Which of the following will have maximum total translational kinetic energy at temperature 300 K …

Q: Which of the following will have maximum total translational kinetic energy at temperature 300 K (Neglect vibrational motion and treat gas as ideal )

(a) 1 kg , H2

(b) 1 kg , He

(c) (1/2) kg H2 + (1/2) kg He

(d) (1/4) kg H2 + (3/4) kg He

Solution :

$\displaystyle U = \frac{f}{2}n R T $ ;

Where f is the degree of freedom . As H2 has greatest degree of freedom i.e. 5 . and number of mole contain in 1 kh is also maximum . Therefore U is maximum for 1 kg of H2 .

Correct option is (a)

As shown schematically in the figure , two vessels contain water solution (temperature T) of potassium permanganate ….

Q: As shown schematically in the figure , two vessels contain water solution (temperature T) of potassium Permanganate (KMnO4) of different concentration n1 and n2 (n1 >n2) molecules per unit volume with Δn =(n1-n2) << n1 . When they are connected by a tube of small length l and cross-sectional area S , KMnO4 starts to diffuse from left to right vessel through the tube . Consider the collection of molecules to behave as dilute ideal gases and the difference in their partial pressure in the two vessels causing the diffusion . The speed v of molecules is limited by the viscous force -β v on each molecule , where β is constant . Neglecting all terms of the order (Δn)^2 , Which of the following is/are correct ? (kB is the Boltzmann constant)

Numerical

(a) the force causing the molecules to move across the tube is Δn kB T S

(b) force balance implies n1βvl = Δn kB T

(c) total number of molecules going across the tube per sec is $(\frac{\Delta n}{l}) (\frac{k_B T}{\beta})S$

(d) rate of molecules getting transferred through the tube does not change with time

Ans: (a,b,c)

Solution: n1 >> (n1-n2) = Δn

$p_1 = \frac{n_1 R T}{N_A}$ and $p_2 = \frac{n_2 R T}{N_A}$

$F = (n_1 -n_2)k_B T S = \Delta n k_B T S$

$V = \frac{\Delta n k_B T S}{\beta} $

$\Delta n k_B T S = l n_1 S \beta v $

$n_1 \beta v l = \Delta n k_B T $

Total number of molecules/second $= \frac{(n_1 v dt)S}{dt}$

$= n_1 v S = \frac{\Delta n K_B T v S}{\beta v l}$

$= (\frac{\Delta n}{l}) (\frac{k_B T}{\beta})S $

As Δn will decrease with time , therefore rate of molecules getting transfer decreases with time

Consider a collection of a large number of particles each with speed v. The direction of velocity is randomly distributed in the collection….

Q: Consider a collection of a large number of particles each with speed v. The direction of velocity is randomly distributed in the collection. What is the magnitude of the relative velocity between a pairs in the collection

(a) 2V/π

(b) V/ π

(c) 8V/ π

(d) 4V/ π

Click to See Answer :
Ans: (d)

 

N molecules each of mass m of gas A and 2 N molecules each of mass 2 m of gas B are contained in the same vessel at temperature T….

Q: N molecules each of mass m of gas A and 2 N molecules each of mass 2 m of gas B are contained in the same vessel at temperature T. The mean square of the velocity of molecules of gas B is v2 and the mean square of x component of the velocity of molecules of gas A is w2 . The ratio w2/v2 is

(a) 1

(b) 2

(c) 1/3

(d) 2/3

Click to See Answer :
Ans: (d)

 

Which one of the following is not an assumption of kinetic theory of gases

Q: Which one of the following is not an assumption of kinetic theory of gases

(a) The volume occupied by the molecules of the gas is negligible

(b) The force of attraction between the molecules is negligible

(c) The collision between the molecules are elastic

(d) All molecules have same speed

Click to See Answer :
Ans: (d)