Kinetic Theory Of Gases MCQ Level – 2 (part - 1) - IIT JAM MCQ

Therefore 
which gives,   T = 20 K
The correct answer is: 20 K

We have
T_c = 647 K
p_c = 22.09 MPa = 22.09 × 10³ kPa
V_c = 0.0566 dm^-3 mol^-1
Thus, 
 = 0.0189 dm³ mol^-1

= 212.3 kPa dm⁶ mol^-2
The correct answer is:  = 212.3 kPa dm⁶ mol^-2, 0.0189 dm³ mol^-1

Since, b = 4 × volume occupied by the molecule in 1 mole of a gas
or  
 therefore 

= 1.335 × 10^-8 cm = 133.5 pm
d = 2r = 267 pm
The correct answer is: 267 pm

Total kinetic energy;

K.E. = (0.5 mol) 
= 1702 J
The correct answer is: 1702 J

Here

T = 300K
Since,   


= 1.281 × 10²³ m^-3
 The correct answer is: 1.281 × 10²³ m^-3

Vander Waals equation
 = nRT
and 
So,  

= 2589.31 kPa – 363.76 kPa
 = 2225.55 kPa.
The correct answer is: 2222.55 kPa

Most probable speed,


Since,  0.005 u_mp = (0.005) (422.09 ms^-1)
= 2.11 ms^-1
Therefore 
 4.22 ms^-1
According to Maxwell distribution law,

 = exp (–1.000) = 0.3679
Hence,

The correct answer is: 8.303 × 10^-3

Density ρ = 009 gdm^-3 = 0.09 kgm^-3
Average speed, 
Also, 

= 1694 ms^-1
The correct answer is: 1694 ms^-1

If dN₁ is the number of molecule in the speed range u_mp to u_mp + du and dN₂ is the corresponding number in the speed range 2u_mp to 2u_mp + du, then according to the Maxwell distribution, we have,

and   

 therefore 

Now, Since, 
therefore 
The correct answer is: 0.199