Study of Gas Laws

The assumptions of the kinetic molecular theory are :

1. Composition of gases — Gases are made of tiny particles moving in all possible directions at all possible speeds. The molecules are negligibly small as compared to the volume of the occupied gas.
2. Gases have neither a fixed volume nor a fixed shape — There is negligible force of attraction between the particles (gas molecules). Therefore, the particles (gas molecules) are free to move in the entire space available to them. Their movement is only restricted by the container, hence it takes the shape of the container.
3. Gases exert pressure in all directions — The moving particles of gas collide with each other and with the walls of the container. Due to these collisions, gas molecules exert pressure. It has been found, that at a given temperature, time and area, the same number of molecules of a gas strike against the walls of the container. Thus, gases exert the same pressure in all directions.
4. Gases are highly compressible — There are large inter-particle (inter-molecular) space between gas molecules, and this accounts for high compressibility of gases.
5. Gases are highly expansible — Gases increase in volume with a decrease in pressure and an increase in temperature.
6. Gases have low density — The number of molecules per unit volume in a gas is very small due to the large intermolecular spaces between their molecules. Therefore, gases have very low density.
7. Gases have a natural tendency for diffusion — Due to large intermolecular spaces, the molecules of two gases mix with each other in such a manner that a homogeneous gaseous mixture is formed.
8. Gases can be liquefied — On cooling, the kinetic energy of the molecules of a gas is reduced and on applying pressure on a cooled gas, the molecules come closer. Thus, the gas liquefies.

According to the kinetic theory of gases, the number of particles and their average kinetic energy in a given mass of gas remain constant. When the volume of a gas is halved, the particles have less space to move, resulting in double the number of molecules striking the container walls per unit area and thus doubling the pressure. Conversely, doubling the volume gives the particles more space, reducing the number of molecules striking the walls per unit area by half and halving the pressure. This demonstrates Boyle's law: as pressure increases, the volume of a gas decreases at a constant temperature.

According to the kinetic theory of gases, the average kinetic energy of gas particles is directly proportional to the absolute temperature. Increasing the temperature causes the particles to move faster and strike the container walls more frequently and forcefully. When pressure is constant, the volume of the gas increases proportionally. Therefore, at constant pressure, the volume of a gas is directly proportional to its temperature (Charles' law).

According to Charles' law, when pressure (P) is constant,

$\text{V} = \text{V}_0 + \dfrac{\text{V}_0}{273}\text{t} \\[1em] \Rightarrow \text{V} = \text{V}_0\Big(\dfrac{273 + \text{t}}{273}\Big)$

∴ Volume at -273°C = $\text{V}_0\Big(\dfrac{273 - 273}{273}\Big)$ = 0

Hence, the volume of a gas would be reduced to zero at -273°C. This temperature is called absolute zero.

This allowed the scientists to establish an absolute temperature scale (known as Kelvin scale) by defining absolute zero as the point where volume of a gas would theoretically be zero.

(a) Boyle's law states that the volume of a given mass of a dry gas is inversely proportional to its pressure at a constant temperature.

(b) (i) Mathematical expression of Boyle's law :

Suppose a gas occupies volume V₁ when its pressure is P₁; then

V₁ α $\dfrac{1}{\text{P}_1}$ or

V₁ = $\dfrac{\text{k}}{\text{P}_1}$ or

P₁V₁ = k = constant

If V₂ is the volume occupied when the pressure is P₂ at the same temperature, then

V₂ α $\dfrac{1}{\text{P}_2}$ or

V₂ = $\dfrac{\text{k}}{\text{P}_2}$ or

P₂V₂ = k = constant

∴ P₁V₁ = P₂V₂ = k; at constant temperature.

(ii) Graphical representation of Boyle's Law:

(a) V vs $\dfrac{1}{\text{P}}$ : a straight line passing through the origin is obtained.

(b) V vs P : a hyperbolic curve in the first quadrant is obtained.

(c) PV vs P : a straight line is obtained parallel to the pressure axis.

(iii) Significance of Boyle's law :
On increasing pressure, volume decreases. The gas becomes denser. Thus, at constant temperature, the density of a gas is directly proportional to pressure.

(a) Charles law states that pressure remaining constant, the volume of a given mass of dry gas increases or decreases by $\dfrac{1}{273}$ of its volume at 0°C for each 1°C increase or decrease in temperature, respectively.

(b) (i) Graphical representation of Charles' law : The relationship between the volume and the temperature of a gas can be plotted on a graph. A straight line is obtained.

(ii) Mathematical expression of Charles' law :

Suppose, a gas occupies V₁ cm³ at T₁ temperature and V₂ cm³ at T₂ temperature, then by Charles' law:

    V₁ α T₁

or V₁ = kT₁ (k is constant)

or $\dfrac{\text{V}_1}{\text{T}_1}$ = k and

    V₂ α T₂

or $\dfrac{\text{V}_2}{\text{T}_2}$ = k

∴ $\dfrac{\text{V}_1}{\text{T}_1}$ = $\dfrac{\text{V}_2}{\text{T}_2}$ = k (at constant pressure)

(iii) Significance of Charles' Law :
Volume of a given mass of a gas is directly proportional to its temperature, hence, density decreases with an increase in temperature.

When the gas is collected over water, it is moist and contains water vapour. The total pressure exerted by moist gas is equal to the sum of the partial pressure of the dry gas and the pressure exerted by water vapour. The partial pressure of water vapour is called as aqueous tension.

P_total = P_gas + P_{water vapour}

P_gas = P_total – P_{water vapour}

Actual pressure of gas = Total pressure – Aqueous tension

Thus, to obtain the actual pressure, the pressure which is due to moisture in the gas has to be subtracted.

The phenomenon is Diffusion.

Diffusion is the process of gradual mixing of two substances, kept in contact, by molecular motion.

The molecules of hydrogen sulphide gas collide with air particles and due to the collisions of the particles, they start moving in a haphazard manner in all possible directions. Due to this, the molecules of hydrogen sulphide gas quickly spread in the surroundings and we can smell the gas from even 50 metres away.

a straight line parallel to X-axis

Reason — The graph of PV vs P is a straight line parallel to the pressure axis.

300 K

Reason — Celsius values can be converted to Kelvin values by adding 273 to degree celsius values. Hence, 27°C + 273 = 300 K

Charles

Reason — Charles' law states that volume of a given mass of a dry gas is directly proportional to its absolute (Kelvin) temperature, if the pressure remains constant.

1⁄2 times

Reason — According to Boyle's law volume of a given mass of a dry gas is inversely proportional to its pressure at constant temperature.

Hence, if pressure is doubled for a fixed mass of a gas, its volume will become 1⁄2 times.

Kelvin

Reason — The S.I. unit of temperature is Kelvin.

373 K

Reason — The steam point on Celsius scale is 100°C and corresponding to that steam point on Kelvin scale is 273 + 100 = 373 K.

Boyle's law

Reason — For graphical verification of Boyle's law a graph of V vs 1/P is a straight line passing through the origin.

Charles' law

Reason — Charles' law states that volume of a given mass of a dry gas is directly proportional to its absolute (Kelvin) temperature, if the pressure remains constant.

20 litres

Reason — Given,

V₁ = 10 l
P₁ = 10 atm
P₂ = 5 atm
V₂ = ?

