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This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
Refer to figure in previous question, `DeltaU_(1)` and `DeltaU_(2)` be the changes in internal energy of the system in the processes `A` and `B` then |
| Answer» Correct Answer - When the system is taken from the state `A` to `B` the change in internal energy is independent of the path followed. Therefore, `DeltaU_(1) = DeltaU_(2)` | |
| 2. |
A gaseos mixture consists of equal number of moles of two ideal gases having adiabatic exponents `gamma_(1)` and `gamma_(2)` and molar speific heats at constant volume `C_(v_(1))` and `C_(v_(2))` respectively. Which of the following statements is/are correct?A. Adiabatic exponent for gaseous mixture is equal to `(gamma_(1)+gamma_(2))/(2)`B. Molar specific heat at constant volume for gaseous mixture is equal to `(C_(v_(1))+C_(v_(2)))/(2)`C. Molar specific heat at constant pressure for gaseous mixture is equal to `(C_(v_(1))+C_(v_(2))+R)/(2)`D. Adiabatic exponent for gaseous mixture is `1 +(2R)/(C_(v_(1))+C_(v_(2)))` |
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Answer» Correct Answer - B::D `C_(v) = (n_(1)C_(v_(1))+n_(2)C_(v_(2)))/(n_(1)+n_(2))` and `C_(P_(eq)) =C_(V_(eq)) +R, gamma =(C_(P_(eq)))/(C_(V_(eq)))` |
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| 3. |
Two moles of ideal helium gas are in a rubber balloon at `30^@C.` The balloon is fully expandable and can be assumed to require no energy in its expansion. The temperature of the gas in the balloon is slowly changed to `35^@C.` The amount of heat required in raising the temperature is nearly (take R `=8.31 J//mol.K`)A. `62J`B. `104J`C. `124J`D. `208J` |
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Answer» Correct Answer - D `DeltaQ = nC_(P) DeltaT =2 ((f)/(2)R+R) DeltaT` `= 2 [(3)/(2)R+R] xx5 = 2 xx (5)/(2) xx 8,31 xx5 = 208 J` |
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| 4. |
In given figure, let `DeltaU_(1)` and `DeltaU_(2)` be change in internal energy in process `ABC` and `CDA` respectively. `DeltaQ` and `W` be the net heta given and net work done by the system in the process `ABC +CDA`, then A. `DeltaU_(1) +DeltaU_(2) =0`B. `DeltaU_(1) - DeltaU_(2) =0`C. `DeltaQ - W = 0`D. `DeltaQ +W =0` |
| Answer» Correct Answer - A::C | |
| 5. |
In following figs. Variation of volume by change of pressure is shown in Fig. A gas is taken along the path `ABCDA`. The change in internal energy of the tgas will be: A. positive in all cases from (1) to (4)B. positive in cases (1),(2) and (3) but zero in case (4)C. negative in cases (1),(2) and (3) but zero in case (4)D. zero in all the four cases. |
| Answer» Correct Answer - D | |
| 6. |
Refer to figure in previous question, `DeltaU_(1)` and `DeltaU_(2)` be the changes in internal energy of the system in the processes `A` and `B` thenA. `DeltaU_(1) gt DeltaU_(2)`B. `DeltaU_(1) = DeltaU_(2)`C. `DeltaU_(1) lt DeltaU_(2)`D. `DeltaU_(1) ne DeltaU_(2)` |
| Answer» Correct Answer - B | |
| 7. |
The following sets of values for `C_(v)` and `C_(p)` of an ideal gas have been reported by different students. The units are cal `"mole"^(-1)K^(-1)`. Which of these sets is most reliable?A. `C_(v) = 3, C_(p) = 5`B. `C_(v) = 4, C_(p) = 6`C. `C_(v) =3, C_(p) = 2`D. `C_(v) = 3, C_(p) = 4.2` |
| Answer» Correct Answer - A::B | |
| 8. |
