1) Two moles of an ideal gas at 1 atm are compressed to 2 atm at 273 K . The enthalpy change of the process will ?

1 ) Two moles of an ideal gas at 1 atm are compressed to 2 atm at 273 K . The enthalpy change of the process will : -

A ) increased ,                                                    B ) decreased ,

C ) remain same ,                                              D ) None of these ,

Correct Answer : -   C ) remain same ,

Explanation : -

Compression of an ideal gas do not involve change in enthalpy . So enthalpy change will be zero or remain same .

 

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

 

2 ) Rate of diffusion of a gas is proportional to : -  √ ( P / d ) .

 

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

 

3 ) The ratio of most probable velocity to that of average velocity is  : -

Solution : -
The most probable velocity = Vmps = √ ( 2RT / M )  .

Average velocity = Vavg = √ ( 8RT / πM ) .

So  ,  Vmps / Vavg = √ ( 2RT / M ) / √ ( 8RT / πM )  ,

Vmps / Vavg = √ ( 2 π / 8 )  =  √ π / 2  ,  Answer . 

OR

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

 

4 ) The factor possible for lower mercury level in capillary tube is : -

A ) high density ,                                       B ) low density ,

C ) high pressure ,                          D ) None of these ,

Correct Answer : -   A ) high density ,

 

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .