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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 251. |
An amount `Q` of heat is added to a monoatomic ideal gas in a process in which the gas performs work `Q/2` on its surrounding. Find the molar heat capacity for the process.A. `2R`B. `3R`C. `4R`D. `6R` |
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Answer» Correct Answer - B First law of thermodynamics `Q = W + Delta U` `Q = Q/2 + Delta U` `implies Delta U = 3nRDeltaT` `implies nCDeltaT = 3nRDeltaT` `C = 3R` |
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| 252. |
A body cools from `50^(@)C` to `46^(@)C` in 5 minutes and to `40^(@)C` in the next 10 minutes. The surrounding temperature is :A. `30^(@)C`B. `28^(@)C`C. `36^(@)C`D. `32^(@)C` |
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Answer» Correct Answer - A `(d theta)/(dt)=K(theta_(av)-theta_(0))` `(4)/(5)=K(48-theta_(0))` . .. (i) `(6)/(10)=K(43-theta_(0))` . .. (ii) `(4)/(5)xx(10)/(6)=((48-theta_(0)))/((43-theta_(0)))` `(4)/(3)=((48-theta_(0)))/((43-theta_(0)))` After solving, `theta_(0)=30^(@)C`. |
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| 253. |
A body cools at the ratio of `1.2^(@)C//`min when its temperature is more than that of the surrounding by `40^(@)C`. The rate of cooling of the body when its temperature is more than that of surrounding by `25^(@)C` will beA. `0.75^(@)C`/minB. `0.25^(@)C`/minC. `1.25^(@)C`/minD. `1^(@)C`/min |
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Answer» Correct Answer - A `R_(1)=K(theta_(1)-theta_(0)) and R_(2)=k(theta_(2)-theta_(0))` `(R_(2))/(R_(1))=((theta_(2)-theta_(0)))/((theta_(1)-theta_(0)))=(25)/(40)=(5)/(8)` `R_(2)=0.75`. |
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| 254. |
A container has a gas that consists of positively charged ions. The gas undergoes a free expansion in which there is no heat exchange with surrounding. Does the temperature of the gas increases or decreases? Why? |
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Answer» Correct Answer - Temperature increases |
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| 255. |
A vessel of volume `8.0xx10^(-3) m^(3)` contains an ideal gas at 300K and 200kPa. Calulate the amount of the gas (in moles) leaked assuming that the tenperature remains constant. |
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Answer» As the gas leaks out, the volume and the temperature of the remaining gas do not change. The number of moles in the vessel is given by `n=(pV)/(RT)`. The number of moles in the vessel before the leakage is `n_(2)=(p_(1)V)/(RT)` and that after the leakage is `n_(2)=(p_(2)V)/(RT)`.`Thus, the amount leaked is `n_(1)-n_(2)=((p_(1)-p_(2))V)/(RT)` ltbgt `=((200-125)xx10^(3) N m^(-2)xx8.0xx10^(-3)m^(3))/((8.3JK^(-1)mol^(-1))xx(300K))` `=0.24mol^(-1).` |
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| 256. |
Consider the quantity`(MkT)/(pV)` of an ideal gas where M is the mass of the gas. It depends on theA. (a)temperature of the gasB. (b)volume of the gasC. (c)pressure of the gasD. (d)nature of the gas |
| Answer» Correct Answer - D | |
| 257. |
The heat capcity of material depends uponA. the structure of a matterB. temperature of matterC. density of matterD. specific heat of matter. |
| Answer» Correct Answer - D | |
| 258. |
The average momentum of a molecule in a sample of an ideal gas depends onA. (a)temperatureB. (b)number of molesC. (c)volumeD. (d)none of these. |
| Answer» Correct Answer - D | |
| 259. |
Internal energy of an ideal gas depends uponA. temperature onlyB. volume onlyC. both volume and temperatureD. neither volume nor temperature |
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Answer» Correct Answer - A The internal energy of ideal gas depends only upon temperature of gas not on other factors. |
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| 260. |
A Carnot engine, whose efficiency is `40%`, takes in heat from a source maintained at a temperature of 500K. It is desired to have an engine of efficiency `60%`. Then, the intake temperature for the same exhaust (sink) temperature must be: |
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Answer» `eta=1-(T_(2))/(T_(1)), 0.4=1-(T_(2))/(500)rArrT_(2)=300K` `0.6=1(T_(2))/(T_(1)^(1))=1-(300)/(T_(1)^(1))=(300)/(0.4)=750K` |
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| 261. |
The above p-v diagram represents the thermodynamic cycle of an engine, operating with an ideal monoatomic gas. The amount of heat, extracted from the source in a single cycle is |
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Answer» `P_(0)V_(0)=nRT_(A), 2P_(0)V_(0nRT_(B), 2P_(0)2V_(0)=nRT_(C)` Heat supplied `H=nC_(v)DeltaT_(AB)+nC_(v)DeltaT_(BC)` `=nC_(v)=((2P_(0)V_(0))/(nR)-(P_(0)V_(0))/(nR))+nC_(P)((4P_(0)V_(0))/(nR)-(2P_(0)V_(0))/(nR))` `=n((3)/(2)R)((P_(0)V_(0))/(nR))+n(5)/(2)R((2P_(0)V_(0))/(nR))=(12)/(2)P_(0)V_(0)` |
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| 262. |
