According to Boyle's law, at constant temperature for a given mass of gas at constant temperature,
P₁V_I = P₂V₂ = P₃V₃
1 atm x 1 dm³ = 2 atm x V₂ = P₃ x 0.25 dm³
\(\therefore\) 2 atm x V₂ = 1 atm x 1 dm³

\(\boxed{V_2=0.5\ dm^3}\)
and P₃ x 0.25 dm³ = 1 atm x 1 dm³

\(\boxed{P_3=4\ atm}\)

The pressure-density relationship can be derived from the Boyle's law as shown below.
P₁V₁ = P₂V₂ (Boyle's law)
\({ P }_{ 1 }\frac { m }{ { d }_{ 1 } } ={ P }_{ 2 }\frac { m }{ { d }_{ 2 } } \)
where "m" is the mass, d ₁ and d₂ are the densities of gases at pressure P₁ and P₂ .
\(\frac { { P }_{ 1 } }{ { d }_{ 1 } } =\frac { { P }_{ 2 } }{ { d }_{ 2 } } \)
In other words, the density of a gas is directly proportional to pressure.

T₁ = 15^oC + 273 T₂= 38 + 273
T₁ = 288 K  T₂ = 311 K
V₁ = 2.58 dm³ V₂ = ?
(P = 1 atm constant)
\(\frac { { V }_{ 1 } }{ { T }_{ 1 } } =\frac { { V }_{ 2 } }{ { T }_{ 2 } } \)
\({ V }_{ 2 }=\left( \frac { { V }_{ 1 } }{ { T }_{ 1 } } \right) \times { T }_{ 2 }\)

V₂ = 2.78 dm³ i.e. volume increased from 2.58 dm³ to 2.78 dm³.

Rate of diffusion \(\text { (r) } \alpha \frac{1}{\sqrt{M}}\)
\(\mathrm{M}_{\mathrm{NH}_{3}}=17 ; \mathrm{M}_{\mathrm{HCl}}=36.5 \)
\(\therefore \mathrm{r}_{\mathrm{NH}_{3}}>\mathrm{r}_{\mathrm{HCl}} \)
Hence white fumes are first formed near HCL.

The passenger in a moving bus falls in the forward direction, when brakes are applied suddenly is in accordance with Newton's first law of motion.
When breaks are applied, the automobile comes to rest but the passenger due to inertia of motion tend to continue to move in forward direction hence fall forward.
The movement of helium balloon in the opposite direction is due to the difference in the density of surrounding air. When the car stops suddenly, the air moves forward due to inertia of motion.
As a result the air in the front of the car is more dense that at the rear of the car creating more pressure on the front side of the balloon. This causes the balloon move towards the rear of the car where air is less dense.
The body is at rest but the car is forward motion, due to inertia the passengers are pushed toward the front of the car on forward acceleration.

On heating incineration might take place due to the increase of pressure

The larger the size of the molecule, the greater will be Van der Waals' attraction. Therefore greater the deviation from ideal behaviour. So bromine will deviate more from ideal behaviour because it has bigger atoms.

a) Hydrogen is the lightest gas in the atmosphere. So it rises up and other gases which are heavier like O₂ & N₂ come down towards the surface of the earth according to Graham's law of diffusions \(\mathrm{r} \alpha \sqrt{\frac{1}{M}}\). Hydrogen diffuses very fast. Its mean speed in greater than the escape velocity from the earth. As a consequence, H₂ would have escaped from the atmosphere long time ago.
b) The acceleration due to gravity 'g' on moon surface is small. The value of escape velocity is also small. The molecules of the atmospheric gases on the moon's surface have thermal velocities greater than the escape velocity.
All the molecules have escaped. So the atmosphere is so thin.

If there is a tiny sticky area on the walls of the container of a gas, then the collision of the molecules will not be elastic. So the gas will behave as a real gas. So the pressure will be less than the calculated value.