Let `n_(1)` and `n_(2)` moles of two different ideal gases be mixed. If adiabatic coeefiecient of the two gases are `gamma_(1)` and `gamma_(2)` respectively, then adiabatic coefficient `gamma` of the mixture is given through the relationA. `(n_(1)+n_(2)) gamma =n_(1)gamma_(1)+n_(2)gamma_(2)`B. `((n_(1)+n_(2)))/(gamma-1) = (n_(1))/(gamma_(1)-1) +(n_(2))/(gamma_(2)-1)`C. `(n_(1)+n_(2)) (gamma)/(gamma-1)= n_(1) (gamma_(1))/(gamma_(1)-1) +n_(2) (gamma_(2))/(gamma_(2)-1)`D. `(n_(1) +n_(2)) (gamma -1) = n_(1) (gamma_(1)-1) +n_(2) (gamma_(2) -1)` |
| Answer» Correct Answer - B::C | |
| 9. |
A thermally insulated chamber of volume `2V_(0)` is divided by a frictionless piston of area `S` into two equal part `A` and `B`. Part `A` has an ideal gas at pressrue `P_(0)` and temperature `T_(0)` and part `B` is vacuum. A massless spring of force constant `K` is connected with the piston and the wall of the container as shown. Initially the spring is unstretched. The gas inside chamber `A` is allowed to expand. Let in equilibrium the spring be compressed by `x_(0)`. Then A. Pressure of the gas at equilibrium is `(Kx_(0))/(S)`B. Work done by the gas is `(1)/(2)Kx_(0)^(2)`C. Increase in internal energy of the gas is `(1)/(2)Kx_(0)^(2)`D. Temperature of the gas is decreased |
| Answer» Correct Answer - A::D | |
| 10. |
A sample of 2 kg of monatomic helium (assumed ideal) is taken through the process ABC another sample of 2 kg of the same gas is taken through the process ADC as shown in Fig. Given molecular mass of helium = 4. a. What is the temperature of helium in each of the states A, B, C and D ? b. Is there any way of telling afterwards which sample of helium went through the process ABC and which went through the process ADC ? Write yes or no. How much is the heat involved in each of the process ABC and ADC ? |
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Answer» Correct Answer - (i) `T_(A) = 120K, T_(B) = 241 K`, `T_(C) = 481 K, T_(D) = 241 K` (ii) No (iii) `DeltaQ_(ABC) = (13)/(4) xx 10^(6)J; DeltaQ_(ADC) = (11)/(4) xx 10^(6)J` Number of gram moles of `He`, `n = (m)/(M) = (2 xx 10^(3))/(4) = 500` (i) `V_(A) = 10 m^(3), (i) P_(A) = 5 xx 10^(4) N//m^(2)` `:. T_(A) = (P_(A)V_(A))/(nR) = ((10)(5xx10^(4)))/((500)(8.31))` or `T_(A) = 120.34 K ~~ 120 K` Similarly, `V_(B) = 10 m^(3), P_(B) = 10 xx 10^(4) N//m^(2)` `:. T_(B) = ((10)(10xx10^(4)))/((500)(8.31))K` `:. T_(B) = 240.68 K ~~ 241 K` `V_(C) = 20 m^(3), P_(C) = 10 xx 10^(4) N//m^(2)` `:. T_(C) = ((20)(10xx10^(4)))/((500)(8.31))K` `T_(C) = 481.36 K ~~ 481 K` and `V_(D) = 20 m^(3),P_(D) = 5 xx 10^(4) N//m^(2)` `V_(D) = ((20)(5xx10^(4)))/((500)(8.31))K` `T_(D) = 240.68 K ~~ 241 K`. (ii) No, if is not possible to tell afterwards which sample went through the process `ABC` or `ADC`. But we can say the process which require more work goes through process `ABC`. (iii) In the process `ABC`, `DeltaU = nC_(v)DeltaT = n ((3)/(2)R) (T_(C)-T_(A))` `= (500) ((3)/(2)) 8.31 (481.36 - 120.34) J` `DeltaU = 2.25 xx 10^(6) J` and `DeltaW = "Area under" BC = (20 - 10) (10) xx 10^(4)J = 10^(6)J` `:. DeltaQ_(ABC) = DeltaU +DeltaW = (2.25 xx 10^(6) +10^(6))J` `DeltaQ_(ABC) = 3.25 xx 10^(6)J` In the process `ADC, DeltaU` will be same (because it depends on initial and final temperature only) `DeltaW = "Area under" AD` `= (20 - 10) (5 xx 10^(4))J` `= 0.5 xx 10^(6)J` `DeltaQ_(ADC) = DeltaU + DeltaW = (2.25 xx 10^(6) +0.5 xx 10^(6))J` `DeltaQ_(ADC) = 2.75 xx 10^(6)J`. |
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| 11. |
The temperature of `5 mol` of gas which was held at constant volume was change from `100^(@)C` to `120^(@)C`. The change in internal energy was found to ve `80 J`. The total heat capacity of the gas at constant volume will be equal to |
| Answer» Correct Answer - `0.8JK^(-1)` | |
| 12. |
