Gas Laws

That is, P  1/V

 For two or more gases at a constant temperature,

Also, density  1/V

 P  d

 

  The graph between volume and temperature at constant pressure are called isotherms.

Example: A balloon containing 20 L of He gas was connected to a empty cylinder of volume 80L .If the initial pressure was 1 atm , what should be the final pressure.

Solution: Applying , Boylesinglequotes Law , as  the temperature of the system is constant:

                 P₁V₁ = P₂V_{2 .}

               P_{1 =} 1 atm

              V1 = 20 L

             V_{2  =} (20 + 80) L  = 100 L

           P₂ = (20 X 1 /100 ) = 0.2 atm.

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Charles Law:  Charles Law states that "At constant pressure the volume of a fixed mass of a gas is directly proportional to its temperature.Temperature being measured in kelvin scale.

Mathematically, V  T or, (V / T ) = constant.

If we measure temperature in ^oC , then charles law can be stated as " at constant pressure the volume of a fixed mass of gas increases linearly with increase in temperature".

Mathematically, it can be written as:

V_t = V₀ + bt , which is an equation of straight line in volume and temperature.

V_t represents volume at t^oC and V₀ represents volume at 0^oC and t is the temperature.

b is a constant , which is the slope of the line.

Experimentally , it is established that the value of b is [V₀ / (273.15 X 1^oC)] i.e., with a change of 1^oC in temperature, volume changes by (V₀ / 273.15 )

Therefore, the given equation becomes:

The constant singlequotebsinglequote  is also called volume coefficient and is defined as change in volume per degree change in temperature.

From , the equation above , if temperature is reduced to -273.15^oC , the volume of the gas becomes zero.This temperature is called absolute zero.It is the temperature at which entropy of all gases become zero and this temperature can not be practically attained by any means.It is the zero of the kelvin scale of temperature.

The relationship between kelvin scale and ^oC of temperature is :
                                                                                 

Plotting the graph between volume and absolute temperature we get the graph as,

While the graph between volume and ^oC is : ( Its also a straight  line with same slope but different origin)

  The graph between volume and temperature at constant pressure are called isobars.

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Example: Imagine there is a system which maintains constant pressure irrespective of the amount of gases inside it.Now suppose at Instant singlequoteAsinglequote, it contains n₁ moles of a gas X at temperature t₁. At instant singlequoteBsinglequote , n₂ moles of other non-reacting gas Y  was injected into the system. Find the overall change in the temperature of the system.

Solution: Since, the pressure of the system is constant therefore, we can apply charlesinglequotes law here.

i.e., V₁/T₁ = V₂/T_{2 .} 

Now, volume of any gas is directly proportional to the number of moles.

Therfore, the above equation can be written as: 

               n_i / T_{1 =}   n_f / T₂

_Now, n_i = n₁ 

and n_f = n₁ + n₂

the equation becomes , n_{1 /} t₁ = (n₁ + n₂₎ / t₂

? t₂ = (n₁ + n₂₎ t₁ / n_1.

Change in temperture ,

ΔT = t₂ -  t₁ .

? ΔT  = (n₁ + n₂₎ t₁ / n₁ -  t₁

?  ΔT  = n₂ t₁ / n₁

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Gay Lussacsinglequotes Law: Gay Lussacsinglequotes law states that at constant volume the pressure of a given mass of gas is directly proportional to its temperature.

That is, P T

? P₁/T₁ = P₂/T₂

Therefore, the graph between pressure and temperature is  a straight line like the graph of volume vs temperature.

Mathematially, P = P₀ + bt.

The constant b is the slope of the line and is called the pressure coefficient which is defined as the change in pressure per degree change in temperature.

The graph between pressure and temperature is as follows:

The graph of pressure vs. temperature at constant volume is called isochore.

Ideal Gas equation: We have already seen Boylesinglequotes law, Charlesinglequotes law and avogadrosinglequotes law.Rewriting them again here:

                                    P  1/V, 

or,                                          V 1/P ............................ (i)

                                       V  T ................................(ii)

                                        V  n ................................(iii)

Combining equations (i),(ii) and (iii) we get,

                                       V  nT/P

Removing the proportionality constant  we get:

                                       V = R nT/P                             

The above equation is called the ideal gas equation or ideal gas law becuase it defines the behaviour of ideal gases.

R in the above equation is called the ideal gas constant. From the above equation we get :

                                     R = PV/nT

Putting the dimensions of all the quantities on the right hand side of the equation we get:

                                  R = force/area X volume X 1/ moles X 1 / temperature

                                                  = work / mol X Degrees (K)

Hence , by the above observation we can say that ,

 R is work done per degree per mole of the gas. Its unit is that of work mol^-1 K^-1. 

