Kinetic Theory of Gases MCQs for NEET — Physics Questions with Answers

For open container, 1/n₁T₁ = 1/n₂T₂ (P and V are constant)

Suppose initial moles of gas is n.

1/nx300 = 1/n₂x400

n₂ = (3/4)n

% of gas remain = (3/4)n/n x 100 = 75%

higher the polarity, easier to liquify.

The ease of liquification of a gas depends on their intermolecular force of attraction which in turn is measured in terms of van der Waals’ constant ‘a’. 
Hence, higher the value of ‘a’, greater the intermolecular force of attraction, easier the liquification. 
In the present case, NH3 has highest ‘a’, can most easily be liquefied.

r ∝ 1/$\sqrt{m}$ ∝ r ∝ n

At higher temperature a/v² is negligible incomparision to 'P'

So, Vander Waal gas equation becomes 

P(V-b) = RT

PV-Pb = RT

PV = RT +Pb

n_H2 = 2/2 = 1mole, n_SO2 = 32/64 = 0.5mole

P_H2 = 1/1.5xP_T = 2/3xP_T

At high pressure, a/V² is negligible in comparision to P. So, Vander Waal equation becomes

P(V-b) = RT

PV - Pb =RT

PV/RT - Pb/RT = RT/RT

z = 1+Pb/RT

P₁V₁ + P₂V₂ =PV

720 x 200 + 750 x 400 =P x 1000 P =444 mm

r_A/r_B = 1/4 , r_B/r_C = 1/3

r_C/r_A = 1/(r_A/r_B) x 1/(r_B/r_C) 

= r_B/r_A x r_C/r_B = 4x3 =12 

$$\begin{gathered} \frac{\sqrt{3{\mathrm{RT}}_1}}{{\mathrm{M}}_1}=\frac{\sqrt{3{\mathrm{RT}}_2}}{{\mathrm{M}}_2} \\ \frac{{\mathrm{T}}_1}{{\mathrm{M}}_1}=\frac{{\mathrm{T}}_2}{{\mathrm{M}}_2} \\ {\mathrm{T}}_1={\mathrm{T}}_2\times \frac{{\mathrm{M}}_1}{{\mathrm{M}}_2} \\ {\mathrm{T}}_1=294\times \frac{44}{28}=462 \\ =189 \end{gathered}$$

Average K.E.  = 3/2nRT

K.E₁ = K.E₂

3/2x0.3xRxT₁ = 3/2x0.4xRx400

T₁ = 1600/3 = 533K