4131 + Interview Questions in THERMODYNAMICS IN GENERAL AWARENESS Page 71 InterviewSolution

3501.	What is the specific heat of a gas in an adiabatic process ?
Answer» The specific heat of a gas in an adiabatic process is zero.

Discussion

3502.	Heat system based on circulation of steam are more efficient in warning a building than those based on circulation of hot water why ?
Answer» Because steam at 100ºC has more heat than water at 100ºC.

Discussion

3503.	A perfect gas goes from a state A to another state B by absorbing 8 × 105 J of heat and doing 6.5 × 105 J of external work. It is now transferred between the same two states in another process in which it absorbs 105 J of heat. In the second process
Answer» Correct Answer - `10xx10^(5)`

Discussion

3504.	Identify the state and path functions out of the following (a) Enthalpy (b) Entropy (c) Heat (d) Temperature (e) Work (f) Free energy
Answer» State function : Enthalpy, entropy, temperature and free energy. Path function : Heat and work.

3504.

Identify the state and path functions out of the following (a) Enthalpy (b) Entropy (c) Heat (d) Temperature (e) Work (f) Free energy

Answer»

State function : Enthalpy, entropy, temperature and free energy.

Path function : Heat and work.

Discussion

3505.	The heat of combustion of methane is `-880 KJ mol^(-1)`. The quantity of heat liberated in the combustion of 3.2 g methane isA. `-88 KJ`B. `+88 KJ`C. `+176 KJ`D. `-176 KJ`
Answer» Correct Answer - D `16 gm rarr -880` `3.2 gm rarr` ?

Discussion

3506.	Which of the following variables are called state variables or state functions?A. `P`B. `V`C. `T`D. All of these
Answer» Correct Answer - D Variable like `P,V` and `T` are called state variable (or state functions) because their values depend only on the state of the system and not on how it is reached. State functions are very useful because a change in their value depends only on the initial and final states of the system and not on how the change is carried out.

Discussion

3507.	Calculate the free enegry change when `1mol` of `NaCI` is dissolved in water at `298K`. Given: a. Lattice enegry of `NaCI =- 778 kJ mol^(-1)` b. Hydration energy of `NaCI - 774.3 kJ mol^(-1)` c. Entropy change at `298 K = 43 J mol^(-1)`
Answer» Correct Answer - `-9.1555 kj` `= 3.7 kJ mol^(-1) = 3700 J mol^(-1)` `Delta H_("dissolution") = Delta h_(("ionisation")) + Delta H_(("hydration")) = 778 - 774.3` `Delta S_("disolution") = 43 J mol^(-1)` `Delta G_("disolution") = Delta H - T Delta S = 3700 - 298 X 43 = -9114 J, Delta G = - 9.114 kJ`

Discussion

3508.	How much heat is required to change `10g` ice at `0^(@)C` to steam at `100^(@)C`? Latent heat of fusion and vaporisation for `H_(2)O` are `80 cl g^(-1)` and `540 cal g^(-1)`, respectively. Specific heat of water is `1cal g^(-1)`.
Answer» Correct Answer - 7200 cal `Delta H_("total") = Delta H_("fusion") + Delta H_("heating") + Delta H_("vaporization")` `10 xx 80 + 10 xx 1 xx 100 + 10 xx 540 = 7200 cal`

Discussion

3509.	How much energy is released when 6 mole of octane is burnt in air ? Given `DeltaH_(f)^(@)` for `CO_(2)(g),H_(2)O(g)` and `C_(8)H_(18)(l)` respectively are `-490,-240` and `+160KJ//mol`A. `- 6.2 MJ`B. `- 37.4 MJ`C. `-35.5 MJ`D. `- 20 MJ`
Answer» Correct Answer - B Thermochemical equation for the combustion of one mole octane is `C_(8) H_(18) (1) + (15)/(2) O_(2) (g) rarr 8 CO_(2) (g) + 9 H_(2) O (1)` Accroding to Eq., we have `Delta_(C ) H^(@) = [(8 mol) Delta_(f) H^(@) (C_(8) H_(18), l) + ((15)/(9) mol) Delta_(f) H^(@) (O_(2), g)]` `[8(-490) + 9 (-240)] - (160)` `= - 3920 - 2160 - 160` `= - 6240 kJ` This is the heat energy released when 1 mol of octane burns. Thus, the amount of heat energy released on burning 6 mol of octane will be `- (6) (6240) kJ = - 37400 kJ` `= - 37.440 xx 10^(3) kJ` `= - 37.440 MJ (1 MJ = 10^(6) J = 10^(3) kJ)`

