2019-03-01_Physics_Times

In the paths b c and da (^) Tconstant
so U remains constant.
In the paths c d and ab (^) Sconstant
In additional process  S 0
As T decreases in c d so V should increase.
Similarly T increases in ab so V should de-
crease.
Adiabatic process TV^1 Constant
2
1
f
 
2
TV Cf
3 (1 )
2 2
V
fR R aRT
C

 
3Re
2 2
fR aRT

2 2
f 3 eaRT

2
TV^3 eaRT Constant
 
 
No option is matching.
Ideal gas equation PV=nRT For isobaric
process
nR
V T
P
 
 
 
Graph between V and T is a straight line.
Slope of line
nR
P
 
 
 
Slope
1
P

slope slope slope 3  2  1
P P P 3   2 1
Consider the
1
th
4
part of the cycle as shown in
the figure. Let c is the centre of the circle.
PV
T
R

When PV is maximum T is also maximum.
 
max max
6 6 36
T T
R R

  
Temperature on the isothermal curve is
6 4 24
T
R R

 
As the given curve is above isothermal curve
max
24
T
R

As max
36 24
T
R R
 
The possible option is
30
R
(^) PV C^2 
nRT V C 2
V
  
 
 
PV nRT 
 TV K
 TV T V1 1 2 2
If V V 2  1 thenT T 2  1.
For monoatomic gas
7.Sol:
8.Sol:
9.Sol:
10.Sol:
11.Sol:
5
, 1
3
 n
PV C^3  , comparing with PV Cx
Here x 3