A Carnot engine, whose efficiency is `40%`, takes in heat from a source maintained at a temperature of 500K. It is desired to have an engine of efficiency `60%`. Then, the intake temperature for the same exhaust (sink) temperature must be:A. efficiency of carnot engine cannot be made larger than 50 %B. 1200 KC. 750 KD. 600 K |
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Answer» Correct Answer - C The efficiency of engine , `eta=1-(T_("sink"))/(T_("source"))` `rArr" " 0.4=1-(T_("sink"))/(500)rArrT_("sink")=0.6xx500=300K` ` 0.6=1-(300K)/(T_("source"))rArrT_("source")=(300K)/(0.4)=750K` |
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| 263. |
If pressure of a gas contained in a closed vessel is increased by `0.4%` when heated by `1^(@)C`, the initial temperature must beA. `250 K`B. `250^(@)C`C. `2500 K`D. `25^(@)C` |
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Answer» Correct Answer - A `PV = nRT` (i) `P(1 + (0.4)/(100))V = nRT(T + 1)` (ii) `(ii)//(i)` `1 + (0.4)/(100) = (T + 1)/(T) = 1 + (1)/(T)` `T = (100)/(0.4) = 250 K` |
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| 264. |
A certain mass of an ideal diatomic gas contained in a closed vessel to heated it is observed that the temperature remains constant. However, half the amount of gas gets dissociated. The ratio of the heat supplied to the gas initial internal energy of the gas will beA. `1:2`B. `1:4`C. `1:5`D. `1:10` |
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Answer» Correct Answer - D `U_(1) = n((5R)/(2))T` and `U_(2) = n/2((5R)/(T))T+(2n)/(2)((3R)/(2))T = n((11R)/(4))T` `U_(2) =U_(1)+Q` `:. Q/(U_1) = (U_2)/(U_1)-1 = 11/10 - 1 = 1/10`. |
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| 265. |
A vessel contains two non-reactive gases neon (monoatomic) and oxygen (diatomic). The ratio of their partial pressures is 3:2. Estimate the ratio of (i) number of molecules, and (ii) mass density of neon and oxygen in the vessel. Atomic mass of neon = 20.2 u, and molecular mass of oxygen = 32.0 u. |
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Answer» Since V and T are common to the two gases, we have `P_(1)V=mu_(1)RT` and `P_(2)V=mu_(2)RT` `therefore (P_(1))/(P_(2))=(mu_(1))/(mu_(2))` where 1 and 2 refer to neon and oxygen respectively Given `(P_(1))/(P_(2))=(3)/(2) therefore (mu_(1))/(mu_(2))=(N_(2))/(N_(A))` where `N_(1)` and `N_(2)`are the number of molecules of 1 and 2 `therefore` The ratio of number of molecules `(N_(1))/(N_(2))=(mu_(1))/(mu_(2))=(3)/(2)` (ii) If `rho_(1)` and `rho_(2)` are mass densities of 1 and 2 respectively, we have `(rho_(1))/(rho_(2))=(m_(1)//V)/(m_(2)//V)=(m_(1))/(m_(2))` But we can also write `mu_(1)=(m_(1))/(M_(1))` and `mu_(2)=(m_(2))/(M_(2))` `therefore (m_(1))/(m_(2))=(mu_(1))/(mu_(2))xx(M_(1))/(M_(2))=(3)/(2)xx(20.2)/(32.0)=0.947` |
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| 266. |
Gaseous hydrogen contained initially under standard conditions in sealed vessel of volume `V = 5 L` was cooled by `Delta T = 55 K`. Find how much the internal energy of the gas will change and what amount of heat will be lost by the gas. |
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Answer» By the first law of thermodynamics `Delta Q = Delta U + Delta W` Here `Delta W = =` as the volume remains constant. `:. Delta Q = Delta U` Now `Delta U = (m)/(M) (R ) C_(V) (- Delta T)` `= - (m)/(M) (R )/(gamma - 1) Delta T` `(as C_(V) = (R )/(gamma - 1))` We have `P_(0)V = (m)/(M) RT_(0)` `= - (1.013 xx 10^(5) xx (5 xx 10^(-3)) xx 55)/(273 (1.4 - 1)) = - 255 J` |
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| 267. |
A monatomic idea gas of `2 mol` is taken through a cyclic process starting from `A` as shown in figure. The volume ratio are `V_(B)//V_(A)=2` and `V_(D)//V_(A)=4`. If the temperature `T_(A)` at `A` is `27^(@)C`, and gas constant is `R`. Calculate. The temperature of the gas at point `B`A. `500K`B. `700K`C. `600 K`D. `300 K` |
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Answer» Correct Answer - C Gas is monatomic, hence `C_(v)=(3)/(2)R` and `C_(p)=(5)/(2)R` Number of moles, `n=2` `T_(A)=27^(@)C=300K` Process `A rarr B` is a striaght line passing through origin, hence it is an isobaric process. `(V_(A))/(T_(A))=(V_(B))/(T_(B))` `T_(B)=((V_(B))/(V_(A)))T_(A)` `=2 xx 300 K=600K` ltbr. Porcess `A rarr B ` is isobaric. `Q_(p)=nC_(p)(T_(B)-T_(A))` `Q_(p)=2((5)/( 2)R)(600-300)=1500R` Heat is added to the system. Process` B rarr C` is isothermal. From the first law of thermodynamics. `Q_(BC)=W_(BC)` `(` as `Delta U=0)` ` W=nRT_(B)1n((V_(C))/(V_(B)))` `=(2)(R)(600)1n((4V_(A))/(2V_(A)))` `=(1200R)1n (2)` Heat is added to the system. Process `C rarr D ` is isochoric. From the first law of thermodynamics, `Q_(CD)=nC_(v)(T_(D)-T_(C))=n((3)/(2)R)(T_(A)-T_(B))` `=2((3)/(2) R)(300-600)=-900R` Negative sign implies that heat is rejected by the system. Process `D rarr A` is isothermal. `Q_(DA)=W_(DA)=nRT_(0)1n((V_(A))/(V_(D)))` `=(2)(R)3001n((V_(A))/(4V_(A)))=600R 1n((1)/(4))` Negative sign implies that heat is rejected by the system. In a cyclic process, `Delta U=0` `Q_(n et)=W_( n et)` `W_(n et)=Q_(AB)+Q_(BC)+Q_(CD)+Q_(DA)` `=1500 R+8.316R-900R-831.6R=600R` |