An ideal monatomic gas is at `P_(0), V_(0)`. It is taken to final volume `2V_(0)` when pressure is `P_(0)//2` in a process which is straight line on `P -V` diagram. Then A. The final temperature is equal to initial temperatureB. There is no change in internal energyC. The work done by the gas is `+(P_(0)V_(0))/(4)`D. The heat is absorbed in the process. |
| Answer» Correct Answer - A::B::D | |
| 13. |
An ideal monoatomic gas is initially in state `1` with pressure `p_(1) = 20 atm` and volume `v_(1) = 1500 cm^(3)`. If is then taken to state `2` with pressure `p_(2) = 1.5 p_(1)` and volume `v_(2) = 2v_(1)`. The change in internal energy from state `1` to state `2` is equal toA. `2000J`B. `3000 J`C. `6000 J`D. `9000 J` |
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Answer» Correct Answer - D `p_(1) = 20 xx 10^(5) N//m^(2), v_(1) = 1500 xx 10^(-6) m^(3)` `p_(2) = 30 xx 10^(5) N//m^(2) , v_(2) = 300 xx 10^(-6) m^(3)` `T_(1) = (p_(1)v_(1))/(nR)` and `T_(2) = (p_(2)v_(2))/(nR)` `dU = nC_(v)dT = n. (3)/(2)R. (T_(2)-T_(1))` `=n.(3)/(2)R ((p_(2)v_(2)-p_(1)v_(1)))/(nR)` `= (3)/(2) (p_(2)v_(2)-p_(1)v_(1)) = 9000 J` |
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| 14. |
`P_(i), V_(i)`are initial pressure and volumes and `V_(f)` is final volume of a gas in a thermodynamic process respectively. If `PV^(n) =` constant, then the amount of work done by gas is`: (gamma = C_(p)//C_(v))`. Assume same, initial states & same final volume in all processes.A. minimum for `n = gamma`B. minimum for `n = 1`C. minimum for `n =0`D. minimum for `n = (1)/(gamma)` |
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Answer» Correct Answer - A For largen n, pressure will be smaller, so work done will be smaller for larger n. |
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| 15. |
A sample of an ideal gas initially having internal energy `U_(1)` is allowed to expand adiabatically performing work W. Heat Q is then supplied to it, keeping the volume constant at its new value, until the pressure raised to its original value. The internal energy is then `U_(3)` (see figure). find the increase in internal energy `(U_(3) - U_(1))`? |
| Answer» Correct Answer - `Q -W` | |
| 16. |
An ideal gas `(C_(p)//C_(v) =gamma)` having initial pressure `P_(0)` and volume `V_(0)` (a) The gas is taken isothermally to a pressure `2P_(0)` and then adiabatically to a pressure `4P_(0)`. Find the final volume. (b) The gas is brought back to its initial state. It is adiabatically taken to a pressure `2P_(0)` and then isothermally to a pressure `4P_(0)`. Find the final volume. |
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Answer» Correct Answer - `(V_(0))/(2^((gamma+1)/(gamma)))` in each cases Initial pressure of an ideal gas `=P` Initial volume of an idela gas `= V_(0)` (a) For isothermal process `P_(2)V_(2) = P_(1)V_(1)` `rArr V_(2) = (P_(1)V_(1))/(P_(2)) = (V_(0))/(2)` For adiabatic process `P_(3)V_(3)^(gamma-1) =P_(2)V_(2)^(gamma)` `rArr V_(3) = ((P_(2))/(P_(3)))^(1//gamma)V_(2) rArr V_(3) = (V_(0))/(2^((gamma+1)/(gamma)))` (b) For adiabatic process `P_(2)V_(2)^(gamma) = P_(1)V_(1)^(gamma) = V_(2) =((P_(1))/(P_(2)))^(1//gamma) V_(1)` `rArr V_(2) = (V_(0))/(2^((1)/(gamma)))` For isothermal process `P_(3)V_(3) = P_(2)V_(2)` `rArr V_(3) = (P_(2)V_(2))/(P_(3)), V_(3)=(V_(0))/(2^((gamma+1)/(gamma)))` |
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| 17. |
A certain mass of an ideal gas is at pressure `P_(1)` and volume `V_(1)`. If is compressed isothermally and then allowed to expand adiabatically untill its pressure returns to `P_(1)`. The gas is then allowed to expand its original volume. Which of the following `P -V` graphs are these process correctly shown?A. B. C. D. |
