Q.

One mole of a monoatomic gas is taken through a cycle ABCDA as shown in the P-V diagram. Column II gives the characteristics involved in the cycle. Match them with each of the processes given in Column I.                     [2011]

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

 

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

Ans.

(1)

(1) Process AB

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 decrease so Q decreases, that means heat is lost.

(2) Process BC

This is an isochoric process, V = constant pressure decreases,

PT so temperature also decreases.

W=0; ΔU=negative

so  ΔQ negative

Hence heat is lost.

(3) Process CD

This is isobaric, Pressure P = constant V increases and VT

so T increases.

Hence ΔW, ΔU and ΔQ are positive, so heat is gained by the gas.

(4) Process DA

Applying PV=nRT

for D  P(9V)=1RTD     TD=9PVR

for A  3P(3V)=1RTA     TA=9PVR

i.e., the process is isothermal.       ΔU=0

Now,  ΔQ=ΔU+W      ΔQ=W

As volume decreases in this process so W is negative i.e., work done on the gas and ΔQ is negative, hence heat is lost.