Gases which obey Boyle's Law and Charle's law or ideal gas equation PV = nRT are called ideal gases.
All gases whose behaviour is consistent with the assumption of kinetic theory of gases under all condition are called ideal gases.

The following table gives the pressure of a gas with increasing temperature at constant volume for two cylinders.

Both the models explain Gay - Lussac's Law. P \(\propto\) T are constant V & n
For 50 L container:
The pressure increases with rise in temperature (at constant volume)
For 75 L container:
The pressure increases with rise in temperature (at constant volume)

Charles law states that "At constant pressure, the volume of a given mass of an ideal gas is directly proportional to its temperature." According to Charles Law, if we were to take a balloon filled with air and increase the temperature of the air inside, the volume of air would increase causing the balloon to expand. This is caused by the heating of the molecules of air inside the balloon causing them to move rapidly. In the same manner if we cooled the balloon in a freezer, the volume of air decreases, making the balloon look partially deflated.

At a given temperature the volume occupied by a fixed mass of a gas is inversely proportional to its pressure.
\(V\alpha \frac { 1 }{ P }\) at constant T& n.
Mathematical form: P₁V₁ = P₂ V₂ = K

We will use the ideal gas equation for this calculation as below:
\({P=nRT\over V}={{2mol\times 0.0821Latm.K^{-1}.mo^l{-1}\times (70+273K)}\over{6dm^3}}\)
= 9.39 atm.

Boyle's law is applicable only at the condition of constant temperature. Since in adiabatic expansion, temperature is lowered Boyle's law cannot be applied.

Gases are very less denser. They have negligible intermolecular forces of attraction between the molecules. They are in continuous kinetic motion. So they won't settle at the bottom due to gravitational forces. The material that is most dense will sink to the bottom; the less denser will go up.

According to Avogadro's hypothesis: V \(\alpha\) n
\(\therefore \frac{V_{1}}{n_{1}}=\frac{V_{2}}{n_{2}}=\text { constant }\)
V₁ & n₂ are the volume and number of moles of a gas and V₂ & n₂ are a different set of values of volume and number of moles of the same gas at same temperature and pressure.

No. of mol of dioxygen \((n_{O_2})=\frac{ mass of O_2}{molar mass of O_2}\)
=\(\frac{70.6 g }{32.0 g mol^{-1}}=2.21 mol\)
No. of mol of neon \((n_{Ne})=\frac{mass of Ne}{molar mass of Ne}=\frac{167.5 g}{20 g mol^{-1}}\)
= 8.375 mol.
mole fraction of di-hydrogen = \(\frac{n_{O_2}}{n_{O_2}+n_{N_e}}=\frac{2.21}{2.21+8.375}=0.21\)
mole fraction of neon = 1 - 0.21 = 0.79
partial pressure of O₂ = mole fraction of O₂ x Total pressure
= 0.21 x 25 bar = 5.25 bar
partial pressure of Ne = mole fraction of Ne x Total pressure
= 0.79 x 25 bar = 19.75 bar

The average value of the constant is 0.0053.

(a) 5.076 atm
(b) 3.495 atm

According to charles law
\(\frac { { V }_{ 1 } }{ { T }_{ 1 } } =\frac { 1.54 }{ 293 } \)
= 0.0052
T₁ = 20 C + 273 = 293 K
T₂ = 40°C
\(\frac { { V }_{ 2 } }{ { T }_{ 2 } } =\frac { 1.65 }{ 40+273 } =\frac { 1.65 }{ 313 } =0.0052\)
T₃ = 100°C + 273 = 373 K
\(\frac { { V }_{ 3 } }{ { T }_{ 3 } } =\frac { 1.95 }{ 373K } =0.0052\)
T₄ = 120°C+273 = 393 K
\(\frac { { V }_{ 4 } }{ { T }_{ 4 } } =\frac { 2.07 }{ 393 } =0.0052\)
The average value of constant of proportionality is 0.0052.