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Answer» Correct Answer - A Self explainatroy |
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| 18. |
When `100J` of heat is given to an ideal gas it expands from `200cm^(3)` to `400cm^(3)` at a constant pressure of `3 xx 10^(5) Pa`. Calculate (a) the change in internal energy of the gas (b) the number of moles in the gas if the initial temperature is `400K`, (c ) the molar heat capacity `C_(p)` at constant pressure and (d) the molar heat capacity `C_(v)` at constant volume. `[R = (25)/(3)J//mol-K]` |
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Answer» Correct Answer - (a) `40J` (b) `(9)/(500) moles` (c ) `(125)/(9) J//mol-K` (d) `(50)/(9)J//mol-K` (a) `DeltaU = DeltaQ - W` `= 100 - 3 xx 10^(5) xx 200 xx 10^(-6) = 100 - 60 = 40J` (b) `n = (PV)/(RT) = (3xx10^(5)xx200xx10^(-6))/((25)/(3)xx400) =(9)/(500)`moles (c ) `PdV = nRdT` `dT = (3xx10^(5)xx200xx10^(-6))/((9)/(500)xx(25)/(3)) rArr dT = 400K` As `C_(P) = (Q)/(ndT) = (100)/((9)/(500)xx400) = (125)/(9) J//mol-K` (d) `C_(P) -C_(V) = R` `C_(V) = C_(P) - R` `= (125)/(9) - (25)/(3) = (50)/(9) J//mol-K`. |
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| 19. |
Internal energy of two moles of an ideal gas at a temperature of `127^(@)C is 1200R`. Then find the molar specific heat of the gas at constant pressure? |
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Answer» Correct Answer - `2.5R` `U = 1200R = nCvT` `1200R = 2 Cv xx 400` `C_(V) = (3)/(2)R :. C_(P) = R +(3)/(2)R = 2.5 R` |
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| 20. |
Find the change in the internal energy of `2 kg` of water as it heated from `0^(0)C to 4^(0)C`. The specific heat capacity of water is `4200 J kg^(-1) K^(-1)` and its densities at `0^(0)C` and `4^(0)C` are `999.9 kg m^(-3)` and `1000 kg m^(-3)` respectively. atmospheric pressure `=10^(5)` Pa. |
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Answer» Correct Answer - `(33600+ 0.02)J` `m = 2kg` `DeltaT = 4k` `C = 4200 J//kg-K` `dU = ? ` `dQ = dU +dW` `(mCdT) = dU +PdV` `2 xx 4200 xx4 =dU +10^(5) (-dV)` `(33600+0.02) = dU` |
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| 21. |
Calculate the value of mechanical equivalent of heat from the following data. Specific heat capacity of air at constant volume and at constant pressure are `4.93 cal//mol-K` and `6.90 cal//mol-K` respectively. Gas constant `R = 8.3 J//mol-K`. |
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Answer» Correct Answer - `4.2J//cal` As `C_(P) - C_(V) = R = 8.3 J//mol-K` `1.97 = 8.3J`. `:.` Mechanical equivalent of heat is `(8.3J)/(1.97cal) = 4.2 J//cal`. |
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| 22. |
Calculate the value of mechanical equivalent of heat from the following data. Specific heat capacity of air at constant volume =`170 cal kg^(-1) K^(-1)` , `gamma = C_p/ C_v = 1.4` and the density of air at STP is `1.29 kg m^(-3)`. Gas constant `R = 8.3 JK^(-1) mol^(-1)`. |
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Answer» Using `pV = nRT`, the volume of `1` mole of air at `STP` is `V = (nRT)/(p) = ((1mol)xx(8.3J//mol-K)xx(273K))/(1.01xx10^(5)N//m^(2))= 0.024 m^(3)`. The mass of `1` mole is, therefore, `(1.29 kg//m^(3)) xx (0.0224 m^(3)) = 0.029 kg`. The number of moles in `1 kg` is `(1)/(0.029)`. The molar heat capacity at constant volume is `C_(v) = (170 cal)/((1//0.029)mol-K) = 4.94 cal//mol-K`. Hence, `C_(p) = gamma C_(v) = 1.4 xx 4.93 cal//mol-K` or, `C_(p) - C_(v) = 0.4 xx 4.93 cal//mol-K` `=1.97 cal//mol-K`. Also, `C_(p) - C_(v) = R = 8.3 J//mol-K`. Thus, `8.3J = 1.97 cal`. The mechanical equivalent of heat is `(8.3J)/(1.97 cal) = 4.2 J//cal`. |
