JEE > I. E. Irodov Solutions for Physics Class 11 & Class 12 > Irodov Solutions: Transport Phenomena- 1

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**Q. 220. Calculate what fraction of gas molecules (a) traverses without collisions the distances exceeding the mean free path λ; (b) has the free path values lying within the interval from λ to 2λ.**

**Solution. 220. **(a) The fraction of gas molecules which traverses distances exceeding the mean free path without collision is just the probability to traverse the distance s = λ without collision.

Thus

(b) This probability is

**Q. 221. A, narrow molecular beam makes its way into a vessel filled with gas under low pressure. Find the mean free path of molecules if the beam intensity decreases η-fold over the distance Δl. **

**Solution. 221. **From the formula

**Q. 222. Let αdt be the probability of a gas molecule experiencing a collision during the time interval dt; α is a constant. Find: (a) the probability of a molecule experiencing no collisions during the time interval t; (b) the mean time interval between successive collisions.**

**Solution. 222. **(a) Let P (f) = probability of no collision in the interval (0, t). Then

or

where we have used P (0) = 1

(b) The mean interval between collision is also the mean interval of no collision. Then

**Q. 223. Find the mean free path and the mean time interval be- tween successive collisions of gaseous nitrogen molecules (a) under standard conditions; (b) at temperature t = 0°C and pressure p = 1.0 nPa (such a pressure can be reached by means of contemporary vacuum pumps). **

**Solution. 223.**

**Q. 224. How many times does the mean free path of nitrogen molecules exceed the mean distance between the molecules under standard conditions? **

**Solution. 224. **The mean distance between molecules is of the order

This is about 18.5 times smaller than the mean free path calculated in 2.223 (a) above.

**Q. 225. Find the mean free path of gas molecules under standard conditions if the Van der Waals constant of this gas is equal to b = 40 ml/mol. **

**Solution. 225. **We know that the Vander Waal’s constant b is four times the molecular volume. Thus

Hence

**Q. 226. An acoustic wave propagates through nitrogen under standard conditions. At what frequency will the wavelength be equal to the mean free path of the gas molecules?**

**Solution. 226. **The volodty of sound in N_{2} is

so,

or,

**Q. 227. Oxygen is enclosed at the temperature 0°C in a vessel with the characteristic dimension l = 10 mm (this is the linear dimension determining the character of a physical process in question). Find: **

**(a) the gas pressure below which the mean free path of the molecules λ > 1; (b) the corresponding molecular concentration and the mean distance between the molecules. **

**Solution. 227. **

Now

(b) The corresponding n is obtained by dividing by kT and is m^{3} = 1.84 uper c.c. and the corresponding mean distance is

**Q. 228. For the case of nitrogen under standard conditions find: (a) the mean number of collisions experienced by each molecule per second; (b) the total number of collisions occurring between the molecules within 1 cm**

**Solution. 228. **

(b) Total number of collisions is

Note, the factor 1/2. When two molecules collide we must not count it twice.

**Q. 229. How does the mean free path λ and the number of collisions of each molecule per unit time v depend on the absolute temperature of an ideal gas undergoing **

**(a) an isochoric process; (b) an isobaric process?**

**Solution. 229. **

d is a constant and n is a constant for an isochoric process so X is constant for an isochoric process.

(b) for an isobaric process.

or an isobaric process.

**Q. 230. As a result of some process the pressure of an ideal gas increases n-fold. How many times have the mean free path λ and the number of collisions 'of each molecule per unit time v changed and how, if the process is (a) isochoric; (b) isothermal?**

**Solution. 230. **(a) In an isochoric process λ is constant and

must decrease n times in an isothermal process and v must increase n times because <v > is constant in an isothermal process.

**Q. 231. An ideal gas consisting of rigid diatomic molecules goes through an adiabatic process. How do the mean free path λ and the number of collisions of each molecule per second v depend in this process on (a) the volume V; (b) the pressure p; (c) the temperature T? **

**Solution. 231. **

Thus

But In an adiabatic

or

Thus

(c)

But

Thus

**Q. 232. An ideal gas goes through a polytropic process with exponent n. Find the mean free path and the number of collisions of each molecule per second v as a function of (a) the volume V; (b) the pressure p; (c) the temperature T. **

**Solution. 232. **In the polytropic process of index n

so

**Q. 233. Determine the molar heat capacity of a polytropic process through which an ideal gas consisting of rigid diatomic molecules goes and in which the number of collisions between the molecules remains constant (a) in a unit volume; (b) in the total volume of the gas. **

**Solution. 233. **(a) The number of collisions between the jnolecules in a unit volume is

This remains constant in the poly process pV^{-3} = constant

Using (Q.122) the molar specific heat for the polytropic process

is

Thus

It can also be written as

(b) In this case

so

It can also be written as

**Q. 234. An ideal gas of molar mass M is enclosed in a vessel of volume V whose thin walls are kept at a constant temperature T. At a moment t = 0 a small hole of area S is opened, and the gas starts escaping into vacuum. Find the gas concentration n as a function of time t if at the initial moment n (0) = n _{0}.**

**Solution. 234. **We can assume that all molecules, incident on the hole, leak out. Then,

or

Integrating

**Q. 235. A vessel filled with gas is divided into two equal parts 1 and 2 by a thin heat-insulating partition with two holes. One hole has a small diameter, and the other has a very large diameter (in comparison with the mean free path of molecules). In part 2 the gas is kept at a temperature η times higher than that of part 1. How will the concentration of molecules in part 2 change and how many times after the large hole is closed? **

**Solution. 235. **If the temperature of the compartment 2 is η times more than that of compartment 1, it must contain 1/η times less number of molecules since pressure must be the same when the big hole is open. If M - mass of the gas in 1 than the mass of the gas in 2 must be M/η . So immediately after the big hole is closed.

where m = mass of each molecule and are concentrations in 1 and 2. A fter the big hole is closed the pressures will differ and concentration will become n_{1} and n_{2} where

On the other hand

Thus

So

**Q. 236. As a result of a certain process the viscosity coefficient of an ideal gas increases α = 2.0 times and its diffusion coefficient β = 4.0 times. How does the gas pressure change and how many times? **

**Solution. 236. **We know

Thus η changing α times implies T changing α^{2} times. On the other hand

Thus D changing P times means changing β times

So p must change

**Q. 237. How will a diffusion coefficient D and the viscosity coefficient η of an ideal gas change if its volume increases n times: (a) isothermally; (b) isobarically? **

**Solution. 237. **

(a) D will increase n times

η will remain constant if T is constant

Thus D will increase if p is constant

**Q. 238. An ideal gas consists of rigid diatomic molecules. How will a diffusion coefficient D and viscosity coefficient η change and how many times if the gas volume is decreased adiabatically n =10 times? **

**Solution. 238. **

In an adiabatic process

Now V is decreased

So D decreases n^{4/5} times and η increase n^{1/5} times.

The document Irodov Solutions: Transport Phenomena- 1 - Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE is a part of the JEE Course I. E. Irodov Solutions for Physics Class 11 & Class 12.

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