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| 268. |
A closed vessel contains a mixture of two diatomic gases `A` and `B`. Molar mass of `A` is 16 times that of `B` and mass of gas `A` contained in the vessel is 2 times that of `B`. Which of the following statements are correct ?A. only (i),(ii) and(iii) are trueB. only (ii),(iii) and (iv) are trueC. only (i),(ii) and (iv) are trueD. All are true |
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Answer» Correct Answer - D `U = n (nfRT)/(2)` `V_(rms) = sqrt((3RT)/(m))` |
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| 269. |
Certain amount of an ideal gas is contained in a closed vessel. The vessel is moving with a constant velcity `v`. The molecular mass of gas is `M`. The rise in temperature of the gas when the vessel is suddenly stopped is `(gamma C_(P) // C_(V))`A. `(Mv^(2) (gamma -1))/(2 R (gamma + 1))`B. `(Mv^(2) (gamma -1))/(2 R)`C. `(Mv^(2))/(2 R (gamma + 1))`D. `(Mv^(2))/(2 R (gamma - 1))` |
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Answer» Correct Answer - B b. If `m` is the mass of the gas, then its kinetic energy `= 1//2 mv^(2)` When the vessel is suddenly stopped, total kinetic energy will increase the temperature of the gas (because process will be adiabatic), i.e., `(1)/(2) mv^(2) = mu C_(v) Delta T` `= (m)/(M) C_(v) Delta T` `implies (m)/(M) (R )/(gamma - 1) Delta T = (1)/(2) mv^(2)` `(As C_(v) = (R )/(gamma- 1))` `implies Delta T = (M v^(2) (gamma - 1))/(2R)` |
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| 270. |
A monatomic idea gas of `2 mol` is taken through a cyclic process starting from `A` as shown in figure. The volume ratio are `V_(B)//V_(A)=2` and `V_(D)//V_(A)=4`. If the temperature `T_(A)` at `A` is `27^(@)C`, and gas constant is `R`. Calculate. Heat absorbed or released by the gas in process `A rarr B`A. `1500, ` addedB. `1200R 1n (2)`, addedC. `900 R,` rejectedD. `1200 R 1n (2)`, rejected |
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Answer» Correct Answer - A Gas is monatomic, hence `C_(v)=(3)/(2)R` and `C_(p)=(5)/(2)R` Number of moles, `n=2` `T_(A)=27^(@)C=300K` Process `A rarr B` is a striaght line passing through origin, hence it is an isobaric process. `(V_(A))/(T_(A))=(V_(B))/(T_(B))` `T_(B)=((V_(B))/(V_(A)))T_(A)` `=2 xx 300 K=600K` ltbr. Porcess `A rarr B ` is isobaric. `Q_(p)=nC_(p)(T_(B)-T_(A))` `Q_(p)=2((5)/( 2)R)(600-300)=1500R` Heat is added to the system. Process` B rarr C` is isothermal. From the first law of thermodynamics. `Q_(BC)=W_(BC)` `(` as `Delta U=0)` ` W=nRT_(B)1n((V_(C))/(V_(B)))` `=(2)(R)(600)1n((4V_(A))/(2V_(A)))` `=(1200R)1n (2)` Heat is added to the system. Process `C rarr D ` is isochoric. From the first law of thermodynamics, `Q_(CD)=nC_(v)(T_(D)-T_(C))=n((3)/(2)R)(T_(A)-T_(B))` `=2((3)/(2) R)(300-600)=-900R` Negative sign implies that heat is rejected by the system. Process `D rarr A` is isothermal. `Q_(DA)=W_(DA)=nRT_(0)1n((V_(A))/(V_(D)))` `=(2)(R)3001n((V_(A))/(4V_(A)))=600R 1n((1)/(4))` Negative sign implies that heat is rejected by the system. In a cyclic process, `Delta U=0` `Q_(n et)=W_( n et)` `W_(n et)=Q_(AB)+Q_(BC)+Q_(CD)+Q_(DA)` `=1500 R+8.316R-900R-831.6R=600R` |
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| 271. |
An ideal gas is taken round a cyclci thermodynamic process ABCA as shown if Fig. If the internal energy of the gas at point A is assumed zero while at B it is 50 J. The heat absorbed by the gas in the process BC is 90 J. (a) What is the internal energy og the gas at point C ? (b) How much heat energy is absorbed by the gas in the process AB? (c ) Find the heat energy rejected or absorbed by the gas in the process CA. (d) What is the net work done by the gas in the complete cycle ABCA ? |
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Answer» Given that `U_(A) = 0, U_(B) = 50 J` and `Q_(BC) = 90 J` Also `P_(A) = P_(B) = 10 Nm^(-2), P_(C) = 300 Nm^(-2), V_(A) = 1m^(3)` and `V_(B) = V_(C ) = 3 m^(3)` a. In process BC as volume of gas remains constant, work done by gas in this process is zero, thus `W_(BC) = 0` Heat abosrbed by the gas is `Q_(BC_ = 90 J`. From the first law of thermodynamics. `(Delta U) B_(C ) = U_(C ) - U_(B) = Q_(BC) - W_(BC) = 90 J - 0 = 90 J` `U_(C ) = (Delta U)_(BC) + U_(B) = 90 J + 50 J = 140 J` b. In process AB, we have `(Delta U)_(AB) = U_(B) - U_(A)` `= 50 - 0 = 50 J` work done is given as `W_(AB)` = area under AB in P - V diagram = area of rectangle ABED `= AB xx AD = (3 m^(3) = 1 m^(3)) xx 10 Nm^(-2)` `= 20 J` Thus heat abosrbed by the system is `Q_(AB) = (Delta U)_(AB) + W_(AB) = 50 + 20 = 70 J` c. For process CA `(Delta U)_(CA) - U_(A) - U_(C ) = 0 - 140 = - 140 J` Work done is given as `W_(CA) = ` area ACED = area of triangle ACB + area of rectangle ABED `1//2 xx AB xx BC + AB xx AD` `= 1//2 xx (3 - 1) m^(3) xx (30 - 10) Nm^(-2) + 20` `= 20 + 20 = 40 J` In this process, the volume decreases, the work is done on the gas, Hence, the work done is negaitve, Thus, `W_(CA) = - 40 J` Thus heat rejected by the gas is `Q_(CA) = (Delta U)_(CA) + W_(CA) = - 140 - 40 = - 180 J` Net work done in the complete cyclic process ABCA is `W = "area of triangle" ABC = (1)/(2) xx 2 xx 20 = 20 J` As the cycle is anticlockwise, net work is done on the gas. |
