Irodov Solutions: Phase Transformations- 1 Notes | EduRev

Physics For JEE

JEE : Irodov Solutions: Phase Transformations- 1 Notes | EduRev

The document Irodov Solutions: Phase Transformations- 1 Notes | EduRev is a part of the JEE Course Physics For JEE.
All you need of JEE at this link: JEE

Q. 185. A saturated water vapour is contained in a cylindrical vessel under a weightless piston at a temperature t = 100°C. As a result of a slow introduction of the piston a small fraction of the vapour Δm = 0.70 g gets condensed. What amount of work was performed over the gas? The vapour is assumed to be ideal, the volume of the liquid is to be neglected. 

Solution. 185. The condensation takes place at constant pressure and temperature and the work done is 

p Δ V

where ΔV is the volume of the condensed vapour in the vapour phase. It is 

Irodov Solutions: Phase Transformations- 1 Notes | EduRev

where M = 18 gm is the molecular weight of water.


Q. 186. A vessel of volume V = 6.0 l contains water together with its saturated vapour under a pressure of 40 atm and at a temperature of 250°C. The specific volume of the vapour is equal to V'c = 50 l/kg under these conditions. The total mass of the system water-vapour equals m = 5.0 kg. Find the mass and the volume of the vapour. 

Solution. 186. The specific volume of water (the liquid) will be written as  Irodov Solutions: Phase Transformations- 1 Notes | EduRev most of the weight is due to water. Thus if ml is mass of the liquid and mv that of the vapour then

Irodov Solutions: Phase Transformations- 1 Notes | EduRev

Irodov Solutions: Phase Transformations- 1 Notes | EduRev

Irodov Solutions: Phase Transformations- 1 Notes | EduRev in the present case. Its volume is Irodov Solutions: Phase Transformations- 1 Notes | EduRev


Q. 187. The saturated water vapour is enclosed in a cylinder under a piston and occupies a volume V0 = 5.0 l at the temperature t = 100°C. Find the mass of the liquid phase formed after the volume under the piston decreased isothermally to V = 1.6 l. The saturated vapour is assumed to be ideal.

Solution. 187. The volume of the condensed vapour was originally V0 - V at temperature T = 373 K. Its mass will be given by

Irodov Solutions: Phase Transformations- 1 Notes | EduRev gm where p = atmospheric pressure


Q. 188. A volume occupied by a saturated vapour is reduced isothermally n-fold. Find what fraction ri of the final volume is occupied by the liquid phase if the specific volumes of the saturated vapour and the liquid phase differ by N times (N > n). Solve the same problem under the condition that the final volume of the substance corresponds to the midpoint of a horizontal portion of the isothermal line in the diagram p, V. 

Solution. 188.  Irodov Solutions: Phase Transformations- 1 Notes | EduRev specific volume of liquid. Irodov Solutions: Phase Transformations- 1 Notes | EduRev specific volume of vapour.

Let V = Original volume of the vapour. Then

Irodov Solutions: Phase Transformations- 1 Notes | EduRev

So  Irodov Solutions: Phase Transformations- 1 Notes | EduRev

In the case when the final volume of the substance corresponds to the midpoint of a horizontal portion of the isothermal line in the p y v diagram, the final volume must be  Irodov Solutions: Phase Transformations- 1 Notes | EduRev per unit mass of the substance. Of this the volume of the liquid is V'1/2 per unit total mass of the substance. 

Thus  Irodov Solutions: Phase Transformations- 1 Notes | EduRev


Q. 189. An amount of water of mass m = 1.00 kg, boiling at standard atmospheric pressure, turns completely into saturated vapour. Assuming the saturated vapour to be an ideal gas find the increment of entropy and internal energy of the system. 

Solution. 189. From the first law of thermodynamics

Irodov Solutions: Phase Transformations- 1 Notes | EduRev

where q is the specific latent heat of vaporization

Now  Irodov Solutions: Phase Transformations- 1 Notes | EduRev

Thus   Irodov Solutions: Phase Transformations- 1 Notes | EduRev

For water this gives  Irodov Solutions: Phase Transformations- 1 Notes | EduRev


Q. 190. Water of mass m = 20 g is enclosed in a thermally insulated cylinder at the temperature of 0°C under a weightless piston whose area is S = 410 cm2. The outside pressure is equal to standard atmospheric pressure. To what height will the piston rise when the water absorbs Q = 20.0 kJ of heat? 

