Ideal gas equation its derivation

A gas at equilibrium has definite value of Pressure(P), Volume(V), Temperature(T) and Composition(n). These are called state Variables and are determined experimentally.
    The state of the gas can be defined by these variables. Boyle's(1662), Charles's(1787) and Avogadro laws give the birth of an equation of state for Ideal gas.

Ideal Gas Equation its derivation

    Boyle's law, V ∝ 1/P When n and T are constant for a gas.
    Charl's Low, V ∝ T When n and P are constant for a gas.
    Avogadro's Low, V ∝ n When P and T are constant for a gas.
    When all the variables are taken into account, The variation rule states that,
V ∝ (1/P) × T × n
or, V = R × (1/P) × T × n
∴ PV = nRT
Ideal gas equation its derivation
Ideal gas equation
    Where R is the Universal gas constant. This is called the ideal gas equation of state for an ideal gas.
    This equation is found to hold most satisfactory when P tense to zero. At ordinary temperature and pressure, the equation is found to deviated about 5%.

Value of universal gas constant

    At NTP 1 mole gas at 1 atm Pressure occupied 22.4 lit of gas.
Thus, R = (PV)/(nT)
Putting the values above equation,
We have, R = ( 1 atm × 22.4 lit)/(1 mol × 273 K)
= 0.082 lit atm mol⁻¹ K⁻¹

Value of R in C.G.S. and S.I. system

P = 1 atm = 76 cm Hg
= 76 cm × 13.6 gm cm⁻² × 981 cm sec⁻²
= 76 × 13.6 × 981 dyne cm⁻²
Thus, R = (76 × 13.6× 981 dyne cm⁻² ×22.4 × 10³ cm³)/(1 mol × 273 K)

= 8.314 × 10⁷ dyne cm² mol⁻¹ K⁻¹

Again, Work (W) = Force(F) × Displacement(d),
So, erg = dyne cm².
Thus, R = 8.314 × 10⁷ erg mol⁻¹ K⁻¹
We Know That 1 J = 10⁷ erg,
Thus the value of R in S.I. Unit,

= 8.314 J mol⁻¹ K⁻¹

Again, 4.18 J = 1 Cal,
hence, R = 8.314 / 4.18 Cal mol⁻¹ K⁻¹
= 1.987 Cal mol⁻¹ K⁻¹

≃ 2 Cal mol⁻¹ K⁻¹

Significance of universal gas constant

    The universal gas constant R = PV/nT
    Thus, it has the units of (Pressure × Volume)/(amount of gas × temperature). Now the dimension of pressure and volume are,
Pressure = (force/area)
= (force/ length²)
= force × length⁻² and Volume
= length³
∴ R = (force×length⁻²×length³)/(amount of gas×Kelvin)
= (force × length)/(amount of gas × kelvin)
= (Work or Energy)/(amount of gas × kelvin)
    Thus, the dimensions of R are energy per mole per kelvin and hence it represents the amount of work or energy) that can be obtained from one mole of gas when its temperature is raised by one kelvin.

Molar mass from the ideal gas equation

The ideal gas equation is,
PV = nRT

or, PV= (g/M)RT
where g = weight of the gas in gm and M = Molar mass of the gas.

Again, P = ( g/V) (RT/M)
We know that, Density (d) = Weight (g)/Volume (V).
∴ P = dRT/M

Number of gas molecule present in an ideal gas

The ideal gas equation for n mole gas is,
PV = nRT

Again, PV= (N/N₀) RT
where N = number of molecules present in the gas and N₀ = Avogadro number.

Thus, P = (N/V) × (R/N₀) × T


∴ P = N′ kT
where N′ = number of molecules present per unit volume and
k = Boltzmann constant = R/N₀ = 1.38 × 10-16 erg molecule⁻¹ K⁻¹

Ideal gas equation problems and solutions

    Problem
    Determine the value of gas constant R when pressure is expressed in Torr and Volume in dm3.
    Answer
    61.54 Torr dm3 mol-1 K-1
    Problem
    Derive the value of R when, (a) pressure is expressed in the atom, and volume in cm3and (b) Pressure in dyne m-2 and volume mm3.
    Answer
    (a) 82.05 atm cm3 mol-1K-1
    (b) 8.314 × 1014 dyne m-2 mm3 mol-1 K-1
    Problem
    Find the Molar mass of ammonia at 5 atm pressure and 300C temperature (Density of ammonia = 3.42 gm lit-1).
    Answer
    17 gm mol-1 For the details solution see
    Problem
    What is the molecular weight of a gas, 12.8 gms of which occupy 10 liters at a pressure of 750 mm and at 27° C?
    Answer
We have, PV = nRT
or, PV = (g/M)RT (molecular wt. of N₂ = 28)
∴ P = (7/28) × (0.082 × 300)/3
= 2.05 atm.
Here R = 0.082 lit atm mol⁻¹ K⁻¹
    Problem
    Calculate the number of molecules present per ml of an ideal gas maintained at a pressure of 7.6 × 10-3 mm of Hg at 0°C.
    Answer
    We have given that, V = 1ml = 10-6 dm3
    P = 7.6 × 10-3 mmHg
    = (7.6 × 10-3 mmHg) (101.235 kPa/760 mmHg)
    = 1.01235 × 10-3 kPa
    Amount of the gas, n = PV/RT 
= (1.01235 × 10⁻³ kPa)(10⁻⁶ dm³)/(8.314 kPa dm³ mol⁻¹ K⁻¹)(273 K)
= 4.46 × 10⁻¹³ mol
    Hence the number of molecules, N = n N0
    = (4.46 × 10-13 mol)(6.023 × 1023 mol-1) = 2.68 × 10-11
    Problem
    Assuming ideal behavior find out the total pressure exerted by 2 gm ethane and 3 gm CO₂ contained in a vessel of 5-liter capacity at 50°C.
    Answer
    No of molecules of ethane(n₁) = 2/30 = 0.0667
    No of molecules of CO₂(n₂) = 3/44 = 0.0682
    So the total moles (n₁ + n₂) = (0.0667 + 0.0682) = 0.1349
Thus the total pressure P = (n₁ + n₂)RT/V
or, P = {0.1349 × 0.082 × (273+50)}/5
= 0.715 atm
    Problem
    At 2 atm constant pressure slope of a one-mole ideal gas in V vs T graph is x L mol⁻¹ K⁻¹. Find out the value of R by x.
    Answer
The ideal gas equation for 1-mole gas is,
PV = nRT
or, V = (nR/P) × T.
Thus the slope of the V vs T graph = nR/P where P = 2 atm.
∴ (R/2 atm) = x L mol⁻¹ K⁻¹
or, R = 2x L atm mol⁻¹ K⁻¹

Ideal gas equation its derivation, values Of Universal gas constant and physical significance, molar mass and molecule present in ideal gas with problems

[Chemical kinetics] [column1]

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