According to Boyle's law, P₁V₁ = P₂V₂

So, 10 x 10 = 5 x V₂

V₂ = $\dfrac{100}{5}$ = 20 l

273 K and 760 mm

Reason — The standard values chosen are 0°C or 273 K for temperature and 1 atm or 760 mm of Hg for pressure. These values are known as standard temperature and pressure (STP).

Both Q and R

Reason — In graph Q and R, the relationship between volume and temperature of gas is plotted at constant pressure. The straight line proves that volume is directly proportional to temperature (at constant pressure) as per Charles' law.

Only Q

Reason — Boyle's law states that volume of a given mass of a dry gas is inversely proportional to its pressure at constant temperature. Hence, the term isoterm, meaning at constant temperature is used to describe the graphs.

Both A and R are true and R is the correct explanation of A.

Explanation— When air is blown into a balloon, volume and pressure inside the balloon increase. Here, Boyle's law is not violated as the law is valid for a definite mass, whereas mass increases when more air is blown into the balloon. Hence, both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).

A is false but R is true.

Explanation— Gases follow charles's law, volume of given mass of gas is directly proportional to its absolute temperature. At 0°C, a gas still behaves mostly like an ideal gas under standard pressure, and its volume is not negligible. Hence, the assertion (A) is false.
Molecular motion increases with rising temp and decreases with lowering temperature. When the temperature is zero on kelvin scale, the molecular motion ceases. Hence, the reason (R) is true.

Both A and R are true and R is the correct explanation of A.

Explanation— Gas exerts uniform pressure on the walls of the containing vessel. The reason is that the particles (molecules) of the gas collide with each other and with the walls of the containing vessel. Since a large number of particles (molecules) suffer collisions with the wall, an appreciable force acts on the wall. The force exerted on a unit area of the wall of the vessel is equal to the pressure of the gas. Hence, both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).

A is true but R is false.

Explanation— Gases are often compared at Standard Temperature and Pressure (STP) to have a common reference point. Hence, converting volumes to STP (typically 0°C and 1 atm) is a standard procedure. Hence, the assertion (A) is true.
For a given mass of a gas, a change in one or more than one variable, i.e., pressure, volume and temperature, results in a change in the remaining variables. Hence, the reason (R) is false.

A is true but R is false

Explanation— The change in one or more than one variable, i.e., pressure, volume and temperature, results in a change in the remaining variables. So, when stating the volume of a gas, the pressure and temperature should also be given. Hence, the assertion (A) is true.
The volume of a gas is directly proportional to temperature and inversly proportional to pressure. Hence, the reason (R) is false.

Both A and R are true and R is the correct explanation of A.

Explanation— Gases have a natural tendency to mix with one another i.e., they diffuse easily and form homogeneous mixtures. Hence, the assertion (A) is true.
Inter-particle (inter-molecular) spaces in a gas are very large. When two gases are brought in contact with each other, their molecules mix with each other in such a manner that a homogeneous gaseous mixture is formed. Hence, the reason (R) is true and it is the correct explanation of assertion (A).

Both A and R are true and R is the correct explanation of A.

Explanation— According to Charles’s Law, gas volume decreases linearly with temperature (in Kelvin). So, theoretically at 0 K, the volume of an ideal gas would be zero. Hence, the assertion (A) is true. As temperature increases, molecular motion increases, and when temperature decreases, molecular motion also decreases. Theoretically, this suggests that when the temperature is zero, molecular motion ceases. Hence, the reason (R) is true and it is the correct explanation of assertion (A).

Both A and R are true and R is the correct explanation of A.

Explanation— The product of pressure and volume of a given mass of a gas is constant if the temperature remains constant. This can be deduced by platting values of P against values of PV at a constant temperature. Hence, the assertion (A) is true. Volume of a given mass of a dry gas is inversly proportional to its pressure at constant temperature. Hence, the reason (R) is true.

Reason (R) states boyle's law, which explains 'the product of volume and pressure of a given masss of a dry gas at constant temperature is constant. Hence, reason (R) is the correct explanation of assertion (A).

A is true but R is false.

Explanation— For a given mass of a gas, a change in one or more than one variable, i.e., pressure, volume and temperature, results in a change in the remaining variables. So it is necessary to mention conditions while stating the volumes of gases. Hence, the assertion (A) is true. Volume of gas depends on changing pressure and temperature. Volume and temperature relate directly whereas, volume and pressure relate inversely. Hence, The reason (R) is false.

(i) Kelvin = °C + 273 = 273 + 273 = 546 K

(ii) °C = Kelvin - 273 = 293 - 273 = 20°C

V₁ = 800 cm³

P₁ (Initial pressure) = 654 mm of Hg

V₂ = reduces by 40% of it's initial value

= 800 - ($\dfrac{40}{100}$ x 800)

= 800 - 320 = 480 cm³

P₂ (Final pressure) = ?

By Boyle's Law:

$\text{P}_1 \text{V}_1 = \text{P}_2 \text{V}_2$

Substituting the values :

$654 \times 800 = \text{P}_2 \times 480 \\[1em] \text{P}_2 = \dfrac{800 \times 654}{480} \\[1em] \text{P}_2 = 1090 \text{ mm of Hg}$

∴ Final pressure of the gas = 1090 mm of Hg.

(a)

(i) Graph of P vs V :

(ii) Graph of P vs 1/V :

(iii) Graph of PV vs P :

(b) By Boyle's Law:

$\text{P}_1 \text{V}_1 = \text{P}_2 \text{V}_2$

P₁V₁ = 12 is a constant, hence, V₂ = ? at P₂ = 2.3 atm

Substituting the values :

$12 = 2.3 \times \text{V}_2 \\[1em] \text{V}_2 = \dfrac{12}{2.3} \\[1em] \text{V}_2 = 5.217 \text{ lit}$

Hence, volume = 5.217 lit

V₁ = 20 lits.

1 lit. = 1000 cm³

∴ 20 lit = 20,000 cm³

P₁ (Initial pressure) = 100 atmos.

P₂ (Final pressure) = 1 atmos.

V₂ = ?

By Boyle's Law:

$\text{P}_1 \text{V}_1 = \text{P}_2 \text{V}_2$

Substituting the values :

$100 \times 20,000 = 1 \times \text{V}_2 \\[0.5em] \text{V}_2 = 20,00,000 \text{ cm}^3$

200 cm³ = capacity of 1 flask
∴ Number of flasks required for 20,00,000 cm³ of gas

$= \dfrac{20,00,000 }{200} = 10,000$

Hence, number of flask that can be filled = 10,000.

V₁ = 20 litres

P₁ = 29 atmospheric pressure

P₂ = 1.25 atmospheric pressure

V₂ = ?

By Boyle's Law:

$\text{P}_1 \text{V}_1 = \text{P}_2 \text{V}_2$

Substituting the values :

$20 \times 29 = 1.25 \times \text{V}_2 \\[0.5em] \text{V}_2 = \dfrac{20\times 29}{1.25} \\[0.5em] \text{V}_2 = 464 \text{ litres}$

∴ Volume occupied = 464 litres.

V₁ = 561 dm³

P₁ = P

V₂ = 748 dm³

P₂ = ?

