
of the gas decreases. Therefore work is done on the gas. 
 $W=P(3V-V)=2PV$
Also V  decreases so temperature at B decreases 
 Internal energy U decreases. 
From,  $Q=U+W$ as U and  W decreases so  Q decreases that means heat is lost. 
B) Process  $B\to C$
This is an isochoric process  V=constant pressure decreases 
$P\propto T$  so temperature also decreases. 
$W=0; \Delta U=$ negative so  $\Delta Q$  negative 
Hence heat is lost. 
C) Process  $C\to D$
This is an isobaric, Pressure P=constant V increases and  $V\propto T$  so T  increases. Hence $\Delta W, \Delta U$  and  $\Delta Q+ve$ so heat gained by the gas.
D) Process  $D\to A$
Applying  $PV=nRT$
For  $DP\left(9V\right)=1R{T}_D\therefore T_D=\frac{9PV}{R}$
For  $A3P\left(3V\right)=1R{T}_A\therefore T_A=\frac{9PV}{R}$
i.e., the process is isothermal   $\therefore \Delta U=0$
now,  $\Delta Q=\Delta U+W\therefore \Delta Q=W$.
As volume decreases in this process so  W negative i.e., 
Wom done on the gas and  $\Delta Q$ negative hence heat is lost.