0.15 dm²

The gaseous state is described completely using the following four variables T,P, V and n and their relationships were governed by the gas laws studied so far.
Boyle's law \(\mathrm{V} \propto \frac{1}{\mathrm{P}}\)
Charles law \(\mathrm{V} \propto T\)
Avogadro's law \(\mathrm{V} \propto n\)
We can combine these equations into the following general equation that describes the physical behaviour of all gases.
\( \mathrm{V} \propto \frac{\mathrm{nT}}{\mathrm{P}} \)
\(\mathrm{V}=\frac{\mathrm{nRT}}{\mathrm{P}}\)
Where, R is the proportionality constant called universal gas constant. The above equation can be rearranged to give the ideal gas equation
PV = nRT

Astronauts must wear space suits filled with air, whenever they leave a space craft and are exposed to the environment of space.
Dangers experienced on moon's space :
1. In space there is no air to breathe and no air pressure.
2. Space is extremely cold and filled with dangerous radiations.
3. If they do not wear space suits, there may be bleeding due to high body pressure.
4. The space suits prevent astronauts from impacts of small bits of space dust.
5. There is no atmosphere on the moon and there are dangers from micro-meteorite impacts.
6. The astronauts are protected from these dangers on wearing space suits.

The van der Waals equation for n moles is
\(\left( P+\frac { { an }^{ 2 } }{ { v }^{ 2 } } \right) \left( v-nb \right) =nRT\) ....(1)
For 1 mole
\(\left( P+\frac { { a }^{ 2 } }{ { v }^{ 2 } } \right) \left( v-b \right) =RT\) ...(2)
From the equation we can derive the values of critical constants P_c, V_c and T_c, in terms of a and b, the van der Waals constants, On expanding the above equation
\(pv+\frac { a }{ v } -pb-\frac { ab }{ { v }^{ 2 } } -RT=0\)    ...(3)
Multiply equation (3) by V² / P
\(\frac { { v }^{ 2 } }{ p } \left( Pv+\frac { a }{ v } -pb-\frac { ab }{ { v }^{ 2 } } -RT \right) =0\)
\({ v }^{ 3 }+\frac { av }{ p } +-{ bv }^{ 2 }-\frac { ab }{ { v }^{ 2 } } -\frac { { RTV }^{ 2 } }{ p } \)    ...(4)
When the above equation is rearranged in powers of Y.
\({ v }^{ 3 }-\left[ \frac { RT }{ P } +b \right] { v }^{ 2 }+\left[ \frac { a }{ p } \right] v-\left[ \frac { ab }{ p } \right] =0\)        ...(5)
The equation (5) is a cubic equation in V. On solving this equation,
we will get three solutions. At the critical point all these three solutions of V are equal to the critical volume V_C. The pressure and temperature becomes Pc and T_c respectively
i.e., V = V_c
V - V_c = 0
(V - V_C)³ = 0
V3 - 3V_CV² + 3V_c²V - V_c³ = 0 .....(6)
As equation (5) is identical with equation (6), we can equate the coefficients of V², V and constant terms in (5) and (6).
\(-3{ v }_{ c }{ v }^{ 2 }=-\left[ \frac { { RT }_{ c } }{ { p }_{ c } } +b \right] { v }^{ 2 }\)
\(3{ v }_{ c }=\frac { { RT }_{ c } }{ { p }_{ c } } +b\)           .....(7)
\(3{ v }_{ c }^{ 2 }=\frac { a }{ { p }_{ c } } \)           ......(8)
\(3{ v }_{ c }^{ 2 }=\frac { ab }{ { p }_{ c } } \)           ...(9)
Divide equation (9) by equation (8)
\(\frac { { v }_{ c }^{ 3 } }{ 3{ v }_{ c }^{ 2 } } =\frac { ab/{ p }_{ c } }{ a/{ p }_{ c } } \)
\(\frac { { V }_{ c } }{ 3 } =b\)
i.e. V_c = 3b          .....(10)
when equation (10) is substituted in (8)
\(3{ v }_{ c }^{ 2 }=\frac { a }{ { p }_{ c } } \)
\({ p }_{ c }=\frac { a }{ 3{ v }_{ c }^{ 2 } } =\frac { a }{ 3\left( { 3b }^{ 2 } \right) } =\frac { a }{ { 3\times 9b }^{ 2 } } =\frac { a }{ { 27 }b^{ 2 } } \)
\({ p }_{ c }=\frac { a }{ { 27 }b^{ 2 } } \)            ...(11)
substituting the values of V_c and P_c in equation (7),
\(3vc=b+\frac { { RT }_{ c } }{ p } \)
\(3\left( 3b \right) =b+\frac { { RT }_{ c } }{ \left( \frac { a }{ { 27b }^{ 2 } } \right) } \)
\(9b-b=\left( \frac { { RT }_{ c } }{ a } \right) 27{ b }^{ 2 }\)
\(8b=\frac { { t }_{ c }R{ 27b }^{ 2 } }{ a } \)
\(\therefore { T }_{ c }=\frac { 8ab }{ 27R{ b }^{ 2 } } =\frac { 8a }{ 27Rb } \)
\({ T }_{ c }=\frac { { 8 }_{ a } }{ 27Rb } \)           ......(12)
The critical constants can be calculated using the values of van der Waals constant of a gas and vice versa.
\(a=3{ V }_{ C }^{ 2 }{ P }_{ C }\quad and\quad b=\frac { { V }_{ C } }{ 3 } \)