By Boyle's Law:

$\text{P}_1 \text{V}_1 = \text{P}_2 \text{V}_2$

Substituting the values :

$561 \times \text{P} = 748 \times \text{P}_2 \\[1em] \text{P}_2 = \dfrac{561\times \text{P}}{748} \\[1em] \text{P}_2 = 0.75 \text{P}$

Change in pressure = P - 0.75P = 0.25P decrease

Percentage decrease in pressure = $\dfrac{\text{Decrease}}{\text{Original}}$ x 100 = $\dfrac{\text{0.25P}}{\text{P}}$ x 100 = 25%

∴ Percentage decrease in the pressure of the gas = 25%

V₁ = 88 cm³

P₁ = 770 mm

P₂ = 880 mm Hg

V₂ = ?

By Boyle's Law:

$\text{P}_1 \text{V}_1 = \text{P}_2 \text{V}_2$

Substituting the values :

$770 \times 88 = 880 \times \text{V}_2 \\[0.5em] \text{V}_2 = \dfrac{770 \times 88}{880} \\[0.5em] \text{V}_2 = 77$

V₂ = 77 cm³

∴ Decrease in volume = 88 - 77 = 11 cm³

By Gas Law:

$\dfrac{\text{P}_1\times\text{V}_1}{\text{T}_1} = \dfrac{\text{P}_2\times\text{V}_2}{\text{T}_2}$

Substituting the values :

$\dfrac{700 \times 143}{300} = \dfrac{280\times\text{\text{V}}_2}{300} \\[1em] \text{V}_2 =\dfrac{700 \times 143 }{280} \\[1em] \text{V}_2 = 357.5 \text{ cm}^3$

∴ Final volume of the gas = 357.5 cm³

V₁ = Initial volume of the gas = 500 cm³
T₁ = Initial temperature of the gas = 273 K
V₂ = Final volume of the gas = reduced by 20% of its original = 500 - ($\dfrac{20}{100}$ x 500) = 500 - 100 = 400 cm³
T₂ = Final temperature of the gas = ?

By Charles' Law:

$\dfrac{\text{V}_1}{\text{T}_1} = \dfrac{\text{V}_2}{\text{T}_2}$

Substituting the values :

$\dfrac{500}{273} = \dfrac{400}{\text{T}_2} \\[1em] \text{T}_2 = \dfrac{400 \times 273}{500} \\[1em] \text{T}_2 = 218.4 \text{ K}$

Final temperature = 218.4 K

By Gas Law:

$\dfrac{\text{P}_1\times\text{V}_1}{\text{T}_1} = \dfrac{\text{P}_2\times\text{V}_2}{\text{T}_2}$

Substituting the values :

$\dfrac{760 \times \text{V}}{273} = \dfrac{ 2 \times 760 \times\text{\text{V}}_2}{3 \times 273} \\[1em] \text{V}_2 =\dfrac{3 }{2}\text{V} \\[1em] \text{V}_2 = 1\dfrac{1}{2} \text{V}$

∴ Final volume of the gas = $1\dfrac{1}{2}$ times the original volume.

By Gas Law:

$\dfrac{\text{P}_1\times\text{V}_1}{\text{T}_1} = \dfrac{\text{P}_2\times\text{V}_2}{\text{T}_2}$

Substituting the values :

$\dfrac{740 \times 30}{288} = \dfrac{ 760 \times\text{\text{V}}_2}{273} \\[1em] \text{V}_2 =\dfrac{740 \times 30\times 273}{288 \times 760} \\[1em] \text{V}_2 = 27.68 \text{ cm}^3$

∴ Volume of carbon dioxide = 27.7 cm³

By Gas Law:

$\dfrac{\text{P}_1\times\text{V}_1}{\text{T}_1} = \dfrac{\text{P}_2\times\text{V}_2}{\text{T}_2}$

Substituting the values :

$\dfrac{1\times \text{V}}{273} = \dfrac{0.5\times\text{2\text{V}}}{\text{T}_2} \\[0.5em] \text{T}_2 = 0.5\times 2 \times 273\\[0.5em] \text{T}_2 = 273\text{ K}$

∴ Final temperature of the gas = 273 K

Initial conditions [S.T.P.] :

P₁ = Initial pressure of the gas = 760 mm Hg
V₁ = Initial volume of the gas = 100 cm³
T₁ = Initial temperature of the gas = 0°C = 273 K

Final conditions :

P₂ (Final pressure) = increased one and a half times of P₁

= (1 + $\dfrac{1}{2}$) of 760

= $\dfrac{3}{2}$ x 760

= 3 x 380

= 1140 mm Hg

V₂ (Final volume) = ?

T₂ (Final temperature) = increased by one-fifth of 273 K

= $\Big(1 + \dfrac{1}{5}\Big)$ of 273

= $\dfrac{6}{5}$ x 273 = $\dfrac{1638}{5}$

By Gas Law:

$\dfrac{\text{P}_1\times\text{V}_1}{\text{T}_1} = \dfrac{\text{P}_2\times\text{V}_2}{\text{T}_2}$

Substituting the values :

$\dfrac{760 \times 100 }{273} = \dfrac{1140 \times \text{V}_2}{\dfrac{1638}{5}} \\[1em] \dfrac{760 \times 100 }{273} = \dfrac{5 \times 1140 \times \text{V}_2}{1638} \\[1em] \text{V}_2 = \dfrac{760 \times 100 \times 1638}{273 \times 5 \times 1140} \\[1em] \text{V}_2 = 80 \text{ cm}^3$

∴ Final volume of the gas = 80 cm³

Initial conditions [S.T.P.] :

P₁ = Initial pressure of the gas = P
V₁ = Initial volume of the gas = V
T₁ = Initial temperature of the gas = -15°C = -15 + 273 = 258 K

Final conditions :

P₂ (Final pressure) = pressure decreases to 60% of its original value

= $\dfrac{60}{100}$ of P

= $\dfrac{3}{5}$P

V₂ (Final volume) = volume increases by 50% of its original value

= 1 + $\dfrac{50}{100}$ of V

= $\dfrac{100 + 50}{100}$ of V

= $\dfrac{150}{100}$ of V

= $\dfrac{3}{2}$V

T₂ (Final temperature) = ?

By Gas Law:

$\dfrac{\text{P}_1\times\text{V}_1}{\text{T}_1} = \dfrac{\text{P}_2\times\text{V}_2}{\text{T}_2}$

Substituting the values :

$\dfrac{\text{PV }}{258} = \dfrac{\dfrac{3}{5} \text{P}\times\dfrac{3}{2}\text{V}}{\text{T}_2} \\[1em] \text{T}_2 = \dfrac{3\times 3\times 258}{5 \times 2} = \dfrac{2322}{10} \\[1em] \text{T}_2 = 232.2 \text{K} \\[1em]$

∴ Final temperature of the gas = 232.2 - 273 = -40.8°C

Initial conditions [S.T.P.] :

P₁ = Initial pressure of the gas = 100 Pascal

V₁ = Initial volume of the gas = 2 Litres

T₁ = Initial temperature of the gas = 27°C = 27 + 273 = 300 K

Final conditions :

P₂ (Final pressure) = 50 Pascal
V₂ (Final volume) = 1 litre
T₂ (Final temperature) = ?

By Gas Law:

$\dfrac{\text{P}_1\times\text{V}_1}{\text{T}_1} = \dfrac{\text{P}_2\times\text{V}_2}{\text{T}_2}$

Substituting the values :

$\dfrac{100 \times 2}{300} = \dfrac{50 \times 1}{\text{T}_2} \\[1em] \text{T}_2 = \dfrac{50\times 3}{2} \\[1em] \text{T}_2 = 75 \text{K} \\[1em]$

∴ Final temperature of the gas = 75 - 273 = -198°C

V₁ = 2500 cm³

P₁ = 760 mm Hg

P₂ = increased by two and a half times

= (1 + $2\dfrac{1}{2}$ ) times of 760 mm Hg

= (1 + $\dfrac{5}{2}$ ) times of 760 mm Hg

= ( $\dfrac{7}{2}$ x 760) mm Hg

V₂ = ?