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| 272. |
Two moles of an ideal mono-atomic gas undergo a cyclic process as shown in the figure. The temperatures in different states are given as `6T_(1) = 3T_(2) = 2T_(4) = T_(3) = 1800 K`. Determine the work done by the gas during the cycle. A. `- 10 kJ`B. `- 20 kJ`C. `- 15 kJ`D. `- 30 kJ` |
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Answer» Correct Answer - A a. `6T_(1) = 3T_(2) = 2T_(4) = T_(3) = 1800 K` `T_(1) = 300 K, T_(2) = 600 K` `T_(4) = 900 K, T_(3) = 1800 K` `4 rarr 1` and `2 rarr 3` are isochroic process in which work done = 0 `W_(12) = P (V_(2) - V_(1)) = nRT (T_(2) - T_(1))` `= 2 xx R (600 - 300) = 600 R` `W_(34) = P (V_(4) - V_(3)) = nRT (T_(4) - T_(3))` `= 2 xx R (900 - 1800) = - 1800 R` `W_(T o t a l) = 600 - 1800 R = - 1200 T = - 10000 J` |
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| 273. |
A closed vessel contains a mixture of two diatomic gases `A` and `B`. Molar mass of `A` is 16 times that of `B` and mass of gas `A` contained in the vessel is 2 times that of `B`. Which of the following statements are correct ?A. Average kinetic energy per molecule of `A` is equal to that of `B`B. Root mean square value of translational velocity of `B` is four time that of `A`C. Pressure exerted by `B` is eight time of the exerted by `A`.D. Number of moleucles of `B`, in the cylinder, is eight times that of `A` |
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Answer» Correct Answer - A::B::C::D Since both the gases are contained in the same vessel, temperature of both the gases is same. Average `KE` per molecule of a diatomic gas is `5//2 KT`. Hence, average `KE` per molecule of both the gases is same. Therefore. Option `(a)` is correct. `v_(mas)=sqrt((2RT)/( M))` Hence, `((u_(rms))_(2))/((v_(rms))_(1))=sqrt((M_(1))/( M_(2)))=sqrt(16)=4` Hence, option `(b)` is correct. Let molar mass of `B` be `M`, then that of `A` will be equal to `16 M`. Let mass of gass `B` ini the vessel be `m`, then that of `A` will be `2m`. The number of moles of a gas, in the vessel will be `n= m/M` . Hence, number of moles of gases `A` and `B` will be `n_(1)=(2m)/( 16 M) ` and `n_(2)=(m)/( M)` Hence `(n_(1))/( n_(2))=(1)/(8)` Hence, option `(d)` is correct. Partial pressure exerted by a gas is `P=(nRT)/( V)` Hence ` (P_(2))/( P_(1))=(n_(2))/( n_(1))=8` Therefore, option `(c)` is also correct. |
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| 274. |
An ideal diatomic gas is expanded so that the amount of heat transferred to the gas is equal to the decrease in its internal energy. The molar specific heat of the gas in this process is given by `C` whose value isA. `-(5R)/(2)`B. `-(3R)/(2)`C. `2R`D. `(5R)/(2)` |
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Answer» Correct Answer - A It is given that there is decrease in the internal energy while the gas expands absorbing some heat, which is numberically equal to the decrease in internal energy. Hence, `dQ=-dU` Thus, `dQ=dW+dU=-uD` `dw=-2dU(` remembering that both `dW` and `dU` are negative `)` `PdV=2CdT(dU=-CdT) ...(i)` where `C` is the molar specific heat of the gas. Since the gas is ideal , we have `P=(RT)/(V) ....(ii)` Hence Eq. (i) gives `(RT)/(V) dV=2CdT` `(dV)/(V)=(2C)/(R)(dT)/( T)` Solving we get `(VT^(2C//R))=` constant ....(iii) Also since `dQ=-dU`, `C=((dQ)/(dT))_("const ant volume")= ((dU)/(dT))=-C_(v)` Hence, `C=-C_(v)=(R)/( gamma-1)=-(R)/( (7)/(5)-1)=-(5R)/(2)` Since the gas is diatomic. |
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| 275. |
Two moles of monatomic ideal gas is taken through a cyclic process shown on `P - T` diagram in Fig. Process `CA` is represented as `PT` = constant. If efficiency of given cyclic process is `1 - (x)/(12 1n 2 + 15)` then find `x`, |
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Answer» For process `AB`, `T_(A) = 300 K, T_(B) = 600 K` `W = nR Delta T = nR (T_(B) - T_(A)) = 300 nT = 600 R` `Q = nC_(P) Delta T = 2 xx (5)/(2) R (300) = 1500 R` For process `BC`, `W = nRT 1n (V_(f))/(V_(i)) = nRT 1n (P_(i))/(P_(f))` `= nRT 1n 2 = 1200 1n 2` For process `CA`, `W = int PdV = int_(600)^(300) (K)/(T) (2 nRT)/(K) dT` `= - 2 nR (300) = - 1200 R` `Q = n C_(V) Delta T + W` `= 2 xx (3)/(2) R (-300) - 1200 R` `= - 900 R - 1200 R = - 2100 R` `eta = (600 R + 1200 R 1n - 1200 R)/(1500 R + 1200 R 1n 2)` `eta = 1 - (21)/(12 1n 2 + 15)` Comparing with given equation, `x = 21`. |
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| 276. |
