2019-03-01_Physics_Times

16.Sol:

18.Sol:

Heat capacity

1 1

R R

c

 x

 

 

(^2) (2)
3
R R
c   c R
 
 
 
12.Sol: For adiabatic process PVConstant
PV^1 dPV 0
dV
  
P dP
V dV

 
Therefore, n 1
13.Sol: The process XY can be expressed by a general
equation
P a bV  (1)
Where a and b are constants
Now PV nRT (2)
From eq’s (1) & (2)
T^1 aV bV^2
nR
   
When T is maximum
 
1
2 0
dT
a bV
dV nR
  
2
a
V
b

Let Z is a point on the line XY where T is maxi-
mum.
From X Z gas absorbs heat and from Z Y
it rejects heat. The correct option is (c).
14.Sol: For isothermal Process
 Q W 12 12
Wcycle W W 12 31
10    W 12 20 W 12 30 J
Hence, 12 12
17.Sol:
  Q W 30 J
15.Sol: The graph between P and V is shown in the
figure. The area covered by the isothermal curve
w.r.to volume axis is more than that of adiabatic
curve.
12.Sol:
13.Sol:
14.Sol:
15.Sol:
17.Sol:
18.Sol:
W Wi a 0
Gas expands aganist vaccum so work done is
W 0
For Insulated container  Q 0
   Q W U
0 0     U U 0
PV C
P T C^1  
   
(^11)
0.28 233 1 T
   
  
7
5
 
   
1 7/5 7/5 1 7/5 7/5
0.28 233 1 T
 
  
 
7/5 7/5 2/5
T 233 0.28

 
 
2/7
233
0.28
T
T 298 K 25 C
As T is more than 25 C
So to cool it an extra air condition is required.
Given that U aV^3
3
2
fnRT
aV (1)
PV = RT (2)
From eq’ s (1) & (2)
P CV^2
2 2
2
V V
V V

WPdV CV dV