By Boyle's Law:

$\text{P}_1 \text{V}_1 = \text{P}_2 \text{V}_2$

Substituting the values :

$760 \times 2500 = \dfrac{7}{2} \times 760 \times \text{V}_2 \\[1em] \text{V}_2 = \dfrac{5000}{7} \\[0.5em]$

∴ Final volume = $\dfrac{5000}{7}$ cm³

V₁ = Initial volume of the gas = $\dfrac{5000}{7}$ cm³
T₁ = Initial temperature of the gas = 273 K

V₂ = Final volume of the gas = 2500 cm³
T₂ = Final temperature of the gas = ?

By Charles' Law:

$\dfrac{\text{V}_1}{\text{T}_1} = \dfrac{\text{V}_2}{\text{T}_2}$

Substituting the values :

$\dfrac{5000}{7 \times 273} = \dfrac{2500}{\text{T}_2} \\[1em] \text{T}_2 = \dfrac{7 \times 273 \times 2500}{5000} \\[1em] \text{T}_2 = \dfrac{7 \times 273}{2} \\[1em] \text{T}_2 = 3.5 \times 273 \\[1em]$

∴ Final temperature = 3.5 times of Kelvin temperature.

(a) P₁ = Initial pressure of the gas = 100 cm Hg

T₁ = Initial temperature of the gas = 273 K

P₂ = Final pressure of the gas = 10 cm Hg

T₂ = Final temperature of the gas = ?

V₁ = V₂ = V [constant volume]

By Gas Law:

$\dfrac{\text{P}_1\times\text{V}_1}{\text{T}_1} = \dfrac{\text{P}_2\times\text{V}_2}{\text{T}_2}$

Substituting the values :

$\dfrac{100 \times \text{V}}{273} = \dfrac{10 \times \text{V}}{\text{T}_2} \\[1em] \text{T}_2 = \dfrac{ 273}{10} \\[1em] \text{T}_2 = 27.3 \text {K} \\[1em]$

∴ When the pressure is 10 cm Hg, the temperature is 27.3 K.

(b) P₁ = Initial pressure of the gas = 100 cm Hg

T₁ = Initial temperature of the gas (ice point) = 273 K

P₂ = Final pressure of the gas = ?

T₂ = Final temperature of the gas = 100°C = 273 + 100 = 373 K

V₁ = V₂ = V [constant volume]

By Gas Law:

$\dfrac{\text{P}_1\times\text{V}_1}{\text{T}_1} = \dfrac{\text{P}_2\times\text{V}_2}{\text{T}_2}$

Substituting the values :

$\dfrac{100 \times \text{V}}{273} = \dfrac{\text{P}_2 \times \text{V}}{373} \\[1em] \text{P}_2 = \dfrac{ 37300}{273} \\[1em] \text{P}_2 = 136.63 \text { cm of Hg} \\[1em]$

∴ When the chamber is brought to 100°C, the pressure is 136.63 cm Hg.

Initial conditions :

P₁ = Initial pressure of the gas = 800 mm of Hg

V₁ = Initial volume of the gas = 10,000 litres

T₁ = Initial temperature of the gas = -3°C = -3 + 273 = 270 K

Final conditions:

P₂ (Final pressure) = 400 mm of Hg

V₂ (Final volume) = 10 litres x n cylinders

T₂ (Final temperature) = 270 K

where n is the number of cylinders = ?

By Gas Law:

$\dfrac{\text{P}_1\times\text{V}_1}{\text{T}_1} = \dfrac{\text{P}_2\times\text{V}_2}{\text{T}_2}$

Substituting the values :

$\dfrac{800 \times 10,000}{270} = \dfrac{400 \times 10 \times n}{270} \\[1em] \text{n} = \dfrac{800 \times 10,000 \times 270}{400 \times 10 \times 270} \\[1em] \text{n} = \dfrac{800\times 10}{4} \\[1em] \text{n} = 2000 \\[1em]$

∴ Number of cylinders = 2000.

P₁ = Initial pressure of the gas = 1 atm

T₁ = Initial temperature of the gas = 27°C = 27 + 273 = 300 K

P₂ = Final pressure of the gas = 3 atm

T₂ = Final temperature of the gas = ?

V₁ = V₂ = V [constant volume]

By Gas Law:

$\dfrac{\text{P}_1\times\text{V}_1}{\text{T}_1} = \dfrac{\text{P}_2\times\text{V}_2}{\text{T}_2}$

Substituting the values :

$\dfrac{1\times\text{V}}{300} = \dfrac{3\times\text{V}}{\text{T}_2} \\[1em] \text{T}_2 = 3 \times 300\\[1em] \text{T}_2 = 900 \text {K} \\[1em]$

∴ The temperature above which the tube will burst is 627°C.

Initial conditions :

P₁ = Initial pressure of the gas = 70 cm of Hg

V₁ = Initial volume of the gas = V

T₁ = Initial temperature of the gas = 273 K

Final conditions:

P₂ (Final pressure) = 80 cm of Hg

V₂ (Final volume) = 2V

T₂ (Final temperature) = ?

By Gas Law:

$\dfrac{\text{P}_1\times\text{V}_1}{\text{T}_1} = \dfrac{\text{P}_2\times\text{V}_2}{\text{T}_2}$

Substituting the values :

$\dfrac{70 \times \text{V}}{273} = \dfrac{80 \times \text{2V}}{\text{T}_2} \\[1em] \text{T}_2 = \dfrac{80 \times \text{2V} \times 273}{70 \times \text{V}} \\[1em] \text{T}_2 = \dfrac{16 \times 273}{7} \\[1em] \text{T}_2 = 624 \text{ K} \\[1em]$

∴ Final temperature of the gas = 624 - 273 = 351°C

Initial conditions [S.T.P.] :

P₁ = Initial pressure of the gas = P
V₁ = Initial volume of the gas = V
T₁ = Initial temperature of the gas = -15°C = -15 + 273 = 258 K

Final conditions :

P₂ (Final pressure) = pressure decreases to 75% of its original value

$= \dfrac{75 }{100} \text{ of P} \\[1em] = \dfrac{3 }{4} \text{P} \\[1em] = 0.75 \text {P}$

V₂ (Final volume) = Volume increases by 50% of its original value

$= \text {1} + \dfrac{50}{100} \text{ of V} \\[1em] = \dfrac{100 + 50}{100} \text{ of V} \\[1em] = \dfrac{150}{100} \text{of V} \\[1em] = \dfrac{3}{2} \text{V} = 1.5 \text{V}$

T₂ (Final temperature) = ?