Discussion

3510.	How much energy is released when 6 mole of octane is burnt in ? Given that the heat of form action of `CO_(2), H_(2)O` and `C_(8)H_(18)` respectively are -390, -240, and +160 KJ/mole.A. `-32.6 MJ`B. `-37.4 MJ`C. `-35.5 MJ`D. `-20.0 MJ`
Answer» Correct Answer - A `C_(8)H+_(18) + (25)/(2)O_(2)rarr 8CO_(2)+9H_(2)O` `Delta H = H_(P)-H_(R)`

Discussion

3511.	A gaseous mixture of propane, acetylene and `CO_(2)` is burnt in excess of air. Total `4800` KJ heat is evolved . The volume of `CO_(2)` formed during combustion is `224` liters at STP. The total evolved heat is used to perform two separate process: `(i)` Vapourising `87.5%` of water (liquid) obtained in the process of burning the original mixture. `(ii)` Forming `3808` litres ethylene measured at STP from its elements. `DeltaH_(H-H) = 435 KJ//mol " " DeltaH_(C-H) = 416 KJ//mol " " DeltaH_(C-C) = 347 KJ//mol` ` DeltaH_(C-C) = 615 KJ//mol, " " DeltaH_(C-C) = 812 KJ//mol" " DeltaH_("sublimation") "of " (C,s) = 718 KJ//mol` `DeltaH_(f)^(@)(C_(2)g) = -394 KJ//mol " "DeltaH_(f)^(@)(H_(2)O, l) = -286 KJ//mol. " " DeltaH_(f)^(@)(H_(2)O,g) = -247 KJ//mol.` `DeltaH_(rxxn)^(@) "for" C_(2)H_(2)(g) + H_(2)g rarr C_(2)H_(4)(g)`A. `-220 kJ//mol`B. `-180 KJ//mol`C. `-250 KJ//mol`D. `-200 KJ//mol`
Answer» Correct Answer - D

Discussion

3512.	Given `Delta_("ioniz")H^(o)(HCN)=45.2 kJ mol^(-1)` and `Delta_("ioniz")H^(o)(CH_(3)COOH)=2.1 kJ mol^(-1)`. Which one of the following facts is ture?A. `pK_(a)(HCN)=pK_(a)(CH_(3)COOH)`B. `pK_(a)(HCN)gtpK_(a)(CH_(3)COOH)`C. `pK_(a)(HCN)ltpK_(a)(CH_(3)COOH)`D. `pK_(1)(HCN)=(45.17//2.07)pK_(a)(CH_(3)COOH)`
Answer» Correct Answer - B

Discussion

3513.	A black body emit heat at the rate of `20 W`, when its tempertaure is `227^(@)C` Another black body emits heat at the rate of `15 W`, when its temperature is `227^(@)C`. Compare the area of the surface of the two bodies, if the surrounding is at `NTP`A. `16:1`B. `1:4`C. `12:1`D. `1:12`
Answer» Correct Answer - D `P_(1)/(P_(2))= (sigmaA_(1)T_(1)^(4))/(sigmaA_(2)T_(2)^(4))` or `(A_(1))/(A_(2))=(P_(1)T_(2)^(4))/(P_(2)T_(1)^(4))= (20)/(15)xx((227+273))/((272+273))=1/2`

Discussion

3514.	In (figure). shows two path that may be taken by a gas to go from a state A to state C In the process AB, `400 J` of heat is added to the system and in process Bc, `100 J` of heat is added to the system. The heat absorbed by the system in the process AC will beA. `500 J`B. `460 J`C. `300 J`D. `380 J`
Answer» Correct Answer - B In cyclic process ABCA `DeltaU_(cyclic)=0` From first law of thermodybamics `Q_(cyclic)= W_(cyclic)+U_(cyclic)` `:. Q_(cyclic)= W_(cyclic)` `Q_(AB)+Q_(BC)+Q_(CA)=` closed loop area `400+100+Q_(CA)= 1/2(2xx10^(-3))xx4xx10^(4)` `500-Q_(AC)= 40 [:. Q_(CA)= - Q_(AC)]` `Q_(AC)= 460 J`

Discussion

3515.	In (figure). shows two path that may be taken by a gas to go from a state A to state C In the process AB, `400 J` of heat is added to the system and in process Bc, `100 J` of heat is added to the system. The heat absorbed by the system in the process AC will beA. `500 J`B. `460 J`C. `300 J`D. `380 J`
Answer» Correct Answer - B In cyclic process ABCA `DeltaU_(cyclic)=0` From first law of thermodynamics `Q_(cyclic)=W_(cyclic)+U_(cyclic)` `:. Q_(cyclic)=W_(cyclic)` `Q_(AB)+Q_(BC)+Q_(CA)=` closes loop area `400+100+Q_(CA)= 1/2(2xx10^(-3))xx4xx10^(4)` `500-Q_(AC)=40 [:. Q_(CA)= -Q_(AC)` `Q_(A)= 4600 J`