Solution. 190. Some of the heat used in heating water to the boiling temperature 

T = 100°C = 373 K The remaining heat

= Q - mc Δ T

(c = specific heat of water, ΔT - 100 K) is used to create vapour. If the piston rises to a height h then the volume of vapour will be ≈ sh(neglecting water). Its mass will be  Irodov Solutions: Phase Transformations- 1 Notes | EduRev and heat of vapourization will be  Irodov Solutions: Phase Transformations- 1 Notes | EduRev To this must be added the work done in creating the saturated vapour Irodov Solutions: Phase Transformations- 1 Notes | EduRev

Irodov Solutions: Phase Transformations- 1 Notes | EduRevIrodov Solutions: Phase Transformations- 1 Notes | EduRev


Q. 191. One gram of saturated water vapour is enclosed in a thermally insulated cylinder under a weightless piston. The outside pressure being standard, m = 1.0 g of water is introduced into the cylinder at a temperature t0 = 22°C. Neglecting the heat capacity of the cylinder and the friction of the piston against the cylinder's walls, find the work performed by the force of the atmospheric pressure during the lowering of the piston.

Solution. 191.  A quantity  Irodov Solutions: Phase Transformations- 1 Notes | EduRev of saturated vapour must condense to heat the w ater to boiling  point T = 373°K

(Here c = specific heat of water, T0 = 295 K = initial water temperature).

The work done in lowering the piston will then be 

Irodov Solutions: Phase Transformations- 1 Notes | EduRev

since work done per unit mass o f the condensed vapour is Irodov Solutions: Phase Transformations- 1 Notes | EduRev


Q. 192. If an additional pressure Δp of a saturated vapour over a convex spherical surface of a liquid is considerably less than the vapour pressure over a plane surface, then Irodov Solutions: Phase Transformations- 1 Notes | EduRev where p0 and Pl are the densities of the vapour and the liquid, α is the surface tension, and r is the radius of curvature of the surface. Using this formula, find the diameter of water droplets at which the saturated vapour pressure exceeds the vapour pressure over the plane surface by η = 1.0% at a temperature t = 27°C. The vapour is assumed to be an ideal gas. 

Solution. 192. Given  Irodov Solutions: Phase Transformations- 1 Notes | EduRev

or   Irodov Solutions: Phase Transformations- 1 Notes | EduRev

F or water a = 73 dynes/cm, M = 18gm, pl = gm/cc, T = 300 K, and with η = 0.01, we get

Irodov Solutions: Phase Transformations- 1 Notes | EduRev


Q. 193. Find the mass of all molecules leaving one square centimetre of water surface per second into a saturated water vapour above it at a temperature t = 100°C. It is assumed that η = 3.6% of all water vapour molecules falling on the water surface are retained in the liquid phase. 

Solution. 193. In equilibrium the number of "liquid" molecules evaporoting must equals the number of "vapour" molecules condensing. By kinetic theory, this number is

Irodov Solutions: Phase Transformations- 1 Notes | EduRev

Its mass is

Irodov Solutions: Phase Transformations- 1 Notes | EduRev

Irodov Solutions: Phase Transformations- 1 Notes | EduRev

where p0 is atmospheric pressure and T = 373 K and M = molecular weight of water.


Q. 194. Find the pressure of saturated tungsten vapour at a temperature T = 2000 K if a tungsten filament is known to lose a mass μ = 1.2-10-13  g/(s•cm2) from a unit area per unit time when evaporating into high vacuum at this temperature.

Solution. 194. Here we must assume that |i is also the rate at which the tungsten filament loses mass when in an atmosphere of its own vapour at this temperature and that η (of the previous problem) ≈  1. Then

Irodov Solutions: Phase Transformations- 1 Notes | EduRev

from the previous problem where p = pressure of the saturated vapour. 


Q. 195. By what magnitude would the pressure exerted by water on the walls of the vessel have increased if the intermolecular attraction forces had vanished? 

Solution. 195. From the Vander Waals equation

Irodov Solutions: Phase Transformations- 1 Notes | EduRev

where V = Volume of one gm mole of the substances.
For water V = 18 c.c. per mole = 1*8 x 10-2 itre per mole

Irodov Solutions: Phase Transformations- 1 Notes | EduRev

If molecular attraction vanished the equation will be 

Irodov Solutions: Phase Transformations- 1 Notes | EduRev

for the same specific volume. Thus

Irodov Solutions: Phase Transformations- 1 Notes | EduRev


Q. 196. Find the internal pressure pi of a liquid if its density p and specific latent heat of vaporization q are known. The heat q is assumed to be equal to the work performed against the forces of the internal pressure, and the liquid obeys the Van der Waals equation. Calculate pi in water. 

