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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.
| 851. |
The value of enthalpy change `(DeltaH)` for the reaction `C_(2)H_(5)OH (l)+3O_(2) (g) rarr 2CO_(2) (g) +3H_(2)O (l)` at `27^(@)C` is `-1366.5 kJ mol^(-1)`. The value of internal energy change for the above reactio at this temperature will beA. `1369.0` KJB. `-1364.0`KJC. `-1361.5 KJ`D. `-1371.5 ` KJ |
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Answer» Correct Answer - B `C_(2)H_(5)OH(l)+3O_(2)(g) rarr 2CO_(2)(g)+3H_(2)O(l)` `Deltan_(g)=2-3 =-1` `DeltaU = DeltaH - Deltan_(g)RT` `" "=-1366.5 -(-1)xx(8.312)/(10^(3))xx 300` `" =-1366.5 + 0.8314xx3 =- 1364 KJ` |
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| 852. |
`C_(2)H_(5)OH(l)+3O_(2)(g)to2CO_(2)(g)+3H_(2)O(g)` During an experiment 10.00 g of ethanol is complately burned in air to release `CO_(2)(g) and H_(2)O(g)` as shown in the equation above .During the combustion ,296.6 Kj of heat energy is released. What Is the molar enthalpy ofcombution ,`DeltaH_(comb)^(@)`?A. `-2966KJxxmol^(-1)`B. `-1364KJxxmol^(-1)`C. `-64.36Kjxxmol^(-1)`D. `-29.66KJ mol^(-1)` |
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Answer» Correct Answer - b |
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| 853. |
In a thermodynamic process, pressure of a fixed mass of a gas is changed in such a manner that the gas release `20 J` of heat and `8 J` of work is done on the gas. If initial internal energy of the gas was `30 J`, what will be the final internal energy?A. `2J`B. `42 J`C. `18J`D. `58J` |
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Answer» Correct Answer - C `U_(f)-U_(i)=dQ-dW , dQ=(U_(f)-U_(i))+dW` |
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| 854. |
In a thermodynamic process, pressure of a fixed mass of a gas is changed in such a manner that the gas release `20 J` of heat and `8 J` of work is done on the gas. If initial internal energy of the gas was `30 J`, what will be the final internal energy?A. 18jB. 9jC. 4.5jD. 36j |
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Answer» Correct Answer - A |
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| 855. |
Assertion: When a bottle of cold carbonated drink is opened, a slight fog forms around the opening. Reason: Adiabatic expansion of the gas causes lowering of temperature and condersation of water vapours.A. If both assertion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explanation of t he assertion.C. If assertion is true but reason is false.D. If the assertion and reason both are false |
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Answer» Correct Answer - A |
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| 856. |
A Carnot engine whose low temperature reservoir is at `7^(@)C` has an efficiency of `50%`. It is desired to increase the efficiency to `70%` By low many degrees should the temperature of the high temperature reservoir be increasedA. 840 kB. 280 kC. 560 kD. 380 k |
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Answer» Correct Answer - D |
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| 857. |
A mono-atomic gas `X` and an diatomic gas `gamma` both initially at the same temperature and pressure are compressed adiabatically from a volume `V` to `V//2`. Which gas will be at higher temperature?A. `X`B. `Y`C. both are sameD. Cannot be determined |
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Answer» `T_(1)V_(1)^(gamma-1) = T_(2)V_(2)^(gamma -1)` `:. (T_(2))/(T_(1)) = ((V_(1))/(V_(2)))^(gamma-1) = 2^(gamma-1)` Since `gamma` is more for the gas `X`, the temperature will also be more for it. |
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| 858. |
One mole of an ideal gas with `gamma=1.4` is adiabatically compressed so that its temperature rises from `27^(@)C` to `34^(@)C` . The change in the internal energy of the gas is `(R=8.3J mol^(-10k^(-1)))`A. `-166J`B. `166J`C. `168J`D. `-168J` |
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Answer» Correct Answer - B `dU=(nR(T_(1)-T_(2)))/(gamma-1)` |
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| 859. |
One gm mol of a diatomic gas `(gamma=1.4)` is compressed adiabatically so that its temperature rises from `27^(@)C` to `127^(@)C` . The work done will beA. `2077.5` joulesB. `207.5` joulesC. `207.5` ergsD. `205.5` joules |
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Answer» Correct Answer - A `W=(nR(T_(2)-T_(1)))/(gamma-1)` |
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| 860. |
One mole of an ideal gas with `gamma=1.4` is adiabatically compressed so that its temperature rises from `27^(@)C` to `34^(@)C` . The change in the internal energy of the gas is `(R=8.3J mol^(-1)k^(-1))`A. `-166j`B. `166j`C. `-168j`D. `168j` |
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Answer» Correct Answer - B |
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| 861. |
One mole of an ideal gas is allowed to expand reversible and adiabatically from a temperatureof`27^(@)C)`if the work done during the process is`3`kJ,the final temperature will be equal to`(C_(v)=20JK^(-1))`A. ` 100 k `B. ` 450 k `C. ` 150 k `D. `400 k` |