Discussion

3516.	`A` cylinder of fixed capacity `67.2` liters contains helium gas at `STP` . Calculate the amount of heat required to rise the temperature of the gas by `15^(@)C` ? `(R=8.314 J mol^(-1)k^(-1))`A. `561.19 J`B. `570.9 J`C. `580.9 J`D. `590.9 J`
Answer» Correct Answer - A We know that `1` mole of an ideal gas at `STP` occupies a volume of `22.4` litres. Thus the cylinder contains `3` moles of helium. Heat required `=nC_(v)DeltaT=3xx(3)/(2)RDeltaT`

Discussion

3517.	`DeltaH^(Theta)` and `DeltaS^(Theta)` for the reaction: `Br_(2)(l) +CI_(2)(g) hArr 2BrCI(g)` at `298K` are `29.3 kJ mol^(-1)` and `104.1 J K^(-1) mol^(-1)`, respectively. Calculate the equilibrium constant for the reaction.
Answer» `DeltaG^(Theta) =DeltaH^(Theta) -T DeltaS^(Theta)` `= 29.3 xx 10^(3) - 298 xx 104.1 =- 1721.8 J mol^(-1)` `log K =- (DeltaG^(@))/(2.303 RT) =- (-1721.8)/(2.303xx8.314xx298) = 0.3018` `K = 2.003`

Discussion

3518.	Calculate `Delta_(r)G^(Theta)` for the following reactions using `Delta_(f)G^(Theta)` values and and predict which reactions are spontaneous. a. `Ca(s) +CI_(2)(g) rarr CaCI_(2)(s)` b. `H_(2)O(s) rarr Hg(l) +(1)/(2)O_(2)(g)` c. `NH_(3)(g) +2O_(2)(g) rarr HNO_(3)(l) +H_(2)O(l)` `Delta_(f)G^(Theta)` value `(kJ mol^(-1))` are: `CaCI_(2)(s) =- 748.1, HgO(s) =- 58.84` `NH_(3)(g) =- 16.45, HNO_(3)(l) = - 80.71`, `H_(2)O(l) =- 237.13`
Answer» `Delta_(r)G^(Theta)`can be calculated by using the relation. `Delta_(r)G^(Theta) = sum Delta_(r)G^(Theta) ("products") -sum Delta_(r)G^(Theta) ("reactants")` a. `Ca(s) +CI_(2) (g) rarr CaCI_(2)(s)` `Delta_(r)G^(Theta) = Delta_(f)G^(Theta) [CaCI_(2)(s)]` `-{Delta_(f)G^(Theta)[Ca(s)] +Delta_(f)G^(Theta) [CI_(2)(g)]}` `=- 748.1 -(0+0) =- 748.1 kJ` Reaction is spontaneous. b. `HgO(s) rarr Hg(l) +(1)/(2)O_(2)(g)` `Delta_(r)G^(Theta) =DeltaG^(Theta) [Hg(l) +(1)/(2)DeltaG^(@) [O_(2)(g)] -Delta_(f)G^(Theta) [HgO(s)]` `=0+0 -(-58.54) = 58.84 kJ` Reaction is not spontaneous c. `NH_(3)(g) +2O_(2)(g) rarr HNO_(3)(l) +H_(2)O(l)` `Delta_(r)G^(Theta) =Delta_(f)G^(Theta) [HNO_(3)(l)] +Delta_(f)G^(Theta) [H_(2)O(l)]` `-{Delta_(f)G^(Theta) [NH_(3)(g)] +2Delta_(f)G^(Theta) [O_(2)(g)]}` `=- 80.71 +(-237.13) -{-16.45+0}` `=- 80.71 -237.13 +16.45 =- 301.39kJ` Reaction is spontanoeus. a. `DeltaG^(Theta) =- 2.303 RT logK`