When a thermodynamic system is taken from an initial state I to a final state F along the path IAF, as shown in Fig. the heat energy absorbed by the system is `Q = 55 J` and work done by the system is `W = 25 J`. If the same system is taken along the path IBF, the value of `Q = 35 J`. (a) If `W = - 15 J` for the curved path FI, how much heat energy is lost by the system along this path ? (b) If `U_(1) = 10 J`, what is `U_(F)` ? (d) If `U_(B) = 20 J`, what is Q for the process BF and IB ? |
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Answer» The first law of thermodynamics states that `Delta Q = Delta U + Delta W` or `Q = (U_(F) - U_(I)) + W` Here `U_(I)` and `U_(F)` are the internal energies in the initial and the final state, respectively. Given that for path `IAF, Q = 55 J` and `W = 25 J`. Therefore, `Delta U = U_(F) - U_(I) = Q - W = 55 - 25 = 30 J` The internal energy is independent of the path, it depends only on the initial and final states of the system. Thus, the internal energy between I and F states is 30 J irrespective of the path followed by the system. a. For path IBF, `Q = 35 J` and `Delta U = 30 J`. Therefore, `W = Q - Delta U = 35 - 30 = 5 J` b. `W = - 15 J`, but `Delta U = - 30 J`. Therefore, `Q = W + Delta U = - 15 - 30 = - 45 J`. Given `U_(I) = 10 J`. Therefore, `U_(F) = Delta U + U_(I) = 30 + 10 = 40 J`. d. The process BF is isochoric, i.e., the volume is constant. Hence, W = 0. Therefore, `Q = (Delta U)_(BF) = U_(F) - U_(B) = 40 - 20 = 20 J`. The process IB is isobaric (constant pressure). Therefore. `Q = (Q)_(IBF) - (Q)_(BF) = 35 - 20 = 15 J`. |
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| 277. |
A mixture of 2 moles of helium gas (`(atomic mass)=4a.m.u`) and 1 mole of argon gas (`(atomic mass)=40a.m.u`) is kept at 300K in a container. The ratio of the rms speeds `((v_(rms)(helium))/((v_(rms)(argon))` isA. `0.32`B. `0.45`C. `2.24`D. `3.16` |
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Answer» Correct Answer - D `((v_(rms))_(He))/((v_(rms))_(Ar)) = sqrt(((M_(0))_(Ar))/((M_(0))_(He)))=sqrt(40/4)=sqrt(10)=3.16` |
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| 278. |
Two moles of helium gas are taken along the path `ABCD` (as shown in Fig.) The work done by the gas is A. `2000R ((1)/(2) + 1n (4)/(3))`B. `500 R (R + 1n 4)`C. `500 R (2 + 1n (16)/(9))`D. `1000 R (1 + 1n (16)/(9))` |
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Answer» Correct Answer - A a. `A rarr B` is isobaric process, `V prop T` so, `Delta W_(AB) = nR Delta T = 2 xx R xx (750 - 250) = 1000 R` `B rarr C` is an isochoric process `:. Delta W_(BC) = 0` and `C rarr D` is an isothermal process `Delta W_(CD) = nRT 1n ((V_(f))/(V_(i)))` ` = 2 xx R xx 1000 1n ((20)/(15)) = 2000 1n ((4)/(3))` Total work done, `Delta W = Delta W_(AB) + Delta W_(BC) + Delta W_(CD)` |
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| 279. |
Four moles of hydrogen, 2 moles of helium and 1 mole of water form an ideal gas mixture. What is the molar specific heat at constant pressure of mixture ?A. `(16)/(7)R`B. `(7 R)/(16 R)`C. RD. `(23)/(7) R` |
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Answer» Correct Answer - D `C_(v)` for hydrogen `=(5R)/(2),C_(v)` for helium = `(3R)/(2)` `C_v` for water vapour = `(6R)/(2)` `:. [C_(v)]_(mix) = (4xx(5R)/(2)+2(3R)/(2)+1xx3R)/(4+2+1) = (16R)(7)` `:. C_(p)+C_(v)+R = (16R)/(7)+R = (23R)/(7)`. |
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| 280. |
Four moles of hydrogen, 2 moles of helium and 1 mole of water form an ideal gas mixture. What is the molar specific heat at constant pressure of mixture ?A. `(16)/(7)R`B. `(7 R)/(16)`C. `R`D. `(23)/(7)R` |
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Answer» Correct Answer - D d. `C_(v)` for hydrogen `= 5 R //2, C_(V)` for helium `= 3 R//2, C_(v)` for water vapour `= 6 R//2` `:. (C_(v))_(max) = (4 xx (5R)/(2) + 2 (3R)/(2) + 1 xx 3R)/(4 + 2 + 1) = (16R)/(7)` `:. C_(P) + C_(v) + R = (16 R)/(7) + R = (23R)/(7)` |
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| 281. |
Internal energy of `n_(1)` mol of hydrogen of temperature `T` is equal to the internal energy of `n_(2)` mol of helium at temperature `2T`. The ratio `n_(1) // n_(2)` isA. `(3)/(5)`B. `(2)/(3)`C. `(6)/(5)`D. `(3)/(7)` |
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Answer» Correct Answer - C c. Internal energy of `n` moles of an ideal gas at temperature ture `T`is given by `U = (f)/(2) mRT` (f = degrees of freedom) `U_(1) = U_(2)` `f_(1) n_(1) T_(1) = f_(2) n_(2) T_(2)` `:. (n_(1))/(n_(2)) = (f_(2) T_(2))/(f_(1) T_(1)) = (3 xx 2)/(5 xx 1) = (6)/(5)` Here, `f_(2) = ` degrees of freedom of `He = 3` and `f_(1) = ` degrees of freedom of `H_(2) = 5` |
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| 282. |
Helium gas goes through a cycle ABCDA (consisting of two isochoric and isobaric lines) as shown in figure Efficiency of this cycle is nearly: (Assume the gas to be close to ideal gas) A. 0.105B. 0.125C. 0.154D. 0.091 |