The van der equation for a real gas is
\(\left( P+{{{am}^{2}}\over{{V}^{2}}} \right)(V-nb)=nRT\)
Pressure, Correction:
The pressure of a gas is directly proportional to the force created by the bombardment of molecules on the walls of the container. The speed of a molecule moving towards the wall of the container is reduced by the attractive forces exerted by its neighbours. Hence, the measured gas pressure is lower than the ideal pressure of the gas. Hence, van der Waals introduced a correction term to this effect.
Van der Waals found out the forces of attraction experienced by a molecule near the wall are directly proportional to the square of the density of the gas.
\(P^{\prime} \propto \rho^{2} ; \quad \rho=\frac{n}{v}\)
where n is the number of moles of gas and
V is the volume of the container
\( \Rightarrow p^{\prime} \alpha \frac{n^{2}}{V^{2}} \)
\(\Rightarrow p^{\prime}=a \frac{n^{2}}{V^{2}}\)
where a is proportionality constant and depends on the nature of gas
Therefore \(P_{\text {ideal }}=P+\frac{\operatorname{an}^{2}}{V^{2}}\)

Volume Correction
As every individual molecule of a gas occupies a certain volume, the actual volume is less than the volume of the container,
V. Van der Waals introduced a correction factor V' to this effect. Let us calculate the correction term by considering gas molecules as spheres.
V = excluded volume
Excluded volume for two molecules
\(=\frac{4}{3} \pi(2 r)^{3}=8\left(\frac{4}{3} \pi r^{3}\right)=8 V_{m}\)
Where V_m it a volume of a single molecule
Excluded volume for single molecule = \(\frac{8 \mathrm{~V}_{\mathrm{m}}}{2}=4 \mathrm{~V}_{\mathrm{m}}\)
Excluded volume for n molecule = n(4V_m) = nb
Where b is van der waals constant which is equal to 4V_m
\( \Rightarrow V^{\prime}=n b \)
\(V_{\text {ideal }}=V-n b\)
Replacing the corrected pressure and volume in the ideal gas equation PV = nRT we get the Van der Waals equation of state for real gases as below,
\(\left(p+\frac{a^{2}}{V^{2}}\right)(V-n b)=n R T\)
The constants a and b are van der Waals constants and their values vary with the nature of the gas. It is an approximate formula for the non-ideal gas.