Discussion

3519.	Given that: i. `C("graphite") +O_(2)(g) rarr CO_(2)(g), DeltaH =- 393.7 kJ` ii. `C("diamond") rarr C("graphite"), DeltaH =- 2.1 kJ` a. Calculate `DeltaH` for buring of diamond of `CO_(2)`. b. Calculate the quantity of graphite that must be burnt to evolve `5000 kJ` of heta.
Answer» a. To calculate `DeltaH` of the reaction `C("diamond") +O_(2)(g) rarr CO_(2)(g), DeltaH = ?` `DeltaH = DeltaH_(1) +DeltaH_(2)` `=- 393.7 - 2.1 =- 395.8 kJ` b. `1`mole of `C`(graphite) i.e. `12 g` on buring evolves heat `= 393.7 kJ` `393.7 kJ` of heat obtained from `- 12 gC` `:. 5000 kJ` of heta is evolved from `= (12 xx 5000)/(393.7)` `= 152.4 g` of graphite

Discussion

3520.	What is the sign of `DeltaG^(Theta)` and the values of `K` for an electrochemical cell for which `E^(Theta)underset(cell)` = 0.80volt`?A. `{:(DeltaG^(Theta),K),(-,gt1):}`B. `{:(DeltaG^(Theta),K),(+,gt1):}`C. `{:(DeltaG^(Theta),K),(+,lt1):}`D. `{:(DeltaG^(Theta),K),(-,lt1):}`
Answer» We know that `DeltaG^(Theta) =- nFE^(Theta)` When `E^(Theta) = 0.80` Then `DeltaG^(Theta) =- ve` and `K = antilog ((-(DeltaG^(@)))/(-2.303RT)) = antilog (+ve)` `:. K gt 1`

Discussion

3521.	The thermochemical equation for solid and liquid rocket fuel are given below: `2AI(s) +1(1)/(2)O_(2)(g) rarr AI_(2)O_(3)(s), DeltaH =- 1667.8 kJ` `H_(2)(g) +(1)/(2)O_(2)(g) rarr H_(2)O(l), DeltaH =- 285.9 kJ` a. If equal masses of aluminium and hydrogen are used, which is a better rocket fuel? b. Determine `DeltaH` for the reaction `AI_(2)O_(3)(s) rarr 2AI(s)+1(1)/(2)O_(2)(g)`
Answer» From the first equation a. `2` moles of `AI (i.e. 2 xx 27g = 54 g)` On combustion given hate `= 1667.8 kJ` `:. 1g of AI` givens `rArr (1667.8)/(54) rArr 39.9 kJ` From the second equation: `1"mole of" H_(2)(=2g)` on combustion gives heat `= 285.9 kJ` `:.1 g of H_(2)` gives `rArr (285.9)/(2) rArr 142.95 kJ` Thus, `H_(2)` is a better rocket fuel. b. Writing the reverse of the first reaction, we have `DeltaH = 1667.8 kJ mol^(-1)`

Discussion

3522.	If `2AI(s) +1(1)/(2) O_(2)(g) rarr AI_(2)O_(3)(s), DeltaH =- 1667.8 kJ` `H_(2)(g) +(1)/(2)O_(2)(g) rarr H_(2)O(l), DeltaH =- 285.9 kJ` Calculate `DeltaH` for the reaction `AI_(2)O_(3)(s) rarr 2AI(s) +1(1)/(2)O_(2)(g)`
Answer» To calculate `DeltaH` of the reaction. `AI_(2)O_(3)(s) rarr 2AI(s) +1(1)/(2)O_(2)(g), DeltaH = ?` `DeltaH = - DeltaH_(1)` `= (-166.78) = 166.78 kJ mol^(-1)`

Discussion

3523.	Statement-1: A reaction which is spontaneous and accompained by decrease of randomness must be exothermic. Statement -2: All exothermic reaction that are accomained by decrease of randomness.A. Statement-1 is true, Statement-2, is true, Statement -2 is a correct explanation for statement-1B. Statement-1 is true, Statement-2, is true, Statement -2 is not a correct explanation for statement-1C. Statement-1 is true, Statement-2 is falseD. Statement-1 is false, Statement-2 is true
Answer» Correct Answer - C

Discussion

3524.	Statement-1: A reaction which is spontaneous and accompained by decrease of randomness must be exothermic. Statement -2: All exothermic reaction that are accomained by decrease of randomness.A. Statement-1 is True, Statement -2 is True, Statement-2 is a correct explanation for Statement-1.B. Statement -1 is True ,Starement -2 is True ,Statement-2 is not a correct explanation for Statement-1C. Statement-1 is True ,Statement-2 is False.D. Statement-1 is False ,Statement-2 is True.
Answer» Correct Answer - c

Discussion

3525.	If theta of reaction for the given acid-base reaction: `HA+NaOHrarrNaA+H_(2)O , Delta H= -4.7 kcal` The heat of dissocitation of HA is ________.
Answer» Correct Answer - 9

