Physics Times – July 2019

 U Q

For on mole

Using equipartition of energy, we have

(^3) RTMCT
(^3)
3 3 8.314
27 10
C R
M 
  

 C 925 /J kgK
5.Sol: Kinetic energy of each molecule,

.^3
2 B

K E K T

Given thatT 273 K

Height attained by the gas molecule, h=?

3 819

. (273)
2 2

K E KB  KB

K.E = P.E

819

2

 KB Mgh^819

2

 h KB

Mg

6.Sol:

3

 

RT

c

M

2

(1930) 3 8.314 300

M

  

3 8.314 300 2

1930 1930

   



M g

The gas is H 2.

7.Sol: From the given information we calculate
theaverage volume occupied by each
molecule (not the volume of molecule). The
average volume occupied or share of each
molecule is equal to the ratio of total volume
divides by total number of molecules. This
volume can be approximated as a cube of side
length ( ). The cube root of the average
volume per molecule is equal to ( ). Here
the value ( ) is also called as mean free path
and it is equal to average distance between
two molecules of a gas as shown in the figure.

Ideal gas equation is

A

PV nRT N RT

N

 

Given that

8 , 300

.

R J T K

K mole

 

6 10^231

NA mole

  

15 5

P 4.0 10 10 Nm 2

  

15 5 23

8 300

A 4.0 10 10 6 10

V RT

N PN 

  

   

3 V 1011 m 3

N

   

  2.15 10^4 0.215mm

8.Sol: In isothermal process temperature

remains constant.

3 3 3

rms  

RT PV PV

v

M Mn m

(i)

Isothermal work done is

2

1

ln ln 2

V

W nRT PV

V

 

Given that W  575 J PVln 2

575

ln(2)

 PV (ii)

From eq’s (i) & (Ii) we get

5.Sol:

6.Sol:

7.Sol:

8.Sol:

9.Sol:

2

3 575 3 575 499 /

rms (ln 2) 10 0.693

v m s

m 

    



9.Sol: In isobaric process the heat given to a

gas is

Q nC T P

The graph between Q and T is a straight

line passing through origin.

slopenCP

As n same for all the gases

We know that

CP( Poly ) CP( Di ) CP(Mono)

a,b,c correspond to P, D and M