Write an equation for the combined gas law, using temperature in degrees celsius. explain why the kelvin scale is convenient.

The combined gas law

There are several gas equations in various processes:  

In the same temperature and pressure, in the same volume conditions, the gas contains the same number of molecules  

So it applies: the ratio of gas volume will be equal to the ratio of gas moles

At a fixed temperature, the gas volume is inversely proportional to the pressure applied  

When the gas pressure is kept constant, the gas volume is proportional to the temperature  

When the volume is not changed, the gas pressure in the tube is proportional to its absolute temperature  

Combined with Boyle's law and Gay Lussac's law  

P1 = initial gas pressure (N / m2 or Pa)  

V1 = initial gas volume (m3)  

P2 = gas end pressure  

V2 = the final volume of gas  

T1 = initial gas temperature (K)  

T2 = gas end temperature  

Can be concluded

Boyle's law at constant T, P = 1 / V ⇒PV = k

Charles's law, at constant P, V = T ⇒V/T = k

Gay Lussac's Law , constant V, P=T⇒P/T = k

Avogadro's law, at constant P and T, V = n  

So an equation for the combined gas law :

Become the general ideal gas equation  

PV = nRT  

PV = NkT  

N = number of gas particles  

n = number of moles  

R = gas constant (8,31.10³ J / kmole K  

k = Boltzmann constant (1,38.10⁻²³)  

n = = N / No  

n = m / M  

n = mole  

No = Avogadro number (6.02.10²³)  

m = mass  

M = relative molecular mass  

Absolute zero point of the Kelvin scale can better indicate the condition of a substance that has frozen, so the Kelvin scale is always positive and easy to calculate

Which equation agrees with the ideal gas law

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Which law relates to the ideal gas law

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Keywords : ideal gas law,the combined gas law,Kelvin scale

the missing options are :

(1) lower boiling point and a lower freezing point(2) lower boiling point and a higher freezing point(3) higher boiling point and a lower freezing point(4) higher boiling point and a higher freezing point

answer : the correct answer is option (3) higher boiling point and a lower freezing point

explanation :

when a non volatile solute is added to a solvent like water, it results in elevation of boiling point.

let us find the increase in boiling point of water when cacl₂ is added to water to make a solution of 1 m

step 1 : find molality of the solution

let us assume we have 1 l of solution.

let us convert this to grams of cacl₂.

molar mass of cacl2 is 111 g/mol

grams of cacl₂ =

mass of solute = 111 g

assuming density of solution same as water which is 1g/ml we have ,

mass of solution = density * volume

mass of solution =

mass of solution = mass of solvent + mass of solute

mass of solute = 1000g - 111g

mass of solute 889 g = 0.889 kg

the molality of the solution can be calculated as

molality = 1.12 m

step 2 : find boiling point elevation.

the elevation in boiling point is calculated as

here δtb = boiling point elevation

i = vant hoff factor which is the number of ions formed by the solute .

cacl₂ dissociates as

it forms 3 ions . therefore,

i = 3

m = molality which is 1.12 m

kb is constant for a given solvent. for water it is 0.512

the boiling point of water increases by 1.7 °c

step 3: find freezing point depression.

when a non volatile solute is added to water, it results in decrease in freezing point.

the formula to find freezing point depression is

kf is constant for water and its value is 1.86

the freezing point of water decreases by 6.2 °c

therefore we can say that, the solution will have a higher boiling point and a lower freezing point