R can have different numerical values depending upon the units in which we express pressure , temperature etc. Some of them along with the unit used are: 

Ideal gas works most accurately for monoatomic gases high temperatures and low pressures, becuase it doesnsinglequotet take into consideration intermolecular attractions and molecular size. Since, at high temperatures and low pressures the thermal and kinetic energy of the gases are high , the relative significance of intermolecular attractions and molecular size diminish , so all gases tend to attain ideal behaviour.

We can calculate the density of the gas using the ideal gas equation. This can be achieved as follows:

                       PV = nRT 

now, n = (given mass (w)/ molar mass (m) )

PV = wRT / m 

?P = wRT / mV

       =   (w/V) X RT/m

Now, the factor w/V is density if the gas, d.

P = dRT/m

d P

and , d 1/T .

Similarly , we can express molar mass in terms of density, pressure and temperature.

? m = dRT/P   

 Grahamsinglequotes law of effusion: This law was formulated by scottish chemist Thomas Graham. It shates that  " The rate of effusion is inversely proportional to square root of the mass of its particles." Mathematically, it can be written as,

                                                      r 1/ (M)^1/2 where, r = rate of effusion and M = molar mass of the gas.

 If we compare rates of effusion of two gases, say gas1 and gas2, having molar mass M₁ and M₂ respectively,  we can write,

                                                                       r₁/r₂    =    (M₂ / M₁)^1/2   

Effusion is different from diffusion in the way that , diffusion is simply the net movement of molecules from higher concentration to lower concentration. While effusion is esacpe of gas moleules from a tiny hole. So, according to Grahamsinglequotes law . the rate of effusion of gases depend upon their molecular weights.The molecules with lower molecular weight effuse more quickly than gases with higher molecular weight.

Since, mass is directly porportional to its density , therefore we can write,

                                                                     r₁/r₂    =    (D₂ / D₁)^1/2   

Since, rate of effusion can also be written as volume of gas effused per second , we can write,

                                                  r = V/t

                                            r₁/r₂    =    (V₁t₂ / V₂t₁)   

Also , if the gases are under different pressures we can write,

                                           r P/ (M)^1/2

Rate of effusion of two gases under different pressures will be:

                                                r₁/r₂    =   (P₁/P₂) X (M₂ / M₁)^1/2   .

If we want to compare the rate of effusion of the same gas then M₁ = M₂ 

                                         r₁/r₂    =   (P₁/P₂)

Applications of Grahamsinglequotes law of effusion: 

(i) It can be used for the determination of molar mass.

(ii) Seperation of isotopes in a gaseous mixture.

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Example: An unknown gas singlequoteA singlequote has molar mass 16 and it effuses at the rate of 60mL s^-1.  10 Litres of another gas took 1000 seconds to effuse. How many molecules would  28.8 grams of this gas contain.

Solution: First we will have to find out the molar mass of the gas . Now , we know that :

                                                                         r₁/r₂    =    (V₁t₂ / V₂t₁)   

and also                                            r₁/r₂    =  (M₂ / M₁)^1/2  

Combining both we get :                             (M₂ / M₁)^1/2  = (V₁t₂ / V₂t₁).

Now, M₁ = 16, V₁ = 60 and t₁ = 1. M₂ = M₂, V₂= 10,000 t₂ = 100 .

Putting all the values we get ,    ?   (M₂ / M₁)^1/2  = (60X 1000/10,000)  = 6

                                                  ? M₂ = 36 X 16 = 576 g mol^-1

Hence , the molar mass is 576 , i.e., 576 grams of the gas contains 1 mole of the gas = N₀ moleules.                       ( N₀ = 6.023 X 10²³)

28.8 grams will contain  (28.8 X N₀ / 576) molecules = N₀/20 molecules. = 3.011 X 10²³ molecules.

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[SolvedExample]

Example: A 0.05 mol sample COCl₂ gas is placed in a  0.3L container at the temperature of 127⁰C. The final pressure in the container is 6 atm. Find the fraction of COCl₂ which has dissociated_. The dissociation reaction is :

                                                                                         COCl₂ (g)? CO(g) + Cl₂(g) .

Solution: First wesinglequotell find out the total number of moles present inside the container. From the ideal gas equation we get :

                                                                                             n = PV/RT .

                                                                                                  = 6 X 0.3 / 0.0821 X 400

                                                                                                  = 0.055

Now, from the dissociation reaction we get :

                                                                                       COCl₂ (g)? CO(g) + Cl₂(g) .

                                                                   0.05 - χ              χ              χ    

, Total number of moles available in the container: 0.05 + χ  = 0.055 (calculated from above)

     χ = 0.005.

Hence, fraction of COCl₂  dissociated is 0.005/ 0.05 = 0.1 or 10% .

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