Discussion

3526.	If heat of dissocitation of `CHCl_(2)COOH` is 0.7 kcal/mole then `Delta H` for the reaction: `CHCl_(2)COOH+KOH rarrCHCl_(2)COOK+H_(2)O`A. `-13` kcalB. `+13` kcalC. `-14.4` kcalD. `-13.7` kcal
Answer» Correct Answer - A

Discussion

3527.	Let `Q and W` denote the amount of heat given to an ideal gas and the work done by it in an adiabatic process.A. `Q=W`B. `Q!=W`C. `Q=0`D. `W=0`
Answer» Correct Answer - B::C In an adiabatic process `Q=0` but `W ne 0, :. Q neW` So option (b) and (c ) are correct.

Discussion

3528.	In changing the state of a gas adiabatically from an equilibrium state A to another equilibrium state B, an amount of work equal to `22.3J` is done on the system. If the gas is taken from State A to B via a process in which the net heat absorbed by the system is `9.35 cal`., How much is the net work done by the system in the later case? (Take `1 cal.= 4.9J`)
Answer» Here, when the change is adiabatic, `DeltaQ= 0, DeltaW= -22.3J` If `DeltaU` is change in internal energy of the system, then as `DeltaQ= DeltaU+DeltaW` `0= DeltaU-22.3 or DeltaU= 22.3J` In the second case, `DeltaQ= 9.35 cal.= 9.35xx4.2J= 39.3J , DeltaW= ?` As `DeltaU+DeltaW=DeltaQ` `:. DeltaW= DeltaQ-DeltaU= 39.3-22.3= 17.0J`

Discussion

3529.	Two moles of helium gas undergo a cyclic process as shown in Fig. Assuming the gas to be ideal, calculate the following quantities in this process (a) The net change in the heat energy (b) The net work done (c) The net change in internal energy
Answer» Correct Answer - A::B::C::D Number of moles, `n=2` Helium is mono-atomic gas, therefore, `C_(v)=(3)/(2)R , C_(P)=(5)/(2) R` The gas undergoes cyclic process. Since, internal energy is proprty of the system, the net change in internal enrgy during the cyclic process is zero. The net change in the heat energy is equal to the net work done. `(DeltaQ)_(Net)=(DeltaQ)_(AB)+(DeltaQ_(BC))+(DeltaQ_(CD))+(DeltaQ_(DA))` `(DeltaQ)_(AB)=nxxC_(P) xx (T_(B)-T_(A))` `=2xx(5)/(2)xx8.32(400-300)=4160J` Since Process `BC` is isothermal, therefore `DeltaU=0` `(DeltaQ)_(BC)=(DeltaW)_(BC)=nRTln(V_(C)/(V_(B)))=nRTln(P_(B)/(P_(C)))` `=2xx8.32xx400ln ,(2/(1))=4613.6J` ` (DeltaQ_(CD))=nC_(P)(T_(D)-T_(C))` `=2xx(5)/(2)xx8.32xx(300-400)-4160J` `(DeltaQ)_(DA)=nRTln , (P_(D)/(P_(A)))` `=2xx8.32xx300ln, ((1)/(2))=-3460.2J` `(DeltaW)_(Net) =4160+4613.6-4160-3460.2` `=1153.4J`

Discussion

3530.	In a insulated vessel, 0.05 kg steam at 373 K and 0.45 kg of ice at 253 K are mixed. Find the final temperature of the mixture (in kelvin.) Given, `L_(fusion) = 80 cal//g = 336 J//g` `L_("vaporization") = 540 cal//g = 2268 J//g` `s_(ice) = 2100 J//kg.K = 0.5 cal//g.K` and `s_("water") = 4200 J//kg.K = 1cal//g.K` .
Answer» Correct Answer - B::C Heat lost by steam during condensation `=M_(S)L_(V)` `Q_(1)=0.05xx2268xx1000=113400J` Heat lost by water of steam to cool to `273 K` `Q_(2)=0.05xx4200xx100=21,000 J ` `:.` Total heat available from steam `Q=134400J` Heat gained by ice to raise temp from `253k` to `273 k = Q_(1)^(1)=0.45xx2100xx(273-253)=18900 J` Heat gained by ice during metting at `273 k = Q_(2)^(1)` `Q_(1)^(2)=0.45xx336xx9000=151200 J` Total heat required `Q_(1)^(1) to Q_(1)^(2)=Q^(1)=170100 J` `Q^(1)gtQ` `:.` Whole of ice will not melt. Equilibrium temp of mixture `=273 k =0^(@)C`