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Answer» Correct Answer - C `Q_(AB)=nCv[(P_0V_0)/(R)],Q_(BC)=nCp[(2P_(0)V_(0))/(R)]` `Q_(CD)=nCv[(2P_(0)v_(0))/(R)],Q_(DA)=nCp[(p_(0)V_(0))/(R)]` `Q_(repeated)=Q_(0)+Q_(DA)` `Q_(absorbed)=Q_(AB)+Q_(BC)` `eta=1-(Q_(rej))/(Q_(ab))=0.154` |
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| 283. |
Helium gas goes through a cycle ABCDA (consisting of two isochoric and isobaric lines) as shown in figure Efficiency of this cycle is nearly: (Assume the gas to be close to ideal gas) |
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Answer» Evidently, work done by the gas during the cyclic process is `W = (2P_(0) - P_(0)) (2V_(0) - V_(0))` unit `= P_(0) V_(0)` unit For the process A to B which is iscohoric, the heat supplied will be given by `Delta Q_(1) = Delta U + Delta W = nC_(V) Delta T + 0` `= n ((3 R)/(2)) [(V_(0) (2 P_(0) - P_(0)))/(nR)] ( :. nR Delta T = V Delta P)` `= (3)/(2) P_(0) V_(0)` unit And for the process B to C which is isobaric, the heat abosrbed will be given by `Delta Q_(2) = n C_(P) Delta T` `= n ((5 R)/(2)) [(2 P_(0) (2V_(0) - V_(0)))/(nR)] ( :. nR Delta T = P (Delta V))` `= 5P_(0) V_(0)` unit For the processes C to D to A, the heat given will be obviously negative, which implies that heat is abstracted from the system. Therefore, efficiency `(eta)` of the cycle `= ("Work done per cycle")/("Total heat given per cycle") %` `= (P_(0) V_(0))/(((3 P_(0) V_(0))/(2)) + 5 P_(0) V_(0)) = (2)/(3) xx 100 = (200)/(13) % = 15.38%` |
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| 284. |
During an adiabatic process, the pressure of a gas is found to be proportional to the cube of its absolute temperature. The ratio `C_P//C_V` for the gas is |
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Answer» For an ideal gas of one mole PV=RT During an adiabatic process `PpropT^(3)` `P=KT^(3),` where K is a constant `P=k((PV)/(R))^(3)rArrP=((k)/(R^(3)))P^(3)V^(3)` `P^(2)V^(2)=` constant , `PV^(3//2)=` constant Comparing it with the equation `PV^(gamma)`=constant for an adiabatic process, we get `gamma=3//2` |
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| 285. |
An ideal monoatomic gas is taken the cycle `ABCDA` as shown in following `P-V` diagram. The work done during the cycle is A. `PV`B. `2PV`C. `4PV`D. zero |
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Answer» Correct Answer - C Work done=Area of curve enclosed `=2V xx 2P=4PV`. |
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| 286. |
An ideal monoatomic gas is taken the cycle `ABCDA` as shown in following `P-V` diagram. The work done during the cycle is A. PVB. 2 PVC. PV/2D. zero |
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Answer» Correct Answer - A w = area of loop = PV |
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| 287. |
An ideal monoatomic gas is taken round the cycle ABCDA as shown in the P-V diagram. The work done during the cycle is A. PVB. 2PVC. 3PVD. 4PV |
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Answer» Correct Answer - A Work done=Area under rectangle |
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| 288. |
An ideal monoatomic gas is taken round the cycle ABCDA as shown in the P-V diagram. Compute the work done in the process. |
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Answer» Total work done =Area under P-V curve (parallelogram) =Base x Height=`(6V_(0)-3V_(0))(5P_(0)-3P_(0))` `=(3V_(0))(2P_(0))=6P_(0)V_(0)` units |
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| 289. |
An ideal gas has a specific heat at constant pressure `C_P=(5R)/2`. The gas is kept in a closed vessel of volume `0.0083m^3`, at a temperature of 300K and a pressure of `1.6xx10^6 N//m^2`. An amount of `2.49xx10^4` Joules of heat energy is supplied to the gas. Calculate the final temperature and pressure of the gas. |
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Answer» As here volume of gas remains constant, we have `(Delta Q)_(v) = mu C_(v) Delta T` Here, `C_(v) = C_(P) - R (5//2) R - R = (3//2) R` `mu = (PV)/(RT) = (1.5 xx 10^(6) xx 8.3 xx 10^(-3))/(8.3 xx 300) = (16)/(3)` `2.49 xx 10^(4) = (16)/(3) xx (3)/(2) xx 8.3 Delta T` i.e., `Delta T = 375` or `T_(2) - T_(1) = 375`, i.e., `T_(2) = (375 + 300)` Now as for a given mass of an ideal gas at constant volume, `P prop T` `(P_(2)// P_(1)) = (T_(1) // T_(2))` `P_(2) = (675)/(300) xx 1.6 xx 10^(6) = 3.6 xx 10^(6) N//m^(2)` |
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| 290. |
An ideal gas at `NTP` is enclosed in an adiabatic vertical cylinder havin an area of cross section `A=27 cm^(2)` between two light movable pistons as shown in figure. Spring with force constant `K=37-- N//m` is in a relaxed state initially. Now the lower position is moved upwards a distance `h//2`, `h` being the initial length of gas column . It is observed that the upper piston moves up by a distance `h//16`. Final temperature of gas` 4//3 xx 273 K`. Take `gamma` for the gas to be `3//2`. The value of `h` isA. `1 m`B. `1.4m`C. `1.6m`D. `2m` |
