The relaxation time `tau` is nearly independent of applied electric field `E` whereas it changes significiantly with temperature `T`. First fact is (in part) responsible for Ohm's law whereas the second fact leads to variation of `p` with temperature. Elaborate why ?
The relaxation time `tau` is nearly independent of applied electric field `E` whereas it changes significiantly with temperature `T`. First fact is (in part) responsible for Ohm's law whereas the second fact leads to variation of `p` with temperature. Elaborate why ?
Text Solution
AI Generated Solution
To elaborate on the relationship between relaxation time (`τ`), electric field (`E`), temperature (`T`), and their effects on Ohm's law and resistivity (`ρ`), we can break down the explanation into two main parts:
### Part 1: Independence of Relaxation Time from Electric Field and Ohm's Law
1. **Understanding Relaxation Time**:
- Relaxation time (`τ`) is the average time between successive collisions of charge carriers (like electrons) in a conductor. It is a measure of how quickly these carriers can respond to an applied electric field.
2. **Independence from Electric Field**:
...
Topper's Solved these Questions
Similar Questions
Explore conceptually related problems
Dependence of Spontaneity on Temperature: For a process to be spontaneous , at constant temperature and pressure , there must be decrease in free energy of the system in the direction of the process , i.e. DeltaG_(P.T) lt 0. DeltaG_(P.T) =0 implies the equilibrium condition and DeltaG_(P.T) gt 0 corresponds to non- spontaneity. Gibbs- Helmholtz equation relates the free energy change to the enthalpy and entropy changes of the process as : " "DeltaG_(P.T) = DeltaH-TDeltaS" ""..."(1) The magnitude of DeltaH does not change much with the change in temperature but the entropy factor TDeltaS change appreciably . Thus, spontaneity of a process depends very much on temperature. For endothermic process, both DeltaH and DeltaS are positive . The energy factor, the first factor of equation, opposes the spontaneity whereas entorpy factor favours it. At low temperature the favourable factor TDeltaS will be small and may be less than DeltaH, DeltaG will have positive value indicated the nonspontaneity of the process. On raising temperature , the factor TDeltaS Increases appreciably and when it exceeds DeltaH, DeltaG would become negative and the process would be spontaneous . For an expthermic process, both DeltaH and DeltaS would be negative . In this case the first factor of eq.1 favours the spontaneity whereas the second factor opposes it. At high temperature , when T DeltaS gt DeltaH, DeltaG will have positive value, showing thereby the non-spontaneity fo the process . However , on decreasing temperature , the factor , TDeltaS decreases rapidly and when TDeltaS lt DeltaH, DeltaG becomes negative and the process occurs spontaneously. Thus , an exothermic process may be spontaneous at low temperature and non-spontaneous at high temperature. When CaCO_(3) is heated to a high temperature , it undergoes decomposition into CaO and CO_(2) whereas it is quite stable at room temperature . The most likely explanation of it, is
Electrical resistance of certain materials, known as superconductors, changes abruptly from a nonzero value of zero as their temperature is lowered below a critical temperature T_(C) (0) . An interesting property of super conductors is that their critical temperature becomes smaller than T_(C) (0) if they are placed in a magnetic field, i.e., the critical temperature T_(C) (B) is a function of the magnetic field strength B. The dependence of T_(C) (B) on B is shown in the figure. . In the graphs below, the resistance R of a superconductor is shown as a function of its temperature T for two different magnetic fields B_1 (solid line) and B_2 (dashed line). If B_2 is larget than B_1 which of the following graphs shows the correct variation of R with T in these fields?
