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 |

*Problem*

- At 1 atm pressure and 300 K temperature, the volume of the gas is 2000 cm

^{3}, 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

(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 |

*Problem*

- Calculate the pressure of 2 moles of Nitrogen gas occupying 10 lit volume at 27

^{0}C. Given, a = 1.4 atm lit

^{2}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

(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

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,

P

= (2 × 0.082 × 300)/10

= 4.92 atm

Thus, P

P

_{ideal}= nRT/V= (2 × 0.082 × 300)/10

= 4.92 atm

Thus, P

_{real}ã„‘ P_{ideal}due to the inter-molecular attraction.
Thus the extent of deviation from ideal behavior,

= (4.92 - 4.89)

= 0.300.

= (4.92 - 4.89)

= 0.300.

*Problem*

- Calculate the wt of O

_{2}necessary to fill up a cylinder of 5 liter capacity at 0

^{0}C and 100 atm. Pressure when the compressibility factor is 0.96.

*Solution*

- We know that, n = g/M gm mol-1 and Z = compressibility factor.

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

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

Thus, Z1 = P1V1/nRT1 and Z2 = P2V2/nRT2

- Where P1, V1, and T1 is the initial pressure, volume, and temperature and P2, V2 and T2 is the final pressure, volume, and temperature.

∴ 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.

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 + an^{2}/V^{2})(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

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

= dyne cm⁴ mol⁻⁴

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

= Newton m⁴ 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 lit

^{2}mol

^{-2}and b=0.04 lit mol

^{-1}.

*Solution*

We know that Boyle temperature,

T

Thus the Boyle temperature of nitrogen gas,

T

= 427 K

T

_{B}= a/RbThus the Boyle temperature of nitrogen gas,

T

_{B}= 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

P (V - b) = RT

or, PV = RT + Pb

Differentiating with respect of pressure at constant temperature,

[d(PV)/dP]

But b ≠ 0 hence [d(PV)/dP]

[d(PV)/dP]

_{T}= bBut 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]

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 O

_{2}is 0.97. Calculate the mass of O

_{2}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.

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

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

= 1600 gm

= 1.6 Kg