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Answer» Correct Answer - C Final pressure, `P_(f)=(P_(0)+(Kh//16)/(A))` `(P_(f)V_(f))/( T_(f))=(P_(0)V_(0))/(T_(0))` `((P_(0)+(kh)/(16A))Axx((h)/(2)+(h)/(16)))/(T_(f))=(P_(0)Ah)/( T_(0))` `((10^(5)+(3700h)/( 16xx27xx10^(-4)))(9)/(16))/((4)/(3)xx273)=(10^(5))/(273)` `(1+0.856h)=(16)/(9)xx(4)/( 3)` Solving, `h=1.6 m ` |
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| 291. |
A cylindrical chamber `A` of uniform cross section is divided into two parts `X` and `Y` by a movalbe piston `P` which can slide without friction inside the chamber. Initially part `X` contains `1 "mol"` of a monochromatic gas and `Y` contains `2 mol` of a diatomic gas, and the volumes of `X` and `Y` are in the ratio `1:2` with both parts `X` and `Y` being at the same temperature `T`. Assuming the gases to be ideal, the work `W` that will be done in moving the piston slowly to the position where the ratio of the volumes of `X` and `Y` is `2:1` will be A. `-5.8T`B. `8.3 T`C. `12.3 T`D. zero |
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Answer» Correct Answer - A a. Since the piston is moved slowly we assume isothermal condition for both the gases as thermal equilibrium is maintained throughout. Let the total of chamber be. `V`. Then volume of gas in `X` increase (expand) from `V//3` to `2V//3`. The work done is positve and given (for 1 mol of monatomic gas by `W_(x) = + RT log_(e) .(2 V//3)/(V//3) = RT log_(e) 2` The volume of gas in `Y` decrease (compressed) from `2V//3` to `V//3`. The work done (isothermally) is negative and given (for 2 mol of diatomic gas) by `W_(x) = 2 RT log_(e) (V//3)/(2V//3) = - RT log_(e) 2` Hence the total work done on the system is `W = W_(x) + W_(Y) = - RT log_(e) 2` Substituting `R = 8.3, log_(2) 2 = 0.6996` and simplifiying `W = - 5.8 T`. |
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| 292. |
One mole of air `(C_(V) = 5R//2)` is confined at atmospheric pressure in a cylinder with a piston at `0^(@)C`. The initial volume occupied by gas is`V`. After the equivalent of `13200 J` of heat is transferred to it, the volume of gas `V` is nearly `(1 atm = 10^(5) N//m^(3))`A. `37 L`B. `22 L`C. `60 L`D. `30 L` |
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Answer» Correct Answer - C c. `P = 1 atm = 10^(5) N//m^(2)` `T = 0^(0) C = 273 K` `V = (nRT)/(P) = (1 xx 8.3 xx 273)/(10^(5)) = 0.0227 m^(3) = 22.7 L` `C_(V) = (5)/(2) R, C_(P) = (7)/(2) R` Heat transferred `Delta Q = nC_(P) Delta T = n(7R)/(2) Delta T = 13200 J` Work done `nR Delta T = (13200 xx 2)/(7)` `= P (V_(f) - V_(i)) = 3771` `V_(f) - V_(i) = 3771 xx 10^(-5) = 0.0377 m^(3)` `V_(f) = V_(i) + 37.7 L` `22.7 L + 33.7 L` `60.4 L ~~ 60 L` |
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| 293. |
Air is contained in a piston-cylinder arrangement as shown in Fig. with a cross-sectional area of `4 cm^(2)` and an initial volume of `20 cc`. The air is initially at a pressure of 1 atm and temperature of `20^(@)C`. The piston is connected to a spring whose spring constant is `k = 10^(4) N//m`, and the spring is initially underformed. How much heat must be added to the air inside cylinder to increase the pressure to 3 atm (for air, `CV = 718 J//kg.^(@)C`, molecular of air 28.97)? |
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Answer» When pressure changes from 1 atm to 2 atm, the change in pressure `P = 2 atm` `= 2 xx 1 xx 10^(5) N//m^(2)` The force exerted on the piston `F = PA = 2 xx 10^(5) xx 4 xx 10^(-4)` `= 80 N` The compression of the spring `x = (F)/(k) = (80)/(10^(4)) = 0.008 m` The change in volume of air due to displacement of piston by `x` `Delta V = Ax = 4 xx 10^(-4) xx 0.008` `= 3.2 xx 10^(-6) m^(3)` `:.` Final volume, `V_(2) = V_(1) + Delta V` `= 20 xx 10^(-6) + 3.2 xx 10^(-6)` By equation of state `(P_(1) V_(1))/(T_(1)) = (P_(2) V_(2))/(T_(2))` `T_(2) = (P_(2) V_(2) T_(1))/(P_(1) V_(1)) = ((3)/(1)) xx ((23.3 xx 10^(6)))/((20 xx 10^(-6))) xx (273 + 20)` `= 1020 K` The change in internal energy air `Delta U = mC_(v) Delta T` `= (2.38 xx 10^(-5)) xx 718 xx (1020 - 293)` `= 12.42 J` Work done in compressing the spring by `x` `W = (1)/(2) kx^(2) = (10^(4))/(2) xx (0.008)^(2) = 0.32 J` From the first law of thermodynamics `Q = Delta U + W = 12.42 + 0.32 = 1274 J` |
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| 294. |
A cylinder contains 3 moles of oxygen at a temperature of `27^(@)C`. The cylinder is provided with a frictionless piston which maintains a constant pressure of 1 atm on the gas. The gas is heated until its temperature rises to `127^(@)C`. a. How much work is done by the gas in the process ? b. What is the change in the internal energy of the gas ? c. How much heat was supplied to the gas ? For oxygen `C_(P) = 7.03 cal mol^(-1) .^(@)C^(-1)`. |
