Phase Transition: Assignment | Kinetic Theory & Thermodynamics - Physics PDF Download

Q.1. The boiling point of L_i is 1620K .

(a) What is the vapor pressure of liquid L_i at 1000K ?

(b) Estimate the melting point of lithium from the data given.

Data:

Enthalpy of evaporation is 156 kj /mol

Vapour pressure above solid L_i is given by:

(a) Integrate the Clausius Clayperon Equation 

(b) At the melting point of L_i P_solid = P_liquid

1.47 =

T = 375K

Q.2. (a) What is the vapor pressure of H₂O at 0 100^oC boiling point of water)?

(b) The water forecast tells me that it is 59^oF (15^oC) outside with a relative humidity of 65%. What is the dew point of that air?

Data: Enthalpy of evaporation for 0

H₂O100^oC = 40.6kJ /mol

(a) The boiling point is where P* = P_atomsphere .So P* H₂O = 1atm at the regular boiling point.

(b) The Clausius-Clayperon equation allows us to find vapor pressure at some temperature from knowledge of the vapor pressure at another temperature

 

We can integrate this between 373K and 288K

 = 0.021 atm

Relative humanity is defined as

In this problem the relative humanity is 65% so we can find the  PH₂O

 = 0.0136 atm

The dew point is the temperature where PH₂O = P*H₂O. Again we integrate the Clausius- Clayperon equation, this time solving for T

T = 281K= 8^oC

Q.3. A long vertical column is closed at the bottom and open at the top; it is partially filled with a particular liquid and cooled to -5^oC . At this temperature the fluid solidifies below a particular level, remaining liquid above this level. If the temperature if further lowered to 0 -5.2^oC the solid-liquid interface moves upward by 40cm . The latent heat (per unit mass) is 2cal / g and the density of the liquid phase is 1g / cm³ . Find the density of the solid phase. Neglect thermal expansion of all materials. 

(Hint: Note that the pressure at the original position of the interface remains constant)

Once again, we start with me Clausius-Clayperon equations in this form 

Integrating we get

Let P₁ be the pressure at the original interface (T₁ =-5^oC ) and P₂ be the pressure at the new interface (T₂ =-5.2^oC)

We can also relate the densities to the volume change as follows: 

Also, we know

Putting this together we get

Solving for the density of the solid,

Putting in the values of the know quantities (being very careful with units)

ρ^s = 2.6g / cm³

Q.4. For a gas obeying van der Walls' equation, the constants are a =1.32litre² atm. mol^-2 and b = 3.12 x 10² litremol^-1 . Calculate the temperature at which 5 moles of the gas at 5atmospheric pressures will occupy a volume of 20 litre. Given R = 8.31 x 10⁷ ergmol^- K^-1

The van der Waals' equation for n moles of the gas is 

By inverting it, we can write 

T =

Here, 

a = 1.32litre² atm. mol^-2

= 1.34x 10dyne cm⁴ mol^-2

b = 3.12 x10^-2 litre mol²

= 3.12 x 10^-2cm³ mol¹

= 3.12 cm³mol^-1

p = 5 atm

= 5 x 1.013 x10⁶ dyne cm^-2

= 5.065 x10⁶ dyne cm^-2

V = 20 litre = 20 x10³ cm³

n = 5, R = 8.31x10⁷ erg mol^-1 K^-1

T =

= 245.9K

Q.5. It is found that a certain liquid boils at a temperature of 95^oC at the top of a hill, whereas it bolls at a temperature of 105^oC at the bottom. The latent heat is 1000 cal /mol. What is the approximately height of the hill?

There are two steps to solving this problem 

(1) What is the pressure difference in terms of the height? 

(2) What is the height of the hill?

Use the following from the Clayperon equation (this is valid for equilibrium between an ideal gas and a solid or liquid ΔH is latent Heat)

Which in this case can be written in terms of the pressure at the top and bottom of the hill, and the respective boiling points?

Now we write the pressure at the bottom in terms of the pressure at the top

Substituting this,

If we expand ln (1 + x ) we get,

Assuming that the air is an ideal gas, we can relate the density and pressure to the temperature using the ideal gas law

Where ( MW ) is the molecular weight of the air? We now have 

Plugging in some numbers ( g = 10m/s² , MW= 28.8g/mol and we dealing with air at 25^oC

h =

h = 2.6km

Q.6. (a) A chamber at 1500K with a volume of one cubic meter contains one gram of Ag .

What is the status of the silver in the chamber? Is it all vapor, all liquid, or part liquid and part vapor? If the latter, what fraction of the silvers is in the vapor phase?

(b) How much heat is required to evaporate one mol of Ag from a very large quantity of a silver-copper liquid solution at 1500K containing  x_Ag = 0.25  . Assume the solution is ideal.

Data for Silver:

Melting point =1235K 

Vapor pressure of liquid given by

Enthalpy of melting for Ag =11.3kJ /mol

Heat capacity for liquid Ag = 33 J /mol - K

First we can find what the vapor pressure over the liquid would be under these conditions 

 =  -7.95

P* =  3.54 x 10^-4 atm = 35.8Pa

If all of the silver was in the vapor phase ( (1gram )

P =  = 115.61Pa

Hence, there must be both liquid and vapor present 

P = P*= 35.8Pa

Now we can find the moles in the vapor phase

Game of vapor 

So the fraction in the vapor phase is 0.31

(b) In general the amount of heat necessary would be the sum of the mixing enthalpy and the enthalpy of vaporization 

However since we are told to assume the solution is ideal, the enthalpy of mixing is zero. (This is related to the assumption that the mixing species do not interact)

Thus

Since we were not given the enthalpy of vaporization or the boiling point in the data, we must find these quantities. To find the boiling point, use the given equation for the vapor pressure and find the temperature where the pressure is equal to 1 atm .

Choosing T = 1500 and 1600 K , the valued calculated for ΔH_vap is 255 kg / mole . (If you this up you get 251 kJ / mole ) Now we can get enthalpy of vaporization at 1500K , which can be done using a loop, since we know enthalpy is a state function

ΔH_vap (1500K )= 267kJ / mol

Q.7. The Vapor pressure of pure B at 1000K is 4 5 10 atm   . If B is in an ideal solution with another species at 1000K , its vapor pressure will be:

Higher than 5 x 10^-4

Lower than 5 x 10^-4

Equal to 5 x 10^-4

Impossible to say

The vapor pressure of pure B at 1000 K is 5 x10^-4 atm . If B is in an ideal solution with another species at 1000 K , its vapor pressure will be:

Higher than 5 x10^-4 

Lower than 5 x10^-4  is answer 

Equal to 5 x10^-4  

Impossible to say

Q.8. One mole of a gas occupies a volume of 0.55 litres at 0^oC . Calculate the pressure it will exert if it behaves as (a) an ideal gas and (b) as a van der Walls' gas.

Given a=  0.37 Nm⁴ mol^-2 ,b = 43 x 10⁶ m3 mol^-1 and R = 8.31Jmol^-1 K^-1

Here V = 0.55 litre mol^-1 = 550cm³mol^-1 = 550 x 10^-6 m³mol^-1

(a) For one mole of an ideal gas, 

P = RT/V

On substituting the values of various quantities, we get 

p =

= 4.12 x10⁶ Nm^-2

(b) For one mole of van der Walls' gas, 

=

=

= (0.48 x 10⁶ -1.22 x 10⁶) Nm^-2

= 3.26 x 10⁶Nm^-2

As expected, the pressure exerted by a van der Waals' gas is less than that exerted by an ideal gas.