Question 1: Thermodynamics & KTG

Question

One mole of a monatomic gas is taken through a cycle ABCDA as shown in the PV 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$ 	o $ B
(P)
Internal energy decreases.
(B)
Process B$ 	o $C
(Q)
Internal energy increase.
(C)
Process C$ 	o $D
(R)
Heat is lost.
(D)
Process D$ 	o $A
(S)
Heat is gained.
(T)
Work is done on the gas.
![](https://app-content.cdn.examgoal.net/fly/@width/image/1l2ml6be1/d9be342b-0b68-412f-bc44-60137815b547/1a5e4190-c8ea-11ec-8c2d-13f33da2fbe1/file-1l2ml6be2.png?format=png)

Accepted Answer

Process A $	o$ B,
It is a isobaric process
P = constant, V $\propto$ T
$\because$ V
B
 V
C
$\Rightarrow$ T
D
> T
C
$\Delta$U = nC
V
$\Delta$T = +ve
Hence internal energy increases.
$\Delta$Q = nC
P
$\Delta$T = +ve
Hence, heat is gained.
$\Delta$W = nR$\Delta$T = +ve
Hence, work is done by the gas.
Process, D $	o$ A
According to ideal gas equation
${{{P_A}{V_A}} \over {{T_A}}} = {{{P_D}{V_D}} \over {{T_D}}} \Rightarrow {{(3P)(3V)} \over {{T_A}}} = {{(P)(9V)} \over {{T_D}}} \Rightarrow {T_D} = {T_A}$
Hence, it is a isothermal process.
$	herefore$ $\Delta$U = 0
$\because$ V
A
< V
D
$\Delta$W = $-$ve, hence work done is done on the gas.
$\Delta$Q = $\Delta$W = $-$ve, hence heat is lost.
![](https://app-content.cdn.examgoal.net/fly/@width/image/1l2ml6be1/d9be342b-0b68-412f-bc44-60137815b547/1a5e4190-c8ea-11ec-8c2d-13f33da2fbe1/file-1l2ml6be2.png?format=png)

Answer

(A)$	o$(P), (R), (T); (B)$	o$(P), R; (C)$	o$(Q), (S); (D)$	o$(R), (T)

Answer

(A)$	o$(P), (T); (B)$	o$(P), (R); (C)$	o$(Q), (S); (D)$	o$(R)

Answer

(A)$	o$(R), (T); (B)$	o$(P), (R); (C)$	o$(S); (D)$	o$(R), (T)

Answer

(A)$	o$(P), (R), (T); (B)$	o$(P), (R); (C)$	o$(Q); (D)$	o$(R), (T)

Thermodynamics & KTG

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