Dependence of Spontaneity on Temperature: For a process to be spontaneous , at constant temperature and pressure , there must be decrease in free energy of the system in the direction of the process , i.e. DeltaG_(P.T) lt 0. DeltaG_(P.T) =0 implies the equilibrium condition and DeltaG_(P.T) gt 0 corresponds to non- spontaneity. Gibbs- Helmholtz equation relates the free energy change to the enthalpy and entropy changes of the process as : " "DeltaG_(P.T) = DeltaH-TDeltaS" ""..."(1) The magnitude of DeltaH does not change much with the change in temperature but the entropy factor TDeltaS change appreciably . Thus, spontaneity of a process depends very much on temperature. For endothermic process, both DeltaH and DeltaS are positive . The energy factor, the first factor of equation, opposes the spontaneity whereas entorpy factor favours it. At low temperature the favourable factor TDeltaS will be small and may be less than DeltaH, DeltaG will have positive value indicated the nonspontaneity of the process. On raising temperature , the factor TDeltaS Increases appreciably and when it exceeds DeltaH, DeltaG would become negative and the process would be spontaneous . For an expthermic process, both DeltaH and DeltaS would be negative . In this case the first factor of eq.1 favours the spontaneity whereas the second factor opposes it. At high temperature , when T DeltaS gt DeltaH, DeltaG will have positive value, showing thereby the non-spontaneity fo the process . However , on decreasing temperature , the factor , TDeltaS decreases rapidly and when TDeltaS lt DeltaH, DeltaG becomes negative and the process occurs spontaneously. Thus , an exothermic process may be spontaneous at low temperature and non-spontaneous at high temperature. The enthalpy change for a certain rection at 300 K is -15.0 K cal mol^(-1) . The entropy change under these conditions is -7.2 cal K^(-1)mol^(-1) . The free energy change for the reaction and its spontaneous/ non-spontaneous character will be
Dependence of Spontaneity on Temperature: For a process to be spontaneous , at constant temperature and pressure , there must be decrease in free energy of the system in the direction of the process , i.e. DeltaG_(P.T) lt 0. DeltaG_(P.T) =0 implies the equilibrium condition and DeltaG_(P.T) gt 0 corresponds to non- spontaneity. Gibbs- Helmholtz equation relates the free energy change to the enthalpy and entropy changes of the process as : " "DeltaG_(P.T) = DeltaH-TDeltaS" ""..."(1) The magnitude of DeltaH does not change much with the change in temperature but the entropy factor TDeltaS change appreciably . Thus, spontaneity of a process depends very much on temperature. For endothermic process, both DeltaH and DeltaS are positive . The energy factor, the first factor of equation, opposes the spontaneity whereas entorpy factor favours it. At low temperature the favourable factor TDeltaS will be small and may be less than DeltaH, DeltaG will have positive value indicated the nonspontaneity of the process. On raising temperature , the factor TDeltaS Increases appreciably and when it exceeds DeltaH, DeltaG would become negative and the process would be spontaneous . For an expthermic process, both DeltaH and DeltaS would be negative . In this case the first factor of eq.1 favours the spontaneity whereas the second factor opposes it. At high temperature , when T DeltaS gt DeltaH, DeltaG will have positive value, showing thereby the non-spontaneity fo the process . However , on decreasing temperature , the factor , TDeltaS decreases rapidly and when TDeltaS lt DeltaH, DeltaG becomes negative and the process occurs spontaneously. Thus , an exothermic process may be spontaneous at low temperature and non-spontaneous at high temperature. For the reaction at 298 K ,2A + B rarr C DeltaH =100 kcal and DeltaS=0.050 kcal K^(-1) . If DeltaH and DeltaS are assumed to be constant over the temperature range, above what temperature will the reaction become spontaneous?