By Gas Law:

$\dfrac{\text{P}_1\times\text{V}_1}{\text{T}_1} = \dfrac{\text{P}_2\times\text{V}_2}{\text{T}_2}$

Substituting the values :

$\dfrac{\text{P}\text{V}}{258} = \dfrac{0.75\text{P}\times\text{1.5V}}{\text{T}_2}$

Solve for T₂:

$\text {T}_2 = 1.125 \times 258 \\[1em] \text {T}_2 = 290.25 \text {K}$

∴ Final temperature of the gas = 290.25 - 273 = 17.25°C

Initial conditions [S.T.P.] :

P₁ = Initial pressure of the gas = 1 atm

V₁ = Initial volume of the gas = 22.4 l

T₁ = Initial temperature of the gas = 273 K

Final conditions :

P₂ (Final pressure) = 4 atm

T₂ (Final temperature) = 27°C = 27 + 273 = 300 K

V₂ (Final volume) = ?

By Gas Law:

$\dfrac{\text{P}_1\times\text{V}_1}{\text{T}_1} = \dfrac{\text{P}_2\times\text{V}_2}{\text{T}_2}$

Substituting the values :

$\dfrac{1 \times 22.4}{273} = \dfrac{4 \times \text{V}_2}{300} \\[1em] \text{V}_2 = \dfrac{300 \times 22.4}{273 \times 4} \\[1em] \text{V}_2 = 6.15 \text{ l} \\[1em]$

∴ Volume occupied = 6.15 l

1.2 lit N₂ at 25°C and 748 mm Hg

Initial conditions :

P₁ = Initial pressure of the gas = 748 mm Hg

V₁ = Initial volume of the gas = 1.2 lit

T₁ = Initial temperature of the gas = 25°C = 25 + 273 = 298 K

Final conditions :

P₂ (Final pressure) = 760 mm of Hg

T₂ (Final temperature) = 273 K

V₂ (Final volume) = ?

By Gas Law:

$\dfrac{\text{P}_1\times\text{V}_1}{\text{T}_1} = \dfrac{\text{P}_2\times\text{V}_2}{\text{T}_2}$

Substituting the values :

$\dfrac{748 \times 1.2}{298} = \dfrac{760 \times \text{V}_2}{273} \\[1em] \text{V}_2 = \dfrac{748 \times 1.2\times 273}{298 \times 760} \\[1em] \text{V}_2 = 1.081 \space \text{lit} \\[1em]$

∴ Volume of N₂ at S.T.P. = 1.081 lit and volume of O₂ at S.T.P. is 1.25 lit. Hence, volume of O₂ at S.T.P. is greater than N₂ at S.T.P.

Initial conditions :

P₁ = 750 - water vapour pressure = 750 - 14 = 736 mm Hg

V₁ = Initial volume of the gas = 50 cm³

T₁ = Initial temperature of the gas = 17°C = 17 + 273 = 290 K

Final conditions :

P₂ (Final pressure) = 760 mm of Hg

T₂ (Final temperature) = 273 K

V₂ (Final volume) = ?

By Gas Law:

$\dfrac{\text{P}_1\times\text{V}_1}{\text{T}_1} = \dfrac{\text{P}_2\times\text{V}_2}{\text{T}_2}$

Substituting the values :

$\dfrac{736 \times 50}{290} = \dfrac{760 \times \text{V}_2}{273} \\[1em] \text{V}_2 = \dfrac{736 \times 50 \times 273}{290 \times 760} \\[1em] \text{V}_2 = 45.58 \text{ cm}^3 \\[1em]$

∴ Volume of dry gas = 45.6 cm³

(a)

By Gas Law:

$\dfrac{\text{P}_1\times\text{V}_1}{\text{T}_1} = \dfrac{\text{P}_2\times\text{V}_2}{\text{T}_2}$

Substituting the values :

$\dfrac{72 \times 380}{364} = \dfrac{760 \times \text{V}_2}{273} \\[1em] \text{V}_2 = \dfrac{72 \times 380 \times 273}{364\times 760} \\[1em] \text{V}_2 = 27 \text{ cm}^3 \\[1em]$

(b)

By Gas Law:

$\dfrac{\text{P}_1\times\text{V}_1}{\text{T}_1} = \dfrac{\text{P}_2\times\text{V}_2}{\text{T}_2}$

Substituting the values :

$\dfrac{72 \times 380}{364} = \dfrac{700 \times \text{V}_2}{273} \\[1em] \text{V}_2 = \dfrac{72 \times 380 \times 273}{364\times 700} \\[1em] \text{V}_2 = 29.314 \text{ cm}^3 \\[1em]$

Initial conditions :

P₁ = 750 - water vapour pressure = 750 - 12 = 738 mm Hg

V₁ = Initial volume of the gas = 28 cm³

T₁ = Initial temperature of the gas = 14°C + 273 = 287 K

Final conditions [S.T.P.] :

P₂ (Final pressure) = 760 mm of Hg

T₂ (Final temperature) = 273 K

V₂ (Final volume) = ?

By Gas Law:

$\dfrac{\text{P}_1\times\text{V}_1}{\text{T}_1} = \dfrac{\text{P}_2\times\text{V}_2}{\text{T}_2}$

Substituting the values :

$\dfrac{738 \times 28 }{287} = \dfrac{760 \times \text{V}_2}{273} \\[1em] \text{V}_2 = \dfrac{738 \times 28 \times 273}{287 \times 760} \\[1em] \text{V}_2 = 25.86 \text{ cm}^3 \\[1em]$

∴ Volume of dry gas = 25.9 cm³

P₁ = Initial pressure of the gas = 12 atm

T₁ = Initial temperature of the gas = 27°C = 27 + 273 = 300 K

P₂ = Final pressure of the gas = 14.9 atm

T₂ = Final temperature of the gas = ?

V₁ = V₂ = V [constant volume]

By Gas Law:

$\dfrac{\text{P}_1\times\text{V}_1}{\text{T}_1} = \dfrac{\text{P}_2\times\text{V}_2}{\text{T}_2}$

Substituting the values :

$\dfrac{12\times\text{V}}{300} = \dfrac{14.9\times\text{V}}{\text{T}_2} \\[1em] \text{T}_2 = \dfrac{14.9 \times 300}{12}\\[1em] \text{T}_2 = 372.5 \text {K} \\[1em]$

∴ The temperature at which the cylinder will explode = 99.5°C.

Initial conditions :

P₁ = 700 mm Hg

V₁ = Initial volume of the gas = 20 litres

T₁ = Initial temperature of the gas = 27°C + 273 = 300 K

Final conditions [S.T.P.] :

P₂ (Final pressure) = 760 mm of Hg

T₂ (Final temperature) = 273 K

V₂ (Final volume) = ?

By Gas Law:

$\dfrac{\text{P}_1\times\text{V}_1}{\text{T}_1} = \dfrac{\text{P}_2\times\text{V}_2}{\text{T}_2}$

Substituting the values :

$\dfrac{700 \times 20 }{300} = \dfrac{760 \times \text{V}_2}{273} \\[1em] \text{V}_2 = \dfrac{7 \times 2 \times 273}{3\times 76} \\[1em] \text{V}_2 = 16.76 \space l \\[1em]$

22.4 l of gas at S.T.P. weighs 70 g

∴ 16.76 l of gas at S.T.P. will weigh

= $\dfrac{70 \times 16.76}{22.4}$ = 52.38 g

∴ Weight of gas = 52.38 g

Gas is a state of matter in which inter-particle attraction is weak and inter-particle space is so large that the particles become completely free to move randomly in the entire available space.

Diffusion is the process of gradual mixing of two substances, kept in contact, by molecular motion.

If a jar of chlorine is opened in one end of the room, the odour of the gas can be detected in the entire room, because its particles start moving in a haphazard manner in all directions, till there is an equal concentration throughout the room.

