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Questions and Answers of Physical Chemistry

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1.Properties Of Gases.
Problem 1:
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.
Answer: 

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)  = 2000 cm³
Problem 2:
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.
Answer:
P = 4.904 atm, Pi = 4.92 atm and Deviation=0.325%
Problem 3:
Find the Units and Dimensions of  Van dar Walls constant 'a' and 'b' from Van der walls Equation.
Answer:
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 cm⁴ mol⁻²
In SI system,unit of 'b' = m³ mol⁻¹
In CGS system,unit of 'b' = cm³mol⁻¹
Problem 4:
Calculate the Boyle temperature of nitrogen gas given a=1.4 atm lit² mol⁻² and b=0.04 lit mol⁻¹.
Answer:  T🇧 = 427 K
Problem 5:
Comment on the possibility of defining a Boyle Temperature if a=0 and a=b=0.
Answer:
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.
Problem 6:
Determine the value of gas constant R when pressure is expressed in Torr and Volume in dm³.
Solution:
By definition, 1.01325 bar = 760 Torr.
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
Problem 7:
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.
Solution:
(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
Problem 8:
Find the Molar mass of ammonia at 5 atm pressure and 30°C temperature (Density of ammonia = 3.42 gm lit⁻¹).
Solution:
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