NEET Practice Questions (MCQs) with Answers & Solutions

(c) KE =$\frac{3}{2}$RT (for one mole of a gas)

As, the kinetic energy of a gaseous molecule depends only on temperature, thus at constant temperature, the kinetic energy of the molecules remains the same.

(d) SATP means 1 bar and 25 $°$C .

(d) Heating effect is noticed on subjecting a gas for Joule-Thompson effect above its inversion temperature

(d) Most probable velocity, U_mp = $\sqrt{\frac{2RT}{\mathrm{\pi }}}$

                  Mean velocity, U_av = $\sqrt{\frac{8RT}{\mathrm{\pi }}}$

 Root mean square velocity, U_rms = $\sqrt{\frac{3RT}{\mathrm{\pi }}}$

      So, U_mp:U_av:U_rms = $\sqrt{\frac{2RT}{\mathrm{\pi }}}$:$\sqrt{\frac{8RT}{\mathrm{\pi }}}$:$\sqrt{\frac{3RT}{\mathrm{\pi }}}$

                                   =$\sqrt{2}:\sqrt{\frac{8}{\pi }}:\sqrt{3}$

(c) According to ideal gas equation,

                         pV = nRT

                          p = $\frac{n}{V}$RT

                    1 atm = 1 mol L^-1 x 0.082 x T

                          T =1/0.082 = 12 K

(b) The compressibility factor (Z) = (Px22.4)/RT = 1 (for ideal gas)

                                           (Z) = (P x V_m)/RT <1   (for real gas)

                                   .^.. 22.4/V_m > 1 or V_m < 22.4

(b) C_p-C_v = R;

c_p=M x C_p and c_v = M x C_v

(d) These are van der Waals' equations for 1 mole (a) and n mole gas (b), (c).

Applying Boyle's Law, P₁V₁=P₂V₂ 

Suppose initial pressure is P and initial volume is V, then 

PV=P2 x 0.95V

P2=(p/0.95)=(100p/95)

Increase in pressure = (100P/95)-P=(5P/95)

% increase in pressure = {(5P/95)/P} x 100 = 500/95 = 5.26%

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%