Physics Times 07.2019

(^2) ( 1)
( )
2



  
Mv
T K K Kelvin
R
6.Sol: Number of moles of first gas
1
A
n
N

Number of of moles of second gas
2
A
n
N

Number of of moles of third gas
3
A
n
N

By conserving total internal energy
UfUi
1 2 3
1 2 3
A A A
n RT n RT n RT
N N N
 ^123 mix
A
n n n RT
N
  
1 1 2 2 3 3
1 2 3
mix
T n T n T n T
n n n
  
 
7.Sol:

.^55
2 2

K E nRT PV

5 5 1 8 10^44

5 10

2 2 4

mP m

J V

d d

   
      
 
8.Sol: The speed of sound in a gas is given by

v RT

M

 

2 2

2

O O He

He O He

v M

v M





  

1.4 (^4) 0.3237
32 1.67
  
(^2460) 1421 /
0.3237 0.3237
O
He
v
   v m s
9.Sol: Molar specific heat '



Q
C
n T
Specific heat
 

C Q
m T
'
C  m M(Molecular weight)
C n
'

C
C
M
'  1
CP CV R
MCP MCV R  P V
C C R
M
For nitrogen M= 28
28
CP CV R
CALORIMETRY [ONLINE QUESTIONS]
1.Sol: According to principle of calorimetry,
QgivenQused
0.2  s 150 40
    0.15 1 40 25 0.025 40 25   
0.175(15)
0.1193 499.46
0.2 110
cal J
s
g C kg C
  
  
2.Sol: As Pt mc T 
P    10 60 mc 1 100
(i) and P  55 60 mL
(ii) Dividing equation (i) by (ii) we get
10 100
55 L

 L 550. / .cal g
3.Sol: Energy given by heater must be equal to
the sum of energy gained by water and energy
lost from the lid.
Pt ms T  energy lost
1000 t    2 4.2 10^3  50 160t
840 8.4 10 50t  ^3
 t = 500 s = 8 min 20 s.
4.Sol: As the surrounding is identical, vessel is
identical time taken to cool both water and
liquid (from 30 to 25 C C) is same 2 minutes,
therefore
6.Sol:
7.Sol:
8.Sol:
9.Sol:
CALORIMETRY[ONLINE QUESTIONS]
1.Sol:
2.Sol:
3.Sol:
4.Sol:
water liquid
dQ dQ
dt dt
   
      
(m cw w W T) (m cl l W T)
t t
    
(W= water equivalent of the vessel)
or, m cw wm cl l
 Specific heat of liquid,
W W
l
l
c m c
m

50 1 0.5 /
100
  kcal kg