Molecular motion is directly proportional to temperature. When temperature increases molecular motion increases and when temperature decreases molecular motion decreases.

The temperature on the Kelvin scale at which molecular motion completely ceases is called absolute zero. Absolute zero is -273°C.

The temperature scale with its zero at -273.15°C and whose each degree is equal to one degree on the Celsius scale, is called the Kelvin or absolute scale of temperature.

(a) The standard values chosen are 0°C or 273 K for temperature, and 1 atmospheric unit (atm) or 760 mm Hg for pressure. These standard values are known as standard temperature and pressure (S.T.P.).

(b) Since volume of a gas changes remarkably with change in temperature and pressure, it becomes necessary to choose standard values of temperature and pressure to which gas volumes can be referred.

(a) Charles' law

(b) Boyle's law

The size of 1 degree on the Kelvin scale is the same as the size of 1 degree on the Celsius scale i.e., unit size on Kelvin scale is equal to the unit size on Celsius scale.
The value on the Celsius scale can be converted to Kelvin scale by adding 273 to it. Kelvin scale values can be converted to degree Celsius values by subtracting 273 from it.

(a) This can be explained on the basis of Gas equation

$\dfrac{\text{P}_1\text{V}_1}{\text{T}_1}$ = $\dfrac{\text{P}_2\text{V}_2}{\text{T}_2}$

When volume is constant we get,

$\dfrac{\text{P}_1}{\text{T}_1}$ = $\dfrac{\text{P}_2}{\text{T}_2}$

Hence, when temperature is doubled, pressure is doubled, as pressure and temperature are directly proportional.

(b) This can be explained on the basis of Boyle's law

P₁V₁ = P₂V₂

Hence, when the volume is made half of its original value keeping the temperature constant, pressure becomes double, as pressure and volume are inversely proportional.

(a) Kelvin scale makes application and use of gas laws simple. More significantly, all values on Kelvin scale are positive.

(b) Boiling point of water on the Kelvin scale is 373 K.

Temperature in °C = K - 273

So, 373 K - 273 = 100°C.

(a) The temperature -273°C is called absolute zero and it is theoretically the lowest temperature that can ever be achieved. The absolute (Kelvin) scale has its zero at -273°C. As temperatures lower than -273°C are not possible, hence, all temperatures in Kelvin scale are in positive figures.

(b) The number of molecules per unit volume in a gas is very small as compared to solids and liquids. Gases have large inter-molecular spaces between their molecules. Therefore gases have lower density compared to that of solids or liquids.

(c) The moving particles of a gas collide with each other and also with the walls of the container. Due to these collisions gas molecules exert pressure. It has been found that at a given temperature, time and area, the same number of molecules of a gas strike against the walls of the container. Thus, gases exert the same pressure in all directions.

(d) Since volume of a gas changes remarkably with change in temperature and pressure, it becomes necessary to choose standard values of temperature and pressure to which gas volumes can be referred.

(e) Boyle's law states that the volume of a dry gas is inversely proportional to its pressure at a constant temperature. When we inflate a balloon, the volume and pressure of the air increase simultaneously, seemingly violating Boyle's law. However, this is not the case because according to the law, the mass of the gas should remain constant. In the balloon scenario, as more air is blown into the balloon, the mass of the gas increases, ensuring Boyle's law is not violated.

(f) Atmospheric pressure is low at high altitudes, so air is less dense. Hence, lesser quantity of oxygen is available for breathing. Thus, mountaineers have to carry oxygen cylinders with them.

(g) Gas is a state of matter in which inter-particle attraction is weak and inter-particle space is so large that the particles become completely free to move randomly in the entire available space. Hence, it fills completely the vessel in which it is kept.

(a) Boyle's law — It states that the volume of a given mass of a dry gas is inversely proportional to its pressure at a constant temperature i.e., P₁V₁ = P₂V₂

(b) P.V. isothermal for Boyle's law is shown below :

(a) Charles' law — It states that pressure remaining constant, the volume of a given mass of dry gas increases or decreases by $\dfrac{1}{273}$ of its volume at 0°C for each 1°C increase or decrease in temperature, respectively.

(b) Let V₀ be the volume of a fixed mass of a gas at 0°C and let V be its volume at temperature t°C at constant pressure. Then according to Charles' law,

V = V_o + $\dfrac{\text{V}_0}{273}$t (when P is constant)

V = V_o (1 + $\dfrac{\text{t}}{273}$)

= V_o ($\dfrac{\text{273 + t}}{273}$)

V = $\dfrac{\text{V}_0}{273}$T where T = 273 + t

For a given mass of a gas,

$\dfrac{\text{V}_0}{273}$ = constant

∴ V = k x T (where k is constant)

or V α T and $\dfrac{\text{V}}{\text{T}}$ = K

Suppose a gas occupies V₁ cm³ at T₁ temperature and V₂ cm³ at T₂ temperature, then by Charles law,

    V₁ α T₁

or V₁ = kT₁ (k is constant)

or $\dfrac{\text{V}_1}{\text{T}_1}$ = k and

    V₂ α T₂

or $\dfrac{\text{V}_2}{\text{T}_2}$ = k

∴ $\dfrac{\text{V}_1}{\text{T}_1}$ = $\dfrac{\text{V}_2}{\text{T}_2}$ = k (at constant pressure)

(c) Volume of a given mass of a gas is directly proportional to its temperature, hence, density decreases with an increase in temperature. This is the reason that hot air is filled into balloons used for meteorological purposes.

Converting temperature to Kelvin Scale :

Graph of V versus absolute T is shown below :

(a) The graph between V and T is a straight line.

(b) Yes, as we can see in the graph, the line passes through the origin.

(c) Charles' law

(a) The average kinetic energy of the molecules of a gas is proportional to the absolute temperature.

(b) The temperature on the Kelvin scale at which molecular motion completely ceases is called absolute zero.

(c) If the temperature is reduced to half, volume would also reduce to half.

(d) The melting point of ice is 273 Kelvin.

(a) The volume of a given mass of a dry gas is inversely proportional to its pressure at constant temperature.

(b) Volume of a fixed mass of a dry gas is directly proportional to its absolute temperature (Kelvin), pressure remaining constant.

(c) 0°C is equal to 273 Kelvin

(d) Standard temperature is 0°C or 273 K.

(e) The boiling point of water is 373 K.

(i) The three variables for gas laws are Volume (V), Pressure (P), Temperature (T)

(ii) S.I. units of these variables:

Volume: Cubic metre (m³)

Pressure: Pascal (Pa)

Temperature: Kelvin (K)

(a) Standard temperature

(i) 0°C (ii) 273 K

(b) Standard pressure

(i) 1 atmospheric pressure (atm)

(ii) 760 mm Hg

(iii) 76 cm of Hg

(iv) 760 torr

(a) Volume of a gas at 0 Kelvin is zero.

(b) Absolute temperature of a gas at 7°C is 7 + 273 = 280 K.

(c) The gas equation is

$\dfrac{\text{P}_1\text{V}_1}{\text{T}_1}$ = $\dfrac{\text{P}_2\text{V}_2}{\text{T}_2}$

(d) Ice point in absolute temperature = 0 + 273 = 273 K

(e) Standard temperature = 0°C = 273 K
Standard pressure = 760 mm Hg = 76 cm Hg = 1 atmosphere (atm)

V₁ = Initial volume of the gas = 80 mL
P₁ = Initial pressure of the gas = 100 cm Hg
V₂ = Final volume of the gas = 160 mL
P₂ = Final pressure of the gas = ?