Discussion

3531.	Two cylinders are connected by a fixed diathermic partition `A` and a removable adiabatic partition `B` is placed adjacent to `A` as shown in the figure. Inititally n moles of an ideal monoatomic gas in present in both the cylinders an normal atmospheric pressure `p_(0)` . Both the gases occupy same volume `V_(0)` , initially. Now the piston of the left cylinder is compressed in adiabatic manner so that volume of the lift portion becomes `(V_(0))/(2)` and then the left piston clamped. Again the adiabatic slider `B` is removed so equilibrium. Assume all other surfaces except `A` to be adiabatic. For this situation, mark out the correct statement (s). A. Just after the removal of adiabatic separator `B` , the pressure in the lift and right chambers are `2^(gamma) p_(0)` and `p_(0)` , respectively.B. `A` fter the removal of adiabatic separtor `B` , the gas in right chamber expands under constant pressure process.C. Workdone by the gas of the right chamber on surroundings during its expansion is `0.22p_(0)V_(0)` .D. During the expansion of gas in right chamber, the energy transferred from the left chamber to right chamber is `0.55 p_(0)V_(0)` where `gamma=(5)/(3)` .
Answer» Correct Answer - A::B::C::D As the gas of the left chamber is compressed adiabatically to half to its initial volume. `p_(0)V_(0)=p_(1)(V_(0)//2)^(gamma) , impliesp_(1)=p_(0)xx2^(gamma)` i.e., pressure in the left chamber after the completion of adiabatic process is `p_(1)` while pressure in the right chamber remains same as it is isolated from surrondings or you say no energy transfer berween the right chamber and the surroundings occur. Due to this adiabatic compression, temperature will also change in the left chamber. `T_(0)V_(0)^(gamma-1)=T_(1)(V_(0)//2)^(gamma-1) , impliesT_(1)=2^(gamma-1)T_(0)` Where `T_(0)=(p_(0)V_(0))/(nR)` When adiabatic separator is removed, temperature on both sides are different, temperature on left is `T_(1)` while temperature on right is `Y_(0)` , so heat transfer takes place from left chamber to right chamber and the gas in cahmber starts expanding, as piston (right side) is free to move, the pressure remains constant as `p_(0)` , So , expansion in right chamber is taking place at constant temperature , During this expansion of gas in right chamber. the volume of gas in left chamber remains constant but temperature and pressure change, For the instant just a fter the removal of adiabatic separator we can take the pessure of gas in left chamber as the same just before the removal i.e., `p_(0)xx2^(gamma)`. Let the final temperature on both the side be `T`, then the energy transferred from left to right is used to incease the temperature of the tight part and to expand the gas. i.e., `nC_(v)(T_(1)-T)=nC_(p)(T-T_(0))` `=nC_(V)(T-T_(0))+W` `impliesT=(5T_(0)+3T_(1))/(8)` `=((5+3xx2^(gamma-1))/(8))T_(0)~=1.22T_(0)` So, the energy transferred is, `Q=nC_(V)(T_(1)-T)=(3nR)/(2)(2^(gamma-1)-1.22)T_(0)` `=0.55 p_(0)V_(0)` Work done, `W=p_(0)(V_(f-V_(i))=nR(T-T_(0))` `=0.22nRT_(0)~=0.22p_(0)V_(0)`

Discussion

3532.	A thermodynamic process is shown in Fig. The pressures and volumes corresponding to some points in the figure are `P_(A) = 3 xx 10^(4) Pa` `V_(A) = 2 xx 10^(-3) m^(3)` `P_(B) = 8 xx 10^(4) Pa` `V_(D) = 5 xx 10^(-3) m^(3)` In the process `AB` `600 J` of heat is added to the system. The change in internal energy of the system in the process `AB` would be A. `560j`B. `800j`C. `600j`D. `640j`
Answer» Correct Answer - A

Discussion

3533.	Adiabatic modulus of elasticity of a gas is `2.1xx10^(5)N//m^(2).` What will be isothermal modulus if elasticity `(C_(p)/C_(v)=1.4)`A. `1.8xx10^(5)N//m^(2)`B. `1.5xx10^(5)N//m^(2)`C. `1.4xx10^(5)N//m^(2)`D. `1.2xx10^(5)N//m^(2)`
Answer» Correct Answer - B

Discussion

3534.	P-V plots for two gases during adiabatic processes are shown in the figure. Plots 1 and 2 should corresponds respectively to A. He and `O_(2)`B. `O_(2)` and HeC. HE and ArD. `O_(2)` and ` N_(2)`
Answer» Correct Answer - B

