Q.

Heat given to process is positive, match the following option of Column I with the corresponding option of Column II :               [2006]

  Column I   Column II
(A) JK (p) ΔW>0
(B) KL (q) ΔQ<0
(C) LM (r) ΔW<0
(D) MJ (s) ΔQ>0

 

1 A-p, B-q, C-r, D-s  
2 A-q, B-p, C-s, D-r  
3 A-r, B-s, C-p, D-q   
4 A-s, B-r, C-q, D-p  

Ans.

(2) 

From the given P-V graph, in process JK volume,

V=constant p is decreasing and PT

Therefore, T should also decrease.

  W=PdV=0,    ΔV=ΔQ<0  (negative)

From  ΔQ=ΔU+ΔW

In process KL P = constant =nCpΔT  V is increasing so temperature should also increase.

  ΔW=PdV>0,    ΔU=nCvΔT>0  and  Q=mCΔT>0

In process LM  V = constant P increases to T increases.

  W=0,    ΔU>0  and   Q>0

In process  MJ

V is decreasing      ΔW<0

T is also decreasing    ΔU<0  and  ΔQ<0