(a) Surface tension: It is defined as the force acting per unit length perpendicular to the line drawn on the surface. It's unit is Nm^-1.
(b) Surface tension of a liquid depends upon following factors.
(i) Temperature: Surface tension decreases with rise in temperature. As the temperature of the liquid increases, the average kinetic energy of the molecules increases. Thus, there is a decrease in intermolecular force of attraction which decreases the surface tension.
(ii) Nature of the liquid: Greater the magnitude of intermolecular forces of attraction in the liquid, greater will be the value of surface tension.

Total pressure of mixture = 1200 torr
Partial pressure of N₂ (P_N2) = 300 torr
Partial pressure of H₂ (P_H2) 1200 - 300 = 900 torr
According to ideal gas equation,
PV = nRT
\({ P }_{ H_{ 2 } }\) = \(\frac { { n }_{ H_{ 2 } }RT }{ V } \)= 900 torr    ...(i)
\({ P }_{ N_{ 2 } }\) = \(\frac { { n }_{ N_{ 2 } }RT }{ V } \)= 300 torr   ...(ii)
Divide (i) by (ii),   \(\frac { { P }_{ H_{ 2 } } }{ { P }_{ H_{ 2 } } } =\frac { { n }_{ H_{ 2 } } }{ { n }_{ H_{ 2 } } } \) = \(\frac{900}{300}=\frac{3}{1}\)\(\therefore\)\({ n }_{ H_{ 2 } }\):\({ n }_{ N_{ 2 } }\)::3:1

(i) Calculation of critical temperature (T_c)
a = 4.10 dm⁶ bar mol^-1
b = 0.035 dm³ mol^-1
R = 0.0821 dm³ bar mol^-1 K^-1.
Now, critical temperature, T_c = \(\frac{8a}{27Rb}\)
Substituting the values, T_c = \(\frac{8\times4.10}{27\times0.0821\times0.035}\)
(ii) Calculation of critical pressure (P_c)
P_c\(\frac{a}{27b^2}\) = \(\frac{4.10}{27\times0.0821\times0.035}\) = 123.96 bar

In summer, due to higher temperature, the average Kinetic energy of the air molecules inside the tyre increases,i.e. molecules start moving faster. Hence, the pressure on the walls of the tube increases. If pressure inside is not kept low at the time of inflation, at a higher temperature, the pressure may become so high that the tyre may burst.

As we go to higher altitudes, the atmospheric pressure decreases. Thus, the pressure outside the balloon decreases. To regain equilibrium with the external pressure, the gas inside expands to decrease its pressure. Hence, the size of the balloon increases.

pV = nRT
For gas A, P_AV = n_ART ................(i)
Similarly for gas B,  P_BV = n_BRT ...........(ii)
Number of mole of gas A; n_A = \(\cfrac { 1 }{ M_{ A } } \) 
(M_A = molar mass of gas A)
Number of mole of gas B; n_B = \(\cfrac { 1 }{ M_{ B } } \) 
(M_B = molar mass of gas B) 
Pressure of gas A, p_A = 2 bar
Total pressure, p_total = p_A + p_B = 3 bar
Pressure of gas, P_B = P_total - P_A =3 - 2 = 1 bar
V, R and T are same for both the gases.
Hence, from Eqs.(i) and Eqs (ii),
\(\cfrac { { p }_{ A } }{ { p }_{ B } } =\cfrac { n_{ A } }{ { n }_{ B } } =\cfrac { 1\times M_{ B } }{ M_{ A }\times 2 } \)
or  \(\cfrac { { M }_{ B } }{ { M }_{ A } } =\cfrac { 2p_{ A } }{ { p }_{ B } } =\cfrac { 2\times 2 }{ 1 } \)   or M_B = 4 M_A

Z = 2.54 x 10^-3

\(Z=\frac { pV }{ nRT } \ i.e \ 0.5=\frac { 100\times V }{ 1\times 0.082\times 273 } \)
or 
\(V=0.1119L\)        
\( \left( P+\frac { a }{ { V }^{ 2 } } \right) (v-b)=RT \ for \ 1 \ mol\)
Neglecting b, \(\left( P+\frac { a }{ { V }^{ 2 } } \right) V=RT\)
or  \(pV+\frac { a }{ V } =RT\)
or \(\frac { pV }{ RT } +\frac { a }{ VRT } =1\)
or​  
\(a=\left( 1-\frac { pV }{ RT } \right) VRT=(1-0.5)0.1119\times 0.082\times 273\)​​​​​​
\( =1.252 \ atm \ L^{ 2 }mol^{ -2 }\)

Compound = SO₂ Mass of one molecule of gas = 1.07 x 10^-22g

P_CH4 = 1.23 atm, P_H2 = 3.079 atm, total pressure = 4.31 atm

(a) Pressure will remain same (p atm).
(b) Temperature will remain same ( t \(^{o}\) C ).
(c) Density will remain same (d g cm^-3).
(d) The number of moles in each compartment will be n/2.

A liquid boils when vapour pressure becomes equal to atmospheric pressure. At hill station, atmospheric pressure is low. Therefore, liquid boils at a lower temperature and cooking is not perfect. In a pressure cooker, the pressure inside the cooker increases and the liquid boils at a higher temperature.

A point 'A' both the curves intersect each other. At this point real gas behaves as an ideal gas. p₁ and V₁ are the pressure and volume which corresponds to this point A.

At low pressure, the real gas shows very small deviation from ideal behaviour because the two curves almost coincide at low pressure.