- Problem 1:

- Explain an ideal gas cannot be liquefied?

- Answer:

- 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

*ideal gas*, it is assumed that there is no intermolecular attraction and hence it can not be liquefied. Further putting the condition of the critical state of the

*ideal gas*equation,

**PV = RT**, it is possible to show that the critical temperature(T

_{C}) for an

*ideal gas*is

**0**.

^{0}K- For

**,**

__ideal gas Low__**PV = RT,**(dP/dV)

_{T}= - RT/V

^{2}At the critical state (dP/dV)

_{T}= 0 and hence (- RT

_{c}/V

_{c}

^{2}) = 0

**or, T**

_{c}= 0^{0}K.- Thus to liquefy an

*ideal gas*, it is necessary to bring the temperature below 0K and it is not possible and so the

*ideal gas*cannot be liquefied.

- Problem 2:

- Find the density of ammonia gas at 100

^{0}C, when confined by the pressure of 1600 mm Hg.

- Answer:

- The

*ideal gas*equation is, 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 3:

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

*ideal gas*.

- Answer:

- For Solution See Problem 1 of Comparison Between Ideal and Real Gases.

- Problem 4:

- The compressibility factor

**Z**of a gas is given by Z = PV/RT. What is the importance of this expression?

- Answer:

- 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„‘P

_{i}V

_{i}Thus the gas becomes more compressible than

*ideal gas*.

- Again when

**Z 〉1**, the gas is less compressible than the

*ideal gas*.

Ideal gas low problems solutions |

- Problem 5:

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

- Answer:

- The equation 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.

- Problem 6:

- Why an

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

- Answer:

- The reduced equation of state is expressed in terms of reduced variables like

__reduced temperature (Î¸)__**and**

__reduced pressure (Ï€)__**.**

__reduced volume (Ï†)__- These variables are defined as,

**Î¸ = T/Tc, Ï€ = P/Pc and Ï† = V/Vc**

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

- The

*ideal gas*is represented by the equation,

**PV = RT for 1 mole.**

- But the critical temperature of the

*ideal gas*is

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

*ideal gas*since

**0°K**is unattainable. Since critical Constants (Pc, Tc, and Vc) are not available for an

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

*ideal gas*.

- Thus reduced equation of state cannot be constructed for an

*ideal gas*.

- Problem 7:

- How the root means square velocity for

**Oxygen**compares with that of the

**Hydrogen**?

- Answer:

- The root means square velocity of Hydrogen is four-times that of Oxygen.

- For Solution See Problem 8 of Kinetic Theory of Gases.

- Problem 8:

- 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.

- Answer:

- 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

- Problem 9:

- 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?**

- Answer:

- 31.91 gm mol⁻¹

- For Solution See Problem 4 of Kinetic Theory of Gases.

**Problem 10:**

- Calculate the total kinetic energy of 0.5 mol of an

**ideal gas**at 273 K.

**Answer:**

- Total kinetic energy,
= n (3RT/2)
= 0.05 {(3/2) × 8.314 × 273}

**= 1702 J**