Dalton's law of partial pressure: When two or more non-reacting gases are enclosed in a vessel, the total pressure of the gaseous mixture is equal to the sum of the partial pressures that each gas will exert when enclosed separately in the same vessel at constant temperature.
P = P₁ + P₂ + P₃
Where, P is the total pressure of the three gases A, B, and C enclosed in a container. P₁ ,P₂ and P₃ are the partial pressures of the three gases when enclosed separately
in the same vessel at a given temperature one by one.
No, the law cannot be applied. Carbon monoxide and oxygen readily combine to form carbon dioxide. The law can be applied only to the non-reacting gases.

It states that, the pressure of a fixed mass of a gas is inversely proportional to its volume if temperature is kept constant.
P\(\alpha\)\(\frac{1}{V}\)
PV = constant (n and T are constant)
P₁V₁ = P₂V₂.
Graphical representation:
        
Fig. Graph of pressure, P vs. Volume,       
V of a gas at different temperatures.

 Fig. Graph of pressure of a gas, P VS. 1N

Between 'b' and 'c' is the portion of isotherm at which liquid CO₂ is in equilibrium with gaseous CO₂.

Sohan needs less fuel as pressure cooker helps in cooking at faster rate.

Let the volume of the gas at \(0^{ \circ }C=V \ mL\)
Thus, we have \({ V }_{ 1 }=V \ mL,{ V }_{ 2 }=2V \ mL\)
\({ T }_{ 1 }=0+273=273K\)
\({ T }_{ 2 }=to \ be \ calculated\)
By Charles law \(\frac { { V }_{ 1 } }{ { T }_{ 1 } } =\frac { { V }_{ 2 } }{ { T }_{ 2 } } \)
Substituting the corresponding value, we have
\(\frac { V }{ 273 } =\frac { 2V }{ { T }_{ 2 } } \Rightarrow { T }_{ 2 }=\frac { 2V\times 273 }{ V } \)
\( =546K\)
\( { T }_{ 2 }=564-273={ 273 }^{ \circ }C\)

Daltons' Law states that, total pressure exerted by the mixture of non-reactive gases is equal to the sum of the partial pressures of individual gases.

(i) Gases, have no definite shape and no definite volume.
(ii) There is no force of attraction existing between the molecules of gases.
(iii) Gases are highly compressible.
(iv) Gases can mix evenly and can spread in whole space.

A gas that follows Boyle's law, Charles' law and Avogadro law strictly, is called an ideal gas.

T = 1513.33K

P₂ = 1.1157 atm and the pressure will increase.

Step I. Calculation of no. of moles of O₂ left in cylinder.
P = 1.24 bar, V = 34 dm³
T = (21 + 273) = 294 K, R = 0.083 dm³ bar K^-1 mol^-1
According to ideal gas equation,
PV = nRT
n = \(\frac { PV }{ RT } \)
n = \(\frac { \left( 1.24bar \right) \times \left( { 34 \ dm }^{ 3 } \right) }{ \left( { 0.083 \ dm }^{ 3 }{ bar \ K }^{ -1 }{ mol }^{ -1 } \right) \times \left( 294 \ K \right) } \)
= 1.727 mol
Step II. Calculation of mass of oxygen released.
Mass of O₂ left in the cylinder = n x M = (1.727 mol) x (32g mol ^-1) = 55.26 g
Mass of O₂ initially present = 212 g
Mass of O₂ released = (212 - 55.26) = 156.74 g

P₁ = 760 mm Hg, V₁ = 600 mL
T₁ = 25 + 273 = 298 K
V₂ = 640 mL and T₂=10 + 273 = 283 K
According to combined gas law,
\(\frac { { P }_{ 1 }V_{ 1 } }{ { T }_{ 1 } } =\frac { { P }_{ 2 }V_{ 2 } }{ { T }_{ 2 } } \)
\(\Rightarrow \)  P₂ = \(\frac { { p }_{ 1 }{ V }_{ 1 }{ T }_{ 2 } }{ { T }_{ 1 }{ V }_{ 2 } } \)
\(\Rightarrow \) P₂ = \(\frac { (760 \ mm \ Hg)\times (600 \ mL)\times (283K) }{ (640mL)\times (298K) } \)
= 676.6 mm Hg

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.

80.25 mol^-1

Number of molecules =  \(3.51+1{ 0 }^{ 16 }\)