By Boyle's Law:

$\text{P}_1 \text{V}_1 = \text{P}_2 \text{V}_2$

Substituting the values :

$100 \times 80 = \text{P}_2 \times 160 \\[0.5em] \text{P}_2 = \dfrac{100}{2} \\[0.5em] \text{P}_2 = 50 \text{ cm of Hg }$

∴ Final pressure of the gas = 50 cm of Hg.

V₁ = Initial volume of the gas = 74 cm³
P₁ = Initial pressure of the gas = 760 mm
P₂ = Final pressure of the gas = 740 mm
V₂ = Final volume of the gas = ?

By Boyle's Law:

$\text{P}_1 \text{V}_1 = \text{P}_2 \text{V}_2$

Substituting the values :

$760 \times 74 = 740 \times \text{V}_2 \\[0.5em] \text{V}_2 = \dfrac{74 \times 760 }{740} \\[0.5em] \text{V}_2 = 76 \text{ cm}^3$

∴ Final volume of the gas = 76 cm³

(a) V₁ = Initial volume of the gas = 80 cm³
P₁ = Initial pressure of the gas = 100 mm of Hg
P₂ = Final pressure of the gas = 125 mm of Hg
V₂ = Final volume = ?

By Boyle's Law:

$\text{P}_1 \text{V}_1 = \text{P}_2 \text{V}_2$

Substituting the values :

$100 \times 80 = 125 \times \text{V}_2 \\[0.5em] \text{V}_2 = \dfrac{100 \times 80 }{125} \\[0.5em] \text{V}_2 = 64 \text{ cm}^3$

∴ x = 64 cm³

(b) V₁ = Initial volume of the gas = 40 cm³
P₁ = Initial pressure of the gas = 200 mm of Hg
V₂ = Final volume of the gas = 32 cm³
P₂ = Final pressure of the gas = ?

By Boyle's Law:

$\text{P}_1 \text{V}_1 = \text{P}_2 \text{V}_2$

Substituting the values :

$200 \times 40 = \text{P}_2 \times 32 \\[0.5em] \text{P}_2 = \dfrac{200 \times 40 }{32} \\[0.5em] \text{P}_2 = 250 \text{ mm of Hg}$

∴ y = 250 mm of Hg

(i) Graph of volume plotted against pressure :

(ii) Graph of PV plotted against pressure :

V₁ = Initial volume of the gas = 400 cm³
P₁ = Initial pressure of the gas = 760 mm
P₂ = Final pressure of the gas

= increased by 25% of 760 mm Hg

= $\dfrac{25}{100}$ of 760 mm Hg

= ($\dfrac{1}{4}$ x 760) + 760

= 190 + 760 = 950 mm

V₂ = Final volume = ?

By Boyle's Law:

$\text{P}_1 \text{V}_1 = \text{P}_2 \text{V}_2$

Substituting the values :

$760 \times 400 = 950 \times \text{V}_2 \\[0.5em] \text{V}_2 = \dfrac{400 \times 760 }{950} \\[0.5em] \text{V}_2 = 320 \text{ cm}^3$

∴ Final volume = 320 cm³

V₁ = Initial volume of the gas = 600 cm³
P₁ = Initial pressure of the gas = 330 cm Hg
V₂ = Final volume = 600 + 300 = 900 cm³
P₂ = Final pressure of the gas = ?

By Boyle's Law:

$\text{P}_1 \text{V}_1 = \text{P}_2 \text{V}_2$

Substituting the values :

$330 \times 600 = \text{P}_2 \times 900 \\[0.5em] \text{P}_2 = \dfrac{600 \times 330 }{900} \\[0.5em] \text{P}_2 = 220 \text{ cm of Hg}$

∴ Final pressure = 220 cm of Hg

V₁ = Initial volume of the gas = V

P₁ = Initial pressure of the gas = 1080 mm of Hg

V₂ = Final volume of the gas = decreased by 40%
= 60% of V₁ = $\dfrac{60}{100}$V = $\dfrac{3}{5}$V

P₂ = Final pressure of the gas = ?

By Boyle's Law:

$\text{P}_1 \text{V}_1 = \text{P}_2 \text{V}_2$

Substituting the values :

$\text{V} \times 1080 = \text{P}_2 \times \dfrac{3}{5}\text{V} \\[0.5em] \text{P}_2 = \dfrac{5\times 1080 }{3} \\[0.5em] \text{P}_2 = 1800 \text{ mm of Hg}$

∴ Final pressure = 1800 mm of Hg

V₁ = Initial volume of the gas = V

P₁ = Initial pressure of the gas = 940 mm of Hg

V₂ = Final volume of the gas = decreased by 20%
= 80% of V₁ = $\dfrac{80}{100}$V = $\dfrac{4}{5}$V

P₂ = Final pressure of the gas = ?

By Boyle's Law:

$\text{P}_1 \text{V}_1 = \text{P}_2 \text{V}_2$

Substituting the values :

$\text{V} \times 940 = \text{P}_2 \times \dfrac{4}{5}\text{V} \\[0.5em] \text{P}_2 = \dfrac{5\times 940}{4} \\[0.5em] \text{P}_2 = 1175 \text{ mm Hg}$

∴ Final pressure = 1175 mm Hg

V₁ = Initial volume of the gas = 800 cm³
P₁ = Initial pressure of the gas = P
P₂ = Final pressure of the gas = 3P
V₂ = Final volume of the gas = ?

By Boyle's Law:

$\text{P}_1 \text{V}_1 = \text{P}_2 \text{V}_2$

Substituting the values :

$\text{P} \times 800 = \text{V}_2 \times 3\text{P} \\[0.5em] \text{V}_2 = \dfrac{1}{3} \times 800 \\[0.5em]$

∴ Final volume is one-third of initial volume.

Given,

V₁ = Initial volume of the gas = 500 dm³
P₁ = Initial pressure of the gas = 1 bar.
V₂ = Final volume = 200 dm³
P₂ = Final pressure of the gas = ?

By Boyle's Law:

$\text{P}_1 \text{V}_1 = \text{P}_2 \text{V}_2$

Substituting the values :

$1 \times 500 = \text{P}_2 \times 200 \\[0.5em] \text{P}_2 = \dfrac{500}{200} \\[0.5em] \text{P}_2 = 2.5 \text{ bar }$

∴ Final pressure of the gas = 2.5 bar.

V₁ = 2 lits.

P₁ (Initial pressure) = 760 mm Hg

V₂ = 4 dm³

P₂ (Final pressure) = ?

By Boyle's Law:

$\text{P}_1 \text{V}_1 = \text{P}_2 \text{V}_2$

Substituting the values :

$760 \times 2 = \text{P}_2 \times 4 \\[1em] \text{P}_2 = \dfrac{760 \times 2}{4} \\[1em] \text{P}_2 = 380 \text{ mm Hg}$

∴ Final pressure of the gas = 380 mm Hg.

(i) °C = Kelvin - 273 = 37 - 273 = -236°C

(ii) °C = Kelvin - 273 = 273 - 273 = 0°C

(iii) Kelvin = °C + 273 = -27 + 273 = 246 K

(iv) Kelvin = °C + 273 = 27 + 273 = 300 K

V₁ = Initial volume of the gas = 20 mL
T₁ = Initial temperature of the gas = -13°C = -13 + 273 = 260 K
T₂ = Final temperature of the gas = 117°C = 117 + 273 = 390 K
V₂ = Final volume of the gas =?

