Questions and Answers of Physical Chemistry

Problems on physical chemistry appear really as problems to most of the students studying in different schools, colleges, and different universities. To overcome this problem 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.
Ideal gas equation related questions answer
Ideal gas equation

  • Problem
    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.
  • Solution
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; T₂ = 600K and V₂ = ?
∴ 1 × 2000/300 = 2 × V₂/600
or, V₂ = (1 × 2000 × 600)/(300 × 2)
= 2000 cm³
Van der Waals equation related questions answer
Van der Waals equation
  • Problem
    Calculate the pressure of 2 moles of Nitrogen gas occupying 10 lit volume at 270C. 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 the ideal gas equation and find the extent of deviation from ideal behavior.
  • Solution
For n mole Real Gas, Van der Walls equation of state is,
(P + an²/V²)(V - nb) = nRT
    In the given problem n = 2 mol, a = 1.4 atm lit² mol⁻², b = 0.04 lit mol⁻¹, V = 10 lit, R = 0.082 lit atm mol⁻¹ K⁻¹ and T = (273 + 27)K = 300K.
Now, P = {nRT/(V - nb)} - (an²/V²)
or, P = {(2×0.082×300)/(10-2×0.04)} - (1.4×22/10)
= (5.45 - 0.56)
= 4.89 atm

If the gas behaves ideally its pressure,
Pideal = nRT/V
= (2 × 0.082 × 300)/10
= 4.92 atm

Thus, Preal ㄑ Pideal due to the inter-molecular attraction.

Thus the extent of deviation from ideal behavior,
= (4.92 - 4.89)
= 0.300.
  • Problem
    Calculate the wt of O2 necessary to fill up a cylinder of 5 liter capacity at 00C and 100 atm. Pressure when the compressibility factor is 0.96.
  • Solution
PV = ZnRT
or, n = PV/ZRT
∴ g/M = PV/ZRT
or, g = PVM/ZRT
∴ Weight of O2 = PVM/ZRT
= (100 × 5 × 32)/(0.96 × 0.082 × 273)
= 744.4 gm
  • Problem
    One liter of a gas at 300 atm and 473 K is compressed to a pressure of 600 atm and 273 K. The compressibility factors found from the curves are 1.072 and 1.375 respectively at the initial and final states. Calculate the final volume.
  • Solution
We have from compressibility factor, Z = PV/nRT
Thus, Z1 = P1V1/nRT1 and Z2 = P2V2/nRT2
∴ Z₂/Z₁ = (P₂V₂/nRT₂)/(P₁V₁/nRT₁)
or, V₂ = (Z₂/Z₁)×(T₁/T₂)×(P₁V₁/P₂)
Thus the final volume,= (1.375/1.072)×(273/473)×(300 × 1/600)
= 370.1 c.c.
  • Problem
    Find the units and dimensions of Van der Walls constant 'a' and 'b' from Van der walls equation.
  • Solution
    For n mole Real Gas, Van der Walls equation of state is,
(P + an2/V2)(V - nb) = nRT
From the above equation,
unit of an²/V² = unit of Pressure.
Thus, the unit of 'a' × (mol²/lit²) = atm
or, the unit of 'a' = atm lit² mol⁻²

In the CGS system, the unit of 'a'= dyne cm⁻² cm⁶ mol⁻²
= dyne cm⁴ mol⁻⁴

In the SI system, the unit of 'a'= Newton m⁻² m⁶ mol⁻²
= Newton m⁴ mol⁻²

In the SI system, the unit of 'b' = m³ mol⁻¹

In CGS system,unit of 'b' = cm³ mol⁻¹

∴ Dimension of 'a' = [M L⁵ T⁻² mol⁻¹] and dimension of 'b' = [L³ mol⁻¹]
  • Problem
    Calculate the Boyle temperature of nitrogen gas given a = 1.4 atm lit2 mol-2 and b=0.04 lit mol-1.
  • Solution
We know that Boyle temperature,
TB = a/Rb
Thus the Boyle temperature of nitrogen gas,
TB = 1.4/(0.082 × 0.04) K
= 427 K
  • Problem
    Comment on the possibility of defining a Boyle Temperature if  a = 0 and a =b = 0.
  • Solution
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, [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
    At 273 K and under pressure of 100 atm the compressibility factor of O2 is 0.97. Calculate the mass of O2 necessary to fill a gas cylinder of 108.5 lit capacity under the given conditions.
  • Solution
    From the given data, we have T = 273 K, Z = 0.97 and P = 100 atm.
From the definition of compressibility factor,
Z = PV/RT where V is the Molar Volume.
Thus the molar volume of O₂ is,
Vm = ZRT/P
= (0.97 × 0.082 × 273 K)/(100)
= 2.17 lit mol⁻¹
    The mass of this molar volume will be equal to the molar mass of oxygen, that is 2.17 lit of oxygen equal to 32 gm. Thus the mass of oxygen required to fill the gas cylinder 108.5 lit under the given condition,
= {(32 gm)/(2.17 lit)} × (108.5 lit)
= 1600 gm
= 1.6 Kg

Questions and Answers of Physical Chemistry: Properties of Gases, and Ven de Walls Equation.

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