__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.**

__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) =

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%

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

∴ 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 c**

**m⁴ 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.

**TðŸ‡§ = 427 K**

__Answer:__

**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

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

__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:__

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

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

__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