Dependence of Spontaneity on Temperature: For a process to be spontaneous , at constant temperature and pressure , there must be decrease in free energy of the system in the direction of the process , i.e. DeltaG_(P.T) lt 0. DeltaG_(P.T) =0 implies the equilibrium condition and DeltaG_(P.T) gt 0 corresponds to non- spontaneity. Gibbs- Helmholtz equation relates the free energy change to the enthalpy and entropy changes of the process as : " "DeltaG_(P.T) = DeltaH-TDeltaS" ""..."(1) The magnitude of DeltaH does not change much with the change in temperature but the entropy factor TDeltaS change appreciably . Thus, spontaneity of a process depends very much on temperature. For endothermic process, both DeltaH and DeltaS are positive . The energy factor, the first factor of equation, opposes the spontaneity whereas entorpy factor favours it. At low temperature the favourable factor TDeltaS will be small and may be less than DeltaH, DeltaG will have positive value indicated the nonspontaneity of the process. On raising temperature , the factor TDeltaS Increases appreciably and when it exceeds DeltaH, DeltaG would become negative and the process would be spontaneous . For an expthermic process, both DeltaH and DeltaS would be negative . In this case the first factor of eq.1 favours the spontaneity whereas the second factor opposes it. At high temperature , when T DeltaS gt DeltaH, DeltaG will have positive value, showing thereby the non-spontaneity fo the process . However , on decreasing temperature , the factor , TDeltaS decreases rapidly and when TDeltaS lt DeltaH, DeltaG becomes negative and the process occurs spontaneously. Thus , an exothermic process may be spontaneous at low temperature and non-spontaneous at high temperature. A reaction has a value of DeltaH =-40 Kcal at 400 k cal mol^(-1) . The reaction is spontaneous, below this temperature , it is not . The values fo DeltaG and DeltaS at 400 k are respectively
Dependence of Spontaneity on Temperature: For a process to be spontaneous , at constant temperature and pressure , there must be decrease in free energy of the system in the direction of the process , i.e. DeltaG_(P.T) lt 0. DeltaG_(P.T) =0 implies the equilibrium condition and DeltaG_(P.T) gt 0 corresponds to non- spontaneity. Gibbs- Helmholtz equation relates the free energy change to the enthalpy and entropy changes of the process as : " "DeltaG_(P.T) = DeltaH-TDeltaS" ""..."(1) The magnitude of DeltaH does not change much with the change in temperature but the entropy factor TDeltaS change appreciably . Thus, spontaneity of a process depends very much on temperature. For endothermic process, both DeltaH and DeltaS are positive . The energy factor, the first factor of equation, opposes the spontaneity whereas entorpy factor favours it. At low temperature the favourable factor TDeltaS will be small and may be less than DeltaH, DeltaG will have positive value indicated the nonspontaneity of the process. On raising temperature , the factor TDeltaS Increases appreciably and when it exceeds DeltaH, DeltaG would become negative and the process would be spontaneous . For an expthermic process, both DeltaH and DeltaS would be negative . In this case the first factor of eq.1 favours the spontaneity whereas the second factor opposes it. At high temperature , when T DeltaS gt DeltaH, DeltaG will have positive value, showing thereby the non-spontaneity fo the process . However , on decreasing temperature , the factor , TDeltaS decreases rapidly and when TDeltaS lt DeltaH, DeltaG becomes negative and the process occurs spontaneously. Thus , an exothermic process may be spontaneous at low temperature and non-spontaneous at high temperature. For the reaction at 25^(@), X_(2)O_(4)(l) rarr 2XO_(2)(g) DeltaH=2.1 Kcal and DeltaS = 20 cal K^(-1) . The reaction would be
An insulting container of volume 2V_(0) is divided in two equal parts by a diathermic (conducting) fixed piston. The left part contains one mole of monoatomic ideal gas whereas right part contains two moles of the same gas. Initial pressue on each side is P_(0) . The system is left for sufficient time so that a steady state is reached. Find (a) the work done by the gas in the left part during the process, (b) initial temperature in the two sides, (c) the final common temperature reached by the gases, (d) the heat given to the gas in the right part and (e) the increase in the internal energy of the gas in the left part.
The fact tht a changing magnetic flux produces an electric field is basic to the operation of many high energy particle accelerators. Since the principle was first successfully applied to the acceleration of electrons (or beta particles) in a device called the betatron, this method of acceleration is often given that name. The general idea involved is shown in Fig. An electromagnet is used to produce a changing flux through a circular loop defined by the doughnut shaped vacuum chamber. We see that there will be an electric field E along the circular length of the doughnut, i.e. circling the magnet poles, given by 2piaE = d(phi)//dt , where 'a' is the radius of the doughnut. Any charged particle inside the vacuum chamber will experience a force qE and will accelerate. Ordinarily, the charged particle would shoot out the vacuum chamber and becomes lost. However, if the magnetic field at the position of the doughnut is just proper to satisfy the relation, Centripetal force = magnetic force or mv^(2)//a = qvB then the charge will travel in a circle within the doughnut. By proper shaping of the magnet pole piece, this relation can be satisfied. As a result, the charge will move at high speed along the loop within the doughnut. Each time it goes around the loop, it has, in effect, fallen through a potential difference equal to the induced emf, namely epsilon = (d(phi)//dt) . Its energy after 'n' trips around the loop will be q(n(epsilon)) . Variable magnetic flux
