Physics Times 07.2019

[OFFLINE QUESTIONS]

1.Sol: Given: Ti  17 273 290K

Tf  27 273 300K

Atmospheric pressure, P 0  1 10^5 Pa

Volume of room, V 0  30 m^3

 A

PVN

N

RT

Given that number of molecules N=n

   0 0  ^11

f i   f i A

P V

n n N

R T T

1 10 30^52311

6.023 10

8.314 300 290

   

     

 

 2.5 10^25

2.Sol: Molar specific heat '







Q

C

n T

Specific heat

 



C Q

m T

'

C  m M(Molecular weight)

C n

'



C

C

M

' '

CP CV R

MCP MCV R

P V

C C R

M

 

For hydrogen M=2

2

P  V

R

C C a

For nitrogen M=28

28

   CP CV R b

 14

a

b

3.Sol: Average time between collisions

rms

Mean free path



v

(^2)
(^1) ;
3
t
d N V
RT
M


CV
t
T
 where 2
3
 
  
  
C M
d N R
2
2
V
T
t
 
Given that TV^1 k
(^21)
2
V V k
t
 ^1
2

 V k
t
1
2

t V^
1
2
q


4.Sol: The pressure inside the tube after closing
the tube is 76 cm of mercury.
We assume that T is constant
PV1 1P V2 2
After pulling the tube up the pressure of the
air is 76 cm - x, where x is the rise in the level
of mercury
(76)(8) (54  x)(76 x)
   x^2130 x 3496 0
x 38 cm
Length of air column= 54- 38 = 16 cm
5.Sol: Here, work done is zero.
So, loss in kinetic energy = Change in internal
energy
of gas
[OFFLINE QUESTIONS]
1.Sol:
2.Sol:
3.Sol:
4.Sol:
5.Sol:
(^12)
2 V 1
R
mv nC T n T

 

(^12)
2 1
m R
mv T
M