Discussion

3535.	Four curves `A, B, C` and `D` are drawn in Fig. for a given amount of gas. The curves which represent adiabatic and isothermal changes A. C and D respectivelyB. D and C reslectivelyC. A and B respectivelyD. B and A respectively
Answer» Correct Answer - C

Discussion

3536.	An ideal monoatomic gas indergoes a pressure `pV_(n)`=constant. The adiabatic constant for gas is `y`. During the process, volume of gas increases from `V_(0)` to `rV_(0)` and pressure decreases for `p_(0) ` to `(p_(0))/(2r)` Based on above information, answer the following questios : The molar heat capacity of the gas for the process isA. `(R(n-gamma))/((n-1)(gamma-1))`B. `(R(n-1))/((n-gamma)(gamma-1))`C. `(R)/(gamma-1)`D. `(R)/(n-1)+(R)/(gamma)`
Answer» Correct Answer - A `C=C_(v)+(R)/(1-n)` `=(R)/(gamma-1)-(R)/(n-1)=(R(n-gamma))/((gamma-1)(n-1))`

Discussion

3537.	An ideal monoatomic gas indergoes a pressure `pV_(n)`=constant. The adiabatic constant for gas is `y`. During the process, volume of gas increases from `V_(0)` to `rV_(0)` and pressure decreases for `p_(0) ` to `(p_(0))/(2r)` Based on above information, answer the following questios : The of n isA. `(2log r)/(log2r)`B. `(log2 r)/(3)`C. `(log2 r)/(log r)`D. `(log2 r)/(3log r)`
Answer» Correct Answer - C `Pv^(n)`=constant Differentiating above equation, we get ` pxxnV^(n-1)dV+V^(n)dp=0 , np , dV=-Vdp` `nint_(V_(0))^(rV_(0)) (dV)/(V)=-int_(p_(0))^(p_(0))(dp)/(p)impliesnlnr =-ln ((1)/(2r))` `impliesn=("ln"2r)/("ln"r)`.

Discussion

3538.	An ideal gas is taken through a quasi-static process described by `P = alphaV^(2)`, with `alpha = 5.00 atm//m^(6)`. The gas is expanded to twice its original volume of `1.00 m^(3)`. How much work is done by the gas in expanding gas in this process?
Answer» Correct Answer - 1.18 MJ

Discussion

3539.	The cyclic process for 1 mole of an ideal gas is shown in the V-T diagram. The work done in AB, BC and CA respectively is A. `O,"RT"_(2) "ln"((V_(1))/(V_(2))),R(T_(1) - T_(2))`B. `R(T_(1) -T_(2)), 0, "RT"_(1)"ln"(V_(1))/(V_(2))`C. `0,"RT"_(2)"ln"((V_(2))/(V_(1))),R(T_(1) - T_(2))`D. `0,"RT"_(2)"ln"((V_(2))/(V_(1))),R(T_(2) - T_(1))`
Answer» Correct Answer - D

Discussion

3540.	If `w_(1).w_(2),w_(3)` and `w_(4)` are work done in isothermal, adiabatic, isobaric, and isochoric reversible expansion for an ideal gas, respectively, thenA. `W_(3) gt W_(1) gt W_(2)`B. `W_(3) gt W_(2) gt W_(1)`C. `W_(2) gt W_(1) gt W_(3)`D. `W_(1) gt W_(2) gt W_(3)`
Answer» Correct Answer - A

Discussion

3541.	The temperature -entropy diagram of a reversible engine cycle is given in the figure. Its efficiency is A. `1/3`B. `2/3`C. `1/2`D. `1/4`
Answer» Correct Answer - A

Discussion

3542.	Which type of motion of the molecules is responsible for internal energy of a mono-atomic gas?
Answer» Transnlational motion of molecules.

Discussion

3543.	If on giving 40 J of heat to a system, work done on the system is 10 J, what will be the change in internal energy of the system?
Answer» Internal energy of the system will increase by (40 +10)J = 50 J.

Discussion

3544.	A liece of lead is hammered. Does its internal energy increase? Does the heat enter the lead from outside?
Answer» Yes, internal energy of lead increases. No, heat energy from outside enters the lead.

Discussion

3545.	A piece of metal is hammered. Does its internal energy increase?
Answer» The work done during hammering gets converted into heat energy. Due to this, its internal energy increases.