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| 23. |
A system absorbs `1000` cal of heat and does `1675J` work. If `J = 4.18 J//cal`, then find the change in internal energy of the system? |
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Answer» Correct Answer - `+2505 J` `dU = dQ - dW` `= 1000 xx 4.18 - 1675 = 2505J` |
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| 24. |
In given figure, when a thermodynamic system is taken from state `A` to state `B` via path `ACB,100` cal of heat given to the system and `60cal` work is done by the gas. Along the path `ADB`, the work done by the gas is `20cal`. Find the heat flowing into the system in this case? |
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Answer» Correct Answer - `60 cal` `dQ = dU +dW` `dU_(1) = dQ_(1) - dW_(1) = 100 - 60 = 40 cal` `dU_(2) = dQ_(2) - dW_(2) = dQ_(2) - 20 cal` `dU_(1) = dU_(2)` `40 = dQ_(2) - 20, dQ_(2) = 60cal` |
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| 25. |
In given figure, gas is slowely heated for sometime. During the process, the increases in internal energy of the gas is `10J` and the piston is found to move out by `25cm`, then find the amount of heat supplied. The area of cross-section of cylinder `= 40cm^(2)` and atmospheric pressure `100 kPa` |
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Answer» Correct Answer - `110J` `dQ = dU +dW, = dU = PdV = dU +Padx` `= 10 xx 10^(5) xx 40 xx 10^(-4) xx 25 xx 10^(-2)` `rArr 10 +10^(5) xx 10^(-4) xx 1000 xx 10^(-2)` `= 10 +100 = 110J, :. dQ = 110J` |
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| 26. |
Discuss whether the following phenomena are reversible ? (i) Whaterfall (ii) Rusting of iron and (iii) Electrolysis |
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Answer» Correct Answer - (i) Waterfall. The falling of water is not a reversible process. During the fall of the water, the major part of its potential energy is converted into kinetic energy of the water. However, on striking the gound a part of potential energy is converted into heat and sound. it is not possible to convert the heat and the sound produced along with the `K.E`. of water into potential energy so that the water may rise back to its initial height. Therefore, waterfall is not a reversible process. (ii) Rusting of iron. In the rusting of iron, the iron gets oxidised by the oxygen from the air. Since it is a chemical change, it is not a reversible process. (iii) Electrolysis. It is reversible process. provided the electrolyte does not offer any resistance to the flow of current. If we reverse the direaction of current, the direaction of motion of ions is alos reversed. |
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| 27. |
Three closed vessels `A, B` and `C` are at the same temperature T and contain gases which obey the Maxwellian distribution of velocities. Vessel A contains only `O_(2), B` only `N_(2)` and `C` a mixture of equal quantities of `O_(2)` and `N_(2)`. If the average speed of the `O_(2)` molecules in vessel `A` is `V_(1)`, that of the `N_(2)` molecules in vessel `B` is `V_(2)`, the average speed of the `O_(2)` molecules in vessel `C` is (where `M` is the mass of an oxygen molecules)A. `(V_(1)+V_(2))/2`B. `V_(1)`C. `(V_(1)V_(2))^(1//2)`D. `(V_(1))/(2)` |
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Answer» Correct Answer - B `V_(av) = sqrt((8RT)/(piM_(0))), V_(AV) alpha sqrt(T)` For same temp in vessel `A,B` and `C`, Average speed of `O_(2)` molecules is same in vessel `A` and `C` and is equal to `V_(1)`. |
