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Questions and Answers of Physical Chemistry

Problems on Physical Chemistry appear really as problems to the most of students studying in different Schools, Colleges and Different universities. To Overcome this Problems to achieve a good performance in Examinations i will discuss some of these problems Topic Wise. This page is Updated Regularly so Keep on Touch With this Page. 

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)  = 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.

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

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 cm⁴ 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.

Answer:  T🇧 = 427 K

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.

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

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

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