The fact tht a changing magnetic flux produces an electric field is basic to the operation of many high energy particle accelerators. Since the principle was first successfully applied to the acceleration of electrons (or beta particles) in a device called the betatron, this method of acceleration is often given that name. The general idea involved is shown in Fig. An electromagnet is used to produce a changing flux through a circular loop defined by the doughnut shaped vacuum chamber. We see that there will be an electric field E along the circular length of the doughnut, i.e. circling the magnet poles, given by 2piaE = d(phi)//dt , where 'a' is the radius of the doughnut. Any charged particle inside the vacuum chamber will experience a force qE and will accelerate. Ordinarily, the charged particle would shoot out the vacuum chamber and becomes lost. However, if the magnetic field at the position of the doughnut is just proper to satisfy the relation, Centripetal force = magnetic force or mv^(2)//a = qvB then the charge will travel in a circle within the doughnut. By proper shaping of the magnet pole piece, this relation can be satisfied. As a result, the charge will move at high speed along the loop within the doughnut. Each time it goes around the loop, it has, in effect, fallen through a potential difference equal to the induced emf, namely epsilon = (d(phi)//dt) . Its energy after 'n' trips around the loop will be q(n(epsilon)) . Magnetic field which keeps the particles in circular path must
The fact tht a changing magnetic flux produces an electric field is basic to the operation of many high energy particle accelerators. Since the principle was first successfully applied to the acceleration of electrons (or beta particles) in a device called the betatron, this method of acceleration is often given that name. The general idea involved is shown in Fig. An electromagnet is used to produce a changing flux through a circular loop defined by the doughnut shaped vacuum chamber. We see that there will be an electric field E along the circular length of the doughnut, i.e. circling the magnet poles, given by 2piaE = d(phi)//dt , where 'a' is the radius of the doughnut. Any charged particle inside the vacuum chamber will experience a force qE and will accelerate. Ordinarily, the charged particle would shoot out the vacuum chamber and becomes lost. However, if the magnetic field at the position of the doughnut is just proper to satisfy the relation, Centripetal force = magnetic force or mv^(2)//a = qvB then the charge will travel in a circle within the doughnut. By proper shaping of the magnet pole piece, this relation can be satisfied. As a result, the charge will move at high speed along the loop within the doughnut. Each time it goes around the loop, it has, in effect, fallen through a potential difference equal to the induced emf, namely epsilon = (d(phi)//dt) . Its energy after 'n' trips around the loop will be q(n(epsilon)) . Working of betatron is not based upon which of the following theories?
NCERT EXEMPLAR ENGLISH-CURRENT ELECTRICITY-Very Short Answer Type Question.
- Is the motion of a charge across junction momentum conserving ? Why or...
Text Solution
|
- The relaxation time tau is nearly independent of applied electric fiel...
Text Solution
|
- What are the advantages of the nll-point method in a Wheatstone bridge...
Text Solution
|
- What is the advantages of using thick metallic strips to join wires in...
Text Solution
|
- For wiring in the home, one uses Cu wires or A1 wires. What considerat...
Text Solution
|
- Why are alloys used for making standard resistance coils?
Text Solution
|
- Power P is to be delivered to a device via transmission cables having ...
Text Solution
|
- AB is a potentiometer wire Fig. If the value of R is increased, in whi...
Text Solution
|
- While doing an experiment with potentiometer (figure) it was found tha...
Text Solution
|
- A cell of emf E and internal resistance r is connected across an exter...
Text Solution
|
- First a set of n equal resistors of R each are connected in series to ...
Text Solution
|
- Let there be n resistors R(1) . . . R(n) with R(max)=max(R(1) . . ....
Text Solution
|
- The circuit in Fig. shows two cells connected in opposition to each o...
Text Solution
|
- Two cells of same emf E but internal resistance r(1) and r(2) are conn...
Text Solution
|
- Two conductors are made of the same material and have the same length....
Text Solution
|
- Suppose there is a circuit consister of only resistance and batteries ...
Text Solution
|
- Two cells of voltage 10Vand 2 V and internal resistance 10 Omega and 5...
Text Solution
|
- A room AC run for 5 hour at a voltage of 220V The wiring of the room c...
Text Solution
|
- In an experiment with a potentiometer, VB=10 V. R is adjusted to be 50...
Text Solution
|
- (a). Consider circuit in figure. How much energy is absorbed by electr...
Text Solution
|