Solution. 196. The internal pressure being  Irodov Solutions: Phase Transformations- 1 Notes | EduRev the work done in condensation is

Irodov Solutions: Phase Transformations- 1 Notes | EduRev

This by assumption is Mq , M being the molecular weight and Vl , Vg being the molar volumes of the liquid and gas.

Thus  Irodov Solutions: Phase Transformations- 1 Notes | EduRev

where p is the density of the liquid. For water pi =  3-3 x 1013 atm


Q. 197. Demonstrate that Eqs. (2.6a) and (2.6b) are valid for a substance, obeying the Van der Waals equation, in critical state.

Instruction. Make use of the fact that the critical state corresponds to the point of inflection in the isothermal curve p (V). 

Solution. 197. The Vandar Waal's equation can be written as (for one mole)

Irodov Solutions: Phase Transformations- 1 Notes | EduRev

A t the critical point Irodov Solutions: Phase Transformations- 1 Notes | EduRev vanish. Thus

Irodov Solutions: Phase Transformations- 1 Notes | EduRev

Solving these simultaneously we get on division

Irodov Solutions: Phase Transformations- 1 Notes | EduRev

This is the critical molar volume. Putting this back

Irodov Solutions: Phase Transformations- 1 Notes | EduRev

Finally    Irodov Solutions: Phase Transformations- 1 Notes | EduRev

From these we see that  Irodov Solutions: Phase Transformations- 1 Notes | EduRev


Q. 198. Calculate the Van der Waals constants for carbon dioxide if its critical temperature Tcr = 304 K and critical pressure pcr = 73 atm. 

Solution. 198. Irodov Solutions: Phase Transformations- 1 Notes | EduRev

Thus   Irodov Solutions: Phase Transformations- 1 Notes | EduRev

and Irodov Solutions: Phase Transformations- 1 Notes | EduRev


Q. 199. Find the specific volume of benzene (C6H6) in critical state if its critical temperature Tcr = 562 K and critical pressure pcr = 47 atm.

Solution. 199.  Specific volume is molar volume divided by molecular weight. Thus

Irodov Solutions: Phase Transformations- 1 Notes | EduRev


Q. 200. Write the Van der Waals equation via the reduced parameters π, v, and τ, having taken the corresponding critical values for the units of pressure, volume, and temperature. Using the equation obtained, find how many times the gas temperature exceeds its critical temperature if the gas pressure is 12 times as high as critical pressure, and the volume of gas is equal to half the critical volume.

Solution. 200.  

Irodov Solutions: Phase Transformations- 1 Notes | EduRev

or Irodov Solutions: Phase Transformations- 1 Notes | EduRev

or Irodov Solutions: Phase Transformations- 1 Notes | EduRev

where  Irodov Solutions: Phase Transformations- 1 Notes | EduRev

or Irodov Solutions: Phase Transformations- 1 Notes | EduRev

When   Irodov Solutions: Phase Transformations- 1 Notes | EduRev


Q. 201. Knowing the Van der Waals constants, find:
 (a) the maximum volume which water of mass m = 1.00 kg can occupy in liquid state;
 (b) the maximum pressure of the saturated water vapour. 

Solution. 201. (a) The ciritical Volume VMCr is the maximum volume in the liquid phase and the minimum volume in the gaseous. Thus

Irodov Solutions: Phase Transformations- 1 Notes | EduRev

(b) The critical pressure is the maximum possible pressure in the vapour phase in equilibrium with liquid phase. Thus

Irodov Solutions: Phase Transformations- 1 Notes | EduRev


Q. 202. Calculate the temperature and density of carbon dioxide in critical state, assuming the gas to be a Van der Waals one. 

Solution. 202.

Irodov Solutions: Phase Transformations- 1 Notes | EduRev

Irodov Solutions: Phase Transformations- 1 Notes | EduRev

Offer running on EduRev: Apply code STAYHOME200 to get INR 200 off on our premium plan EduRev Infinity!

Related Searches

mock tests for examination

,

past year papers

,

Irodov Solutions: Phase Transformations- 1 Notes | EduRev

,

practice quizzes

,

Previous Year Questions with Solutions

,

Exam

,

pdf

,

Objective type Questions

,

Summary

,

shortcuts and tricks

,

Semester Notes

,

study material

,

Irodov Solutions: Phase Transformations- 1 Notes | EduRev

,

Extra Questions

,

Important questions

,

ppt

,

Free

,

Irodov Solutions: Phase Transformations- 1 Notes | EduRev

,

Viva Questions

,

video lectures

,

MCQs

,

Sample Paper

;