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| 28. |
If one mole of a monatomic gas `(gamma=5/3)` is mixed with one mole of a diatomic gas `(gamma=7/5),` the value of gamma for mixture is |
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Answer» Correct Answer - `(3)/(2)` `gamma_(mix) = (C_(Pmix))/(C_(Vmix)) = (1xx(5)/(2)R+1xx(7)/(2)R)/(1xx(3)/(2)+1xx(5)/(2)R)=(3)/(2)` |
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| 29. |
What is the kinetic interpretation of temperature? |
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Answer» Correct Answer - We know, `(1)/(2) mv^(bar2) = (3)/(2) kT` or `(1)/(2) mv^(bar2) prop T` Therefore, higher the temperature of the gas, more will be the average kinetic energy possessed by the gas molecules, thus, the temperature of the gas given the measure of the average kinetic energy of the gas molecules. If `T = 0`, then `(1)/(2) mv^(bar2) = 0` Therefore, absoulte zero is that temperature at which the molecules motion stops. |
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| 30. |
When a gas expands at constant temperature, why does the pressure decrease? Explanin it on the basis of kinetic theory of gases. |
| Answer» Correct Answer - When the gas expands at constant temperature, the average kinetic energy of the gas molecules remains the same. However, due to increase in volume of the gas, separation between the molecules increases. As a result the number of molecules colliding per second aganist the walls decreases. consequently less momentum is transferred to the walls of the container per seond. Hence the pressure exerted by the gas decreases. | |
| 31. |
One mole of a gas expands with temperature T such thaht its volume, V=`KT^(2)`, where K is a constant. If the temperature of the gas changes by `60^(@)C` then the work done by the gas is |
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Answer» Correct Answer - `1000J` `W = int PdV , V = kT^(2)` (given) `PdV = 2nRdT :. W = int PdV = 2nR int dT` `W = 2nR DeltaT = 2 xx 1 xx R xx 60` `W = 120 R= 120 xx 25//3 = 1000J` |
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| 32. |
A mass of an ideal gas undergoes a reversible isothermal compression. Its molecules will then have compared with initial state, the same (i) rrot mean square velocity (ii) mean mometum (iii) mean kinetic energyA. (i),(ii),(iii) correctB. (i),(ii) correctC. (ii),(iii)correctD. (i)correct |
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Answer» Correct Answer - A As initial and final states are same `:. T_(i) = T_(F) As V_(rms) , vec(P)_(av)` and `vec(K)_(av)` depends on temperature `:.` all are equal. |
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| 33. |
Find the expression for the work done by a system undergoing isothermal compression (or expansion) form volume `V_(1)` to `V_(2)` at temperature `T_(0)` for a gas which obeys the van der wattws equation of state. `(P +an^(2) //V^(2)) (V-bn) =nRT`? |
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Answer» Correct Answer - `nRT_(0)ln ((V_(2)-nb)/(V_(1)-nb)) +an^(2) ((V_(1)-V_(2))/(V_(1)V_(2)))` `p = (nRT)/((V-bn)) - (an^(2))/(V^(2))` work done by ststem `= int pdV` `= overset(V_(2))underset(V_(1))int ((nRT_(0))/(V-bn)-(an^(2))/(V^(2)))dV` `= nRT_(0) overset(V_(2))underset(V_(1)) int (1)/(V-bn) dV - an^(2) overset(V_(2))underset(V_(1))int (dV)/(V^(2))` `W = nRT_(0) ln ((V_(2)-nb)/(V_(1)-nb)) +an^(2)((V_(1)-V_(2))/(V_(1)V_(2)))` |
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