### Study of physical chemistry in chemical science

Study physical chemistry in many universities and the board has been restructured with a greater emphasis on the conceptual and theoretical methodology and the applications of underlying basic concepts and principals.Naturally, the students find themselves unduly constrained when they are forced to refer to various books or online study materials to collect the necessary reading materials.

It is primarily to help the students. This page provides systematic and comprehensive coverage of the theory as well as of the illustration of application and questions answers.

#### Measurement of pressure, the volume of gas molecules

QuestionCalculate 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 law and find the extent of deviation from ideal behavior.

Answer'

Van der Waals equation for n moles real gases

Van der Waals equation |

mole number = 2 mol

a = 1.4 atm lit² mol⁻²

b = 0.04 lit mol⁻¹,

volume = 10 lit

universal constant = 0.082 lit atm mol⁻¹ K⁻¹

temperature = (273 + 27)K = 300K.

The pressure of the ideal gas.

P

= (2 × 0.082 × 300)/10

= 4.92 atm

Preal ã„‘ Pideal due to the intermolecular attraction.

a = 1.4 atm lit² mol⁻²

b = 0.04 lit mol⁻¹,

volume = 10 lit

universal constant = 0.082 lit atm mol⁻¹ K⁻¹

temperature = (273 + 27)K = 300K.

∴ P = {(2×0.082×300)/(10-2×0.04)} - (1.4×22/10)

= (5.45 - 0.56)

= 4.89 atm

= (5.45 - 0.56)

= 4.89 atm

The pressure of the ideal gas.

P

_{ideal}= nRT/V= (2 × 0.082 × 300)/10

= 4.92 atm

Preal ã„‘ Pideal due to the intermolecular attraction.

The extent of deviation from ideal behavior.

= (4.92 - 4.89)

= 0.300.

= (4.92 - 4.89)

= 0.300.

Question

At 1 atm pressure and 300 K temperature, the volume of the gas 2000 cm³. What are the volume of this gas at 600 K temperature and 2 atm pressure?

Answer

Combination of Boyle's and Charles's law.

(P₁V₁)/T₁ = (P₂V₂)/T₂

P₁ = 1 atm, V₁ = 2000 cm³, T₁ = 300K, P₂ = 2, T₂ = 600K and V₂ = ?

∴ 1 × 2000/300 = 2 × V₂/600

or, V₂ = (1 × 2000 × 600)/(300 × 2)

= 2000 cm³

(P₁V₁)/T₁ = (P₂V₂)/T₂

P₁ = 1 atm, V₁ = 2000 cm³, T₁ = 300K, P₂ = 2, T₂ = 600K and V₂ = ?

∴ 1 × 2000/300 = 2 × V₂/600

or, V₂ = (1 × 2000 × 600)/(300 × 2)

= 2000 cm³

Question

One liter of a gas at 300 atm and 473 K 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. What is the final volume of the gas?

Answer

Compressibility factor,

Z = PV/nRT.

Z₁ = P₁V₁/nRT₁

and Z₂ = P₂V₂/nRT₂.

Z = PV/nRT.

Z₁ = P₁V₁/nRT₁

and Z₂ = P₂V₂/nRT₂.

where P₁, V₁, and T₁ are the initial pressure, volume, and temperature and P₂, V₂ and T₂ are the final pressure, volume, and temperature.

∴ Z₂/Z₁ = (P₂V₂/nRT₂)/(P₁V₁/nRT₁)

or, V₂ = (Z₂/Z₁)×(T₁/T₂)×(P₁V₁/P₂)

Final volume of the gas

= (1.375/1.072)×(273/473)×(300 × 1/600)

= 370.1 c.c.

or, V₂ = (Z₂/Z₁)×(T₁/T₂)×(P₁V₁/P₂)

Final volume of the gas

= (1.375/1.072)×(273/473)×(300 × 1/600)

= 370.1 c.c.

#### Weight of the gas molecules

QuestionAt 273 K and under pressure of 100 atm the compressibility factor of oxygen 0.97. Calculate the mass of oxygen necessary to fill a gas cylinder of 108.5 lit capacities under the given conditions.

Answer

T = 273 K, Z = 0.97 and P = 100 atm.

Compressibility factor,

Z = PV/RT

where V is molar volume.

The molar volume of oxygen.

Vm = ZRT/P

= (0.97 × 0.082 × 273 K)/(100)

= 2.17 lit mol⁻¹

Z = PV/RT

where V is molar volume.

The molar volume of oxygen.

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, which 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

Question= 1600 gm

= 1.6 Kg

Calculate the weight of oxygen necessary to fill up a cylinder of 5-liter capacity at 0°C and 100-atmosphere pressure when the compressibility factor 0.96.

Answer

Number of moles (n) = g/M gm mol⁻¹ and Z = compressibility factor.

PV = Z × nRT

or, n = PV/ZRT

∴ g/M = PV/ZRT

or, g = PVM/ZRT.

∴ Weight of oxygen = 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 oxygen = PVM/ZRT

= (100 × 5 × 32)/(0.96 × 0.082 × 273)

= 744.4 gm.

#### Van der Waals equation for real gas molecules

QuestionWhat are the CGS and SI units and dimensions of Van der Waals constant 'a' and 'b'?

Answer

For n mole real gas, Van der Waals equation of state.

(P + an²/V²)(V - nb) = nRT.

Unit of an²/V² = unit of the pressure of the gas.

∴ Unit of 'a' × (mol²/lit²) = atm

or, Unit of 'a' = atm lit² mol⁻².

Unit of an²/V² = unit of the pressure of the gas.

∴ Unit of 'a' × (mol²/lit²) = atm

or, Unit of 'a' = atm lit² mol⁻².

CGS unit of 'a'= dyne cm⁻² cm⁶ mol⁻²

= dyne cm⁴ mol⁻⁴.

= dyne cm⁴ mol⁻⁴.

SI unit of 'a'= Newton m⁻² m⁶ mol⁻²

= Newton m⁴ mol⁻².

= Newton m⁴ mol⁻².

SI unit of 'b' = m³ mol⁻¹.

CGS unit of 'b' = cm³ mol⁻¹.

∴ Dimension of 'a' = [M L⁵ T⁻² mol⁻¹]

and dimension of 'b' = [L³ mol⁻¹].

Questionand dimension of 'b' = [L³ mol⁻¹].

What is the Boyle temperature of nitrogen gas given a = 1.4 atm lit² mol⁻² and b=0.04 lit mol⁻¹?

Answer

Boyle temperature,

T

∴ Boyle temperature for nitrogen gas.

T

= 427 K.

QuestionT

_{B}= a/Rb.∴ Boyle temperature for nitrogen gas.

T

_{B}= 1.4/(0.082 × 0.04) K= 427 K.

Comment on the possibility of defining a Boyle temperature if a = 0 and a =b = 0.

Answer

When, a = 0,

P (V - b) = RT

or, PV = RT + Pb.

P (V - b) = RT

or, PV = RT + Pb.

Differentiating with respect of P at constant T,

[d(PV)/dP]

But b ≠ 0 hence [d(PV)/dP]

The gas has no Boyle temperature since at any temperature, [d(PV)/dP][d(PV)/dP]

_{T}= bBut b ≠ 0 hence [d(PV)/dP]

_{T}≠ 0._{T}≠ 0.

Again when a = b = 0,

Van der Waals equation.

PV = RT

or, [d(PV)/dP]

_{T}= 0