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__Properties of Gases__
At 1 atm pressure and 300 K temperature, the volume of the gas is 2000 cm³, then calculate the volume of this gas at 600 K temperature and 2 atm pressure.

From the

__Combination of Boyle's and Charl's Law__is,
(

**P₁V₁**)/**T₁**= (**P₂V₂**)/**T₂**
Here, P₁ = 1 atm; V₁ = 2000 cm³ and T₁ = 300K and P₂= 2 atm; V₂=❓ and T₂=600 K

∴ 1×2000/300 = 2×V₂/600

or, V₂ = (1×2000×600)/(300×2) =

P = 4.904 atm, Pi = 4.92 atm and Deviation=0.325%

∴ 1×2000/300 = 2×V₂/600

or, V₂ = (1×2000×600)/(300×2) =

__2000____cm³__
Calculate the pressure of 2 moles of Nitrogen gas occupying 10 lit volume at 27°C. Given, a = 1.4 atm lit² mol⁻² and b=0.04 lit mol⁻¹ using Van der Waals equation. Also calculate the pressure of the gas using ideal gas equation and find the extent of deviation from ideal behavior.

Find the Units and Dimensions of Van dar Walls constant 'a' and 'b' from Van der walls Equation.

__In SI system,unit of 'a'__= N m⁻² m⁶ mol⁻² =**N m⁴ mol⁻²**__In CGS system,unit of 'a'__= dyne cm⁻² cm⁶ mol⁻² =

**dyne c**

**m⁴ mol⁻²**

__In SI system,unit of 'b'__=

**m³ mol⁻¹**

__In CGS system,unit of 'b'__=

**cm³mol⁻¹**

Calculate the Boyle temperature of nitrogen gas given a=1.4 atm lit² mol⁻² and b=0.04 lit mol⁻¹.

**TðŸ‡§ = 427 K**

__Answer:__
Comment on the possibility of defining a Boyle Temperature if a=0 and a=b=0.

When, a = 0, Van der Waals Equation,

P (V - b) = RT

or, PV = RT + Pb

__Differentiating with respect of pressure at constant temperature,__

[d(PV)/dP]T = b, but b ≠ 0 hence [d(PV)/dP]T ≠ 0

__The gas has no Boyle temperature since at any temperature since at any__

__temperature,__[d(PV)/dP]T ≠ 0

Again when a = b = 0, Van der Walls equation becomes PV = RT

or, [d(PV)/dP]T = 0

__That is at any temperature this becomes zero and so all the temperature are Boyle Temperatures.__

Determine the value of gas constant R when pressure is expressed in Torr and Volume in dm³.

Hence R = PV/nT = {(1 bar) × (760 Torr/1.01325 bar) × (22.400 dm³)}/{(1 mol) × (273 K)}.

∴ R = 61.54 Torr dm³ mol⁻¹ K⁻¹

Derive the value of R when, (a) pressure is expressed in atom, and volume in cm³and (b) Pressure in dyne m⁻² and volume mm³.

(a) P in atm and V in cm³.

Thus, R = PV/nT = (1 atm × 22400 cm³)/(1 mol × 273 K) = 82.05 atm cm³ mol⁻¹K⁻¹.

(b) P in dyne m⁻² and V in mm³.

Again P = 1 atm = 101325 N m⁻² = 101325 × 10⁵ dyne m⁻² = 1.01325 × 10¹⁰ dyne m⁻² and V = 22400 cm³ = 22400 × 10³ mm³.

Thus, R = PV/nT = (1.01325 × 10¹⁰ dyne m⁻² × 22400 × 10³ mm³)/(1 mol × 273 K)

= 8.314 × 10¹⁴ dyne m⁻² mm³ mol⁻¹ K⁻¹

Let the Molar mass of ammonia = M gm mol⁻¹

From the above equation,P = dRT/M; or, M = dRT/P

Here, P =1 atm, d = Density of ammonia= 3.42 gm mol⁻¹, R = 0.082 lit atm mol⁻¹ K⁻¹ and T = (273+30) K = 303K.

Thus, M = (3.42 gm lit⁻¹ × 0.082 lit atm mol⁻¹ K⁻¹ × 303 K)/(5 atm) = 17 gm mol⁻¹

Find the Molar mass of ammonia at 5 atm pressure and 30°C temperature (Density of ammonia = 3.42 gm lit⁻¹).

From the above equation,P = dRT/M; or, M = dRT/P

Here, P =1 atm, d = Density of ammonia= 3.42 gm mol⁻¹, R = 0.082 lit atm mol⁻¹ K⁻¹ and T = (273+30) K = 303K.

Thus, M = (3.42 gm lit⁻¹ × 0.082 lit atm mol⁻¹ K⁻¹ × 303 K)/(5 atm) = 17 gm mol⁻¹