By Charles' Law:

$\dfrac{\text{V}_1}{\text{T}_1} = \dfrac{\text{V}_2}{\text{T}_2}$

Substituting the values :

$\dfrac{\text{20}}{260} = \dfrac{\text{V}_2}{390} \\[1em] \text{V}_2 = \dfrac{20 \times 390}{260} \\[1em] \text{V}_2 = 30 \text{ mL}$

∴ Final volume = 30 mL

V₁ = Initial volume of the gas = V
T₁ = Initial temperature of the gas = 0°C = 273 K
V₂ = Final volume of the gas = 2V
T₂ = Final temperature of the gas = ?

By Charles' Law:

$\dfrac{\text{V}_1}{\text{T}_1} = \dfrac{\text{V}_2}{\text{T}_2}$

Substituting the values :

$\dfrac{\text{V}}{273} = \dfrac{2\text{V}}{\text{T}_2} \\[1em] \text{T}_2 = 2\times 273 \\[1em] \text{T}_2 = 546 \text{ K}$

∴ Final temperature in °C = 546 - 273 = 273°C

1 lit = 1000 cm³
∴ 0.4 lit = 400 cm³

V₁ = Initial volume of the gas = 400 cm³
T₁ = Initial temperature of the gas = 250 K
T₂ = Final temperature of the gas = 27°C = 27 + 273 = 300 K
V₂ = Final volume of the gas = ?

By Charles' Law:

$\dfrac{\text{V}_1}{\text{T}_1} = \dfrac{\text{V}_2}{\text{T}_2}$

Substituting the values :

$\dfrac{400}{250} = \dfrac{\text{V}_2}{300} \\[1em] \text{V}_2 = \dfrac{400 \times 300}{250} \\[1em] \text{V}_2 = 480 \text{ cm}^3$

Volume of air expelled = 480 - 400 = 80 cm³

V₁ = Initial volume of the gas = 3 litres
T₁ = Initial temperature of the gas = 27°C = 27 + 273 = 300 K
T₂ = Final temperature of the gas = -73°C = -73 + 273 = 200 K
V₂ = Final volume of the gas = ?

By Charles' Law:

$\dfrac{\text{V}_1}{\text{T}_1} = \dfrac{\text{V}_2}{\text{T}_2}$

Substituting the values :

$\dfrac{3}{300} = \dfrac{\text{V}_2}{200} \\[1em] \text{V}_2 = \dfrac{3 \times 2}{3} \\[1em] \text{V}_2 = 2 \text{ lit}$

∴ Final volume = 2 litres.

V₁ = Initial volume of the gas = V

T₁ = Initial temperature of the gas = 300 K

V₂ = Final volume of the gas = $\dfrac{1}{3}$V

T₂ = Final temperature of the gas = ?

By Charles' Law:

$\dfrac{\text{V}_1}{\text{T}_1} = \dfrac{\text{V}_2}{\text{T}_2}$

Substituting the values :

$\dfrac{\text{V}}{300} = \dfrac{\dfrac{1}{3}\text{V}}{\text{T}_2} \\[1em] \text{T}_2 = \dfrac{300}{3} \\[1em] \text{T}_2 = 100 \text{ K}$

∴ Final temperature = 100 K

V₁ = Initial volume of the gas = V

T₁ = Initial temperature of the gas = 273 K

T₂ = Final temperature of the gas = 273°C = 273 + 273 = 546 K

V₂ = Final volume of the gas = ?

By Charles' Law:

$\dfrac{\text{V}_1}{\text{T}_1} = \dfrac{\text{V}_2}{\text{T}_2}$

Substituting the values :

$\dfrac{\text{V}}{273} = \dfrac{\text{V}_2}{546} \\[1em] \text{V}_2 = \dfrac{\text{V}\times 546}{273} \\[1em] \text{V}_2 = 2\text{V} \\[1em]$

∴ Volume of a gas at 273°C is twice its volume at 273 K

V₁ = Initial volume of the gas = 3 litres

T₁ = Initial temperature of the gas = 0°C = 273 K

T₂ = Final temperature of the gas = -20°C = -20 + 273 = 253 K

V₂ = Final volume of the gas = ?

By Charles' Law:

$\dfrac{\text{V}_1}{\text{T}_1} = \dfrac{\text{V}_2}{\text{T}_2}$

Substituting the values :

$\dfrac{3}{273} = \dfrac{\text{V}_2}{253} \\[1em] \text{V}_2 = \dfrac{3 \times 253}{273} \\[1em] \text{V}_2 = 2.78 \text{ litres}$

∴ Final volume = 2.78 litres.

V₁ = Initial volume of the gas = V cc
T₁ = Initial temperature of the gas = 240 K
T₂ = Final temperature of the gas = 127°C = 127 + 273 = 400 K
V₂ = Final volume of the gas = ?

By Charles' Law:

$\dfrac{\text{V}_1}{\text{T}_1} = \dfrac{\text{V}_2}{\text{T}_2}$

Substituting the values :

$\dfrac{\text{V}}{240} = \dfrac{\text{V}_2}{400} \\[1em] \text{V}_2 = \dfrac{400 \times \text{V}}{240} \\[1em] \text{V}_2 = \dfrac{5\times \text{V}}{3}$

Increase in vol. = $\dfrac{5\text{V}}{3}$ - V = $\dfrac{2\text{V}}{3}$

Percentage increase in vol. = $\dfrac{\text{Increase}}{\text{Original}}$ x 100 = $\dfrac{2\text{V}}{3\text{V}}$ x 100 = 66.67%

∴ Percentage increase in the volume of the gas = 66.67%

(a) To find the temperature so that volume gets doubled :

V₁ = Initial volume of the gas = 0.4 litre
T₁ = Initial temperature of the gas = 17°C = 17 + 273 = 290 K
V₂ = Final volume of the gas = double of initial = 2 x 0.4 = 0.8 litre
T₂ = Final temperature of the gas = ?

By Charles' Law:

$\dfrac{\text{V}_1}{\text{T}_1} = \dfrac{\text{V}_2}{\text{T}_2}$

Substituting the values :

$\dfrac{\text{0.4}}{290} = \dfrac{\text{0.8}}{\text{T}_2} \\[1em] \text{T}_2 = \dfrac{0.8 \times 290}{0.4} \\[1em] \text{T}_2 = 580 \text{ K}$

∴ Final temperature in °C = 580 - 273 = 307°C

(b) To find the temperature so that volume gets reduced to half :

V₁ = Initial volume of the gas = 0.4 litre
T₁ = Initial temperature of the gas = 17°C = 17 + 273 = 290 K
V₂ = Final volume of the gas = reduced to half of initial = 0.2 litre
T₂ = Final temperature of the gas = ?

By Charles' Law:

$\dfrac{\text{V}_1}{\text{T}_1} = \dfrac{\text{V}_2}{\text{T}_2}$

Substituting the values :

$\dfrac{\text{0.4}}{290} = \dfrac{\text{0.2}}{\text{T}_2} \\[1em] \text{T}_2 = \dfrac{0.2\times 290}{0.4} \\[1em] \text{T}_2 = 145$

∴ Final temperature in °C = 145 - 273 = -128°C