- The gases are characterized by the lack of definite volume or shape. In the gaseous state, the matter has the properties of filling completely any available space to a uniform density.

- Low density and high compressibility are also pronounced characteristics of gases.

Ideal gas laws |

*Problem*

- Explain an

**cannot be liquefied?**

*ideal gas**Solution*

- For gas to liquefy, it is necessary for the molecules to come close to each other. For that, the gas should have the intermolecular attraction.

- But in an

**, it is assumed that there is no intermolecular attraction and hence it can not be liquefied.**

*ideal gas*- Further putting the condition of the critical state of the

**equation, PV = RT, it is possible to show that the critical temperature(Tc) for an**

*ideal gas***is 0⁰ K.**

*ideal gas*For

PV = RT

or, (dP/dV)

At the critical state (dP/dV)

hence (- RTc/Vc²) = 0

or, Tc = 0⁰ K.

**,***ideal gas law*PV = RT

or, (dP/dV)

_{T}= - RT/V²At the critical state (dP/dV)

_{T}= 0hence (- RTc/Vc²) = 0

or, Tc = 0⁰ K.

- Thus to liquefy an

**, it is necessary to bring the temperature below 0⁰ K and it is not possible and so the**

*ideal gas***cannot be liquefied.**

*ideal gas**Problem*

- Find the density of ammonia gas at 100⁰ C, when confined by the pressure of 1600 mm Hg.

*Solution*

The

PV = nRT

or, PV = (g/M)RT

or, PV = (g/V)(RT/M)

or, P = dRT/M

or, d = PM/RT

In the above problem molecular weight M = 17 gm mol⁻¹,

Pressure P = 1600 mm Hg = 1600/760 atm,

R = 82.1 cc atm mol⁻¹ K⁻¹ and T = (273 +100) K

Thus density (d) = (17 × 1600)/(760 × 82.1 × 373)

= 0.00117 gm/cc

**is,***ideal gas equation*PV = nRT

or, PV = (g/M)RT

or, PV = (g/V)(RT/M)

or, P = dRT/M

or, d = PM/RT

In the above problem molecular weight M = 17 gm mol⁻¹,

Pressure P = 1600 mm Hg = 1600/760 atm,

R = 82.1 cc atm mol⁻¹ K⁻¹ and T = (273 +100) K

Thus density (d) = (17 × 1600)/(760 × 82.1 × 373)

= 0.00117 gm/cc

*Problem*

- Express the coefficient of thermal expansion(Î±) of a gas. Show that Î± depends on the temperature for an

**.**

*ideal gas**Solution*

abcdCoefficient of thermal expansion(Î±) is defined as, Î± = (1/V)[dV/dT]

Hence [dV/dT]

Thus Î± = (1/V) × (R/P) = (R/PV) = 1/T

This means all the gases have the same coefficient of thermal expansion.

_{P}**for 1 mole gas is,PV = RT***Ideal gas equation*Hence [dV/dT]

_{P}= R/PThus Î± = (1/V) × (R/P) = (R/PV) = 1/T

This means all the gases have the same coefficient of thermal expansion.

*Problem*

- The compressibility factor Z of a gas is given by Z = PV/RT. What is the importance of this expression?

*Solution*

- The expression of Z provides the measures of non-ideality of a gas. It is the single parameter by the extent of non-ideality of gas could be measured.

- Thus when Z = 1, the gas behaves ideally.

- But when Z ≠ 1, the gas behaves non-ideal and departure of the value Z from one measures the extent of non-ideality of the gas.

When Z ã„‘1, PV ã„‘RT

That is, PVã„‘Pi Vi

Thus the gas becomes more compressible than

Again when Z 〉1,

the gas is less compressible than the

That is, PVã„‘Pi Vi

Thus the gas becomes more compressible than

**.***ideal gas*Again when Z 〉1,

the gas is less compressible than the

**.***ideal gas**Problem*

- Show that Z is always greater than 1 for a gas obeying the equation P(V - b) = RT. What does the result signify?

*Solution*

The equation is given as,

P(V - b) = RT

or, PV = RT + bP

or, (PV/RT) = 1 + (Pb/RT)

or, Z = 1 + (Pb/RT)

From the above relation, Pb/RT is (+)ve thus Z 〉1.

P(V - b) = RT

or, PV = RT + bP

or, (PV/RT) = 1 + (Pb/RT)

or, Z = 1 + (Pb/RT)

From the above relation, Pb/RT is (+)ve thus Z 〉1.

*Problem*

- Why an

**cannot have a reduced equation of state?**

*ideal gas**Solution*

- The reduced equation of state is expressed in terms of reduced variables like reduced temperature (Î¸) reduced pressure (Ï€) and reduced volume (Ï†).

These variables are defined as, Î¸ = T/Tc

Ï€ = P/Pc

Ï† = V/Vc

where Tc, Pc, and Vc are the critical temperature, pressure, and volume respectively.

The

PV = RT for 1 mole.

Ï€ = P/Pc

Ï† = V/Vc

where Tc, Pc, and Vc are the critical temperature, pressure, and volume respectively.

The

**is represented by the equation,***ideal gas*PV = RT for 1 mole.

- But the critical temperature of the

**is 0° K, hence the critical state is not attained by the**

*ideal gas***since 0°K is unattainable.**

*ideal gas*- Since critical Constants (Pc, Tc, and Vc) are not available for an

**, so the reduced variables (Ï€, Î¸, Ï†) are also not available for an**

*ideal gas***.**

*ideal gas*- Thus reduced equation of state cannot be constructed for an

**.**

*ideal gas**Problem*

- How the root means square velocity for oxygen compares with that of the hydrogen?

*Solution*

- The root means square velocity of hydrogen is four-times that of oxygen.

*Problem*

- Calculate the pressure excreted on the walls of a 3 liter of the flask when 7 gms of nitrogen are introduced into the same at 27°C.

*Solution*

We have, PV = nRT

or, PV = (g/M)RT (molecular wt. of N₂ = 28)

∴ P = (7/28) × (0.082 × 300)/3

= 2.05 atm

Here R = 0.082 lit atm mol⁻¹ K⁻¹

or, PV = (g/M)RT (molecular wt. of N₂ = 28)

∴ P = (7/28) × (0.082 × 300)/3

= 2.05 atm

Here R = 0.082 lit atm mol⁻¹ K⁻¹

*Problem*

- What is the molecular weight of a gas, 12.8 gms of which occupy 10 liters at a pressure of 750 mm and at 27° C?

*Solution*

- 31.91 gm mol⁻¹

*Problem*

- Calculate the total kinetic energy of 0.5 mol of an

**at 273 K.**

*ideal gas**Solution*

Total kinetic energy, = n (3RT/2)

= 0.05 {(3/2) × 8.314 × 273}

= 1702 J

= 0.05 {(3/2) × 8.314 × 273}

= 1702 J