Discussion

3546.	Refer to figure `DeltaU_(1)` and `DeltaU_(2)` be the changes in internal energy in the system in process `A+B` and `DeltaW` be the net work done by the system in the process `A+B`, A. `DeltaQ-DeltaW=0`B. `DeltaQ+DeltaW=0`C. `DeltaU_(1)-DeltaU_(2)=0`D. `DeltaU_(1)+DeltaU_(2)=0`
Answer» Correct Answer - A::D As the system finally returns to its initial state, change in internal energy must be zero , i.e., `DeltaU=DeltaQ-DeltaW=0 or DeltaU=DeltaU_(1)+DeltaU_(2)=0`

Discussion

3547.	Statement-1: Two gases at same pressure, volume & temperature are compressed to same volume. The first isothermally and second adiabatically, greater work is done on the gas for an adiabatic process. Statement-2: Final temperature for gas going under isothermal process is more than gas going under adiabatic process.A. Statement-1 is true, statement-2 is true, statement-2 is a correct explanation for statement-1B. Statement-1 is true, statement-2 is true, statement-2 is NOT a correct explanation for statement-1C. Statement-1 is true, statement-2 is falseD. Statement-1 is false, statement-2 is true
Answer» Correct Answer - C

Discussion

3548.	A certain mass of nitrogen was compressed `eta = 5.0` times (in terms of volume), first adiabatically, and then isothermally. In both cases the initial state of the gas was the same. Find the ratio of the respective works expended in each compression.
Answer» Let `(p_0,V_0,T_0)` be the initial state of the gas. We know `A_("adia") = (-v R Delta T)/(gamma - 1)` (work done by the gas) But from the equation `TV^(gamma - 1) `= constant, we get `Delta T = T_0 (eta^(gamma - 1) - 1)` Thus `A_("adia") = (-vRT_0(eta^(gamma - 1) - ))/(gamma - 1)` On the other hand, we know `A_(iso) = vRT_0 1 n((1)/(eta)) = -v R T_0 1n eta` (work done by the gas ) Thus `(A_(adia))/(A_(iso)) = (eta^(gamma - 1) - 1)/((gamma - 1)1 n eta) = (5^(0.4) -1)/(0.4 xx 1n 5) = 1.4`.

Discussion

3549.	Find the rate `v` with which helium flows out of a thermally insulated vessel into vacuum through a small hole. The flow rate of the gas inside the vessel is assumed to be negligible under these conditions. The temperature of helium in the vessel is `T = 1.000 K`.
Answer» From energy conservation as in the derivation of Bernoullis theorem it reads `(p)/(rho)+(1)/(2) v^2 + gz + u + Q_d = constant` ….(1) In the Eq.(1) `u` is the internal energy per unit mass and in this case is the thermal energy per unit mass of the gas. As the gas vessel is thermally insulated `Q_d = 0`, also in our case. Just inside the vessel `u = (C_VT)/(M) = (RT)/(M(gamma - 1)) also (p)/(rho) = (RT)/(M)`. inside the vessel `v = 0` also. Just outside `p = 0`, and `u = 0`. Ingeneral `gz` is not very significant for gases. Thus applying Eq. (1) just inside and outside the hole, we get `(1)/(2) v^2 = (p)/(rho) + u` =`(RT)/(M) + (RT)/(M(gamma - 1)) = (gammaa RT)/(M(gamma - 1))` Hence `v^2 = (2 gamma RT)/(M(gamma - 1))` or, `v = sqrt((2 gamma RT)/(M(gamma - 1)))= 3.22 km//s`. The velocity here is the velocity of hydrodynamic flow of the gas into vaccum.This requires that the diameter of the hole is not too small (`D gt` mean free `l`). In the opposite case `(D lt lt l)` the flow is called effusion. Then the above result does not apply and kinetic theory methods are needed.

Discussion

3550.	When `NH_(4)NO_(2)(s)` decomposes at373K , it forms `N_(2)(g)` and `H_(2)O(g)`. The `DeltaH` . For the reaction at one atmospheric pressure and 373 K is -223.6 kJ `mol^(-1)`of `NH_(4)NO_(2)(s)` decomposed. What is the value of `DeltaU` for the reaction under the same conditions ? ( Given `R = 8.31 J K^(-1) mol^(-1))`
Answer» Correct Answer - `-232. 9kJ mol^(-1)` `NH_(4)NO_(2)(s) rarr N_(2)(g) + 2H_(2)O(g) , Delta n_(g) = 3-0 =3` `DeltaU = DeltaH - Deltan_(g) RT = - 223.6 kJ mol^(-1) - ( 3 mol) xx ( 8.314 xx 10^(-3) k J K^(-1) mol^(-1)) xx ( 373K)` `= - 232 kJ mol^(-1)`

Discussion

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