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Answer» For oxygen `C_(P) = 7.03 xx cal//mol .^(@)C`. `Delta Q = nC_(P) Delta T = 3 xx 7.03 xx (127 - 27)` `= 2109 cal (n = 3)` Since `gamma = C_(P)//C_(v)` and `gamma = 1.4` for oxygen as it is diatomic, `Delta U = nC_(v) Delta T = 3 xx (7.03)/(1.4) xx 100 = (2109)/(1.4) cal = 1506 cal` By the first law of thermodynamics, `Delta Q = Delta U + Delta W` `:. Delta W = Delta Q - Delta U = 2109 = 1506 = 603 cal` `= 603 xx 4.184 = 2523 J` a. work done by the gas `= 2523 J` b. Change in internal energy `= 1506 cal` c. Heat added to the gas ` = 2109 cal` |
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| 295. |
Two moles of a diatomic gas at 300 K are kept in a nonconducting container enclosed by a piston . Gas is now compressed to increase the temperature from 300 K to 400 K. Find word done by the gas |
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Answer» `Delta Q = Delta u + Delta w` Since container is nonconducting, therefore `Delta Q = 0 Delta u + Delta w` `implies Delta W = - Delta u = - n (f)/(2) R Delta T` `= - 2 xx (5)/(2) R (400 - 300)` `= - 5 xx 8.314 xx 100 J = - 5 xx 831.4 J xx - 4175 J` |
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| 296. |
`n` moles of gas in a cylinder under a piston is transferred infinintley slowly from a state with a volume of `V_(0)` and a pressure `3 P_(0)` to a state with a volume of `3 V_(0)` and a pressure `P_(0)` as shown in Fig. The maximum temperature that gas will reach in this process is A. `(P_(0) V_(0))/(nR)`B. `(3 P_(0) V_(0))/(nR)`C. `(4 P_(0) V_(0))/(nR)`D. `(2 P_(0) V_(0))/(nR)` |
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Answer» Correct Answer - C c. `(P - P_(1))/(P_(1) - P_(2)) = (V - V_(2))/(V_(1) - V_(2))` `(P - P_(1)) (V_(1) - V_(2)) = (V_(1) - V_(2)) = (V - V_(1)) (P_(1) - P_(2))` `(P - 3 P_(0)) (V_(0) - 3 V_(0)) = (V - V_(0)) (3P_(0) - P_(0))` `(P - 3 P_(0)) (- 2 V_(0)) = (V - V_(0)) (2 P_(0))` `- 2 V_(0) P + 6 P_(0) V_(0) = 2 V P_(0) - 2 P_(0) V_(0)` `2 V P_(0) + 2 V_(0) P - 8 P_(0) V_(0) = 0` `V P_(0) + (V_(0) nRT)/(V) - 4 P_(0) V_(0) = 0` `V^(2) P_(0) - 4 P_(0) V_(0) + V_(0) nRT = 0` `T = (P_(0) (V^(2) + 4 V V_(0)))/(V_(0) nR)` For maximum of minimum value of `T`, `(dT)/(dV) = - 2V + 4 V_(0) = 0 implies V = 2 V_(0)` `(d^(2) T)/(dV^(2)) = -2` It is negative so `T` is maximum at `V = 2 V_(0)` `T_(max) = (P_(0) (- V_(0)^(2) + 8 V_(0)^(2)))/(V_(0) nR) = (4 P_(0) V_(0))/(nR)` |
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| 297. |
A gas expands in a piston-cylinder device from volume `V_(1)` to `V_(2)`, the process being described by `P = a//V + b`, a and b are constants. Find the work done in the process. |
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Answer» Work done, `W = int_(C_(1))^(V_(2)) PdV` `= int_(V_(1))^(V_(2)) ((a)/(V) + b) dV` `| a 1n V + b V|_(V_(1))^(V_(2))` `(a 1n V_(2) + v V_(2)) - (a 1n V_(1) + b V_(1))` `a 1n ((V_(2))/(V_(1))) + b (V_(2) - V_(1))` |
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| 298. |
Piston cylinder device initially contains `0.5m^(3)` of nitrogen gas at `400 kPa` and `27^(@)C`. An electric heater within the device is turned on and is allowed to apss a current of `2 A` for `5 mi n` from a `120 V` source. Nitrogen expands at constant pressure and a heat loss of `2800 J` occurs during the process. `R=25//3 kJ //k mol - K` Number of moles of nitrogen gas isA. `0.8 k mol`B. `0.08 k mol`C. `0.8 mol`D. `0.08 mol` |
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Answer» Correct Answer - B Number of moles of nitrogen gas `n=(PV)/( RT)=((40kPa)(0.5m^(3)))/(((25)/(3)kJ//k mol-K)(300K))=0.08 k mol` |
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| 299. |
Piston cylinder device initially contains `0.5m^(3)` of nitrogen gas at `400 kPa` and `27^(@)C`. An electric heater within the device is turned on and is allowed to apss a current of `2 A` for `5 mi n` from a `120 V` source. Nitrogen expands at constant pressure and a heat loss of `2800 J` occurs during the process. `R=25//3 kJ //k mol - K` Electric work done on the nitrogen gas isA. `72 kJ`B. `36 kJ`C. `118 kJ`D. `9kJ` |
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Answer» Correct Answer - A Electric work done on the nitrogen `W_(e)=-Delta VIDeltaT` `=-120 xx 2 xx 5 xx 60=72 xx 10^(3)J` `=-72 kJ` `(` Negative sign is added because the work is done on the system `)`. |
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| 300. |
`1 cm^(3)` of water at its boiling point absorbs `540 cal` of heat to become steam with a volume of `1671 cm^(3)`. If the atmospheic pressure is `1.013 xx 10^(5) N//m^(2)` and the mechanical equivalent of heat `= 4.19 J//cal`, the energy spent in this process in overcoming intermolecular forces isA. `540 cal`B. `40 cal`C. `500 cal`D. Zero |
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Answer» Correct Answer - C c. Energy spent in overcoming inter-molecular forecs. `Delta U = Delta Q - Delta W` `Delta Q - P(V_(2) - V_(1))` `= 540 - (1.013 xx 10^(5) (1671 - 1) xx 10^(-6))/(4.2)` `implies 500 cal` |
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