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Answer» Correct Answer - C `W = - n C_(v) (T_(2)-T_(1)) rArr T_(2) = 150 K` |
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| 862. |
An insulated container of gas has two chambers separated by an insulating partition. One of the chambers has volume `V_1` and contains ideal gas at pressure `P_1` and temperature `T_1`.The other chamber has volume `V_2` and contains ideal gas at pressure `P_2` and temperature `T_2`. If the partition is removed without doing any work on the gas, the final equilibrium temperature of the gas in the container will beA. `(T_(1)T_(2) (P_(1)V_(1) + P_(2)V_(2)))/(P_(1) V_(1) T_(2) + P_(2)V_(2)T_(1))`B. `(P_(1) V_(1)T_(1) + P_(2)V_(2)T_(2))/(P_(1)V_(1) +P_(2)V_(2))`C. `(P_(1) V_(1)T_(2) + P_(2)V_(2)T_(1))/(P_(1)V_(1) +P_(2)V_(2))`D. `(T_(1)T_(2) (P_(1)V_(1) + P_(2)V_(2)))/(P_(1) V_(1) T_(1) + P_(2)V_(2)T_(2))` |
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Answer» Correct Answer - A `P_(1)V_(1)T_(1) P_(2), V_(2), T_(2)` As container is insulated `q = 0 ` `q_(1) + q_(2) = 0` `n_(1)C_(m) (T- T_(1)) + n_(2) C_(m) (T - T_(2)) = 0` `T =(n_(1)T_(1)+n_(2)T_(2))/(n_(1)+ n_(2))=((P_(1)V_(1) + P_(2)V_(2)) T_(1)T_(2))/(P_(1)V_(1)T_(2) + P_(2)V_(2)T_(1))` |
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| 863. |
In which of the following process entropy increases? (a) Rusting of iron (b) Vapourisatin of Camphor (c) Crystallisation of sugar from syrup (d) Atomisation of dihydrogenA. a and bB. b and cC. b and dD. Only d |
| Answer» Correct Answer - C | |
| 864. |
If change in internal energy -80 J. The work done by system is +40 J. Calculate the heat change. |
| Answer» Correct Answer - `-40 J` | |
| 865. |
160 J of work is done on the system and at the same time 100 J of heat is given out. What is the change in the internal energy ? |
| Answer» Correct Answer - `+ 60 J` | |
| 866. |
The work done in an ischoric process is |
| Answer» Correct Answer - A | |
| 867. |
Give an example of some familiar process in which heat is added to an object, without changing its temperature. |
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Answer» 1. Melting of ice 2. Boiling of water. |
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| 868. |
An ideal gas goes through a cycle consisting of ischoric adiabatic and isothermal lines. The isothermal process is perform at minimum temperature. If the absolute temperature varies `K` times with tn the cycle then find out its effcincy. A. `1-(ln K)/(l)`B. `1+(ln K)/(l)`C. `(ln K)/(l)`D. `(l)/(ln K)` |
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Answer» Correct Answer - A Process `2rarr3` is adiabatic expansion and `3rarr1` is isothermal compression. From process `2rarr3` , the internal energy continuosly decrease (I.e., temperature continuously decrease) Point `2` is at highest temerature `Q_(g)=Nc_(v)(KT-T)`, `Q_(r)=nRT"In"(V_(3))/(V_(1))` Process `2rarr3(KT)V_(1)^(gamma-1)=TV_(3)^(gamma-1)implies(V_(3))/(V_(1))=(K)^(1//(gamma-1))` `eta=(Q_(g)-Q_(r))/(Q_(8))=1-(Q_(r))/(Q_(g))=1-("In"K)/(l)` |
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| 869. |
Why should a Carnot cycle have two isothermal two adiabatic processes? |
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Answer» With two isothermal and two adiabatic processes, all reversible, the efficiency of the Carnot engine depends only on the temperatures of the hot and cold reservoirs. [Note : This is not so in the Otto cycle and Diesel cycle.] |
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| 870. |
Figure 12.5 shows the P-V diagram of an ideal gas undergoing a change of state from A to B. Four different parts I, II, III and IV as shown in the figure may lead to the same change of state.(a) Change in internal energy is same in IV and III cases, but not in I and II.(b) Change in internal energy is same in all the four cases.(c) Work done is maximum in case I(d) Work done is minimum in case II. |
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Answer» (b) Change in internal energy is same in all the four cases. (c) Work done is maximum in case I |
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| 871. |
Explain why (a) Two bodies at different temperature `T_(1) and T_(2)` if brought in thermal contact do not necessarily settle to the mean temperature `(T_(1)+T_(2))//2` ? (b) The coolant in a chemical or nuclear plant (i.e., the liquid used to prevent different parts of a plant from getting too hot)should have high specific heat. Comment. (c) Air pressure in a car tyre increases during driving . Why? (d) The climate of a harbour town is more temperature (i.e., without extremes of heat and cold) than that of a town in a desert at the same latitude. Why? |
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Answer» When two bodies at different temperatures `T_(1)" and "T_(2)` are brought in thermal contact, heat flows from the body at the higher temperature to the body at the lower temperature till equilibrium is achieved, i.e., the temperatures of both the bodies become equal. The equilibrium temperature is equal to the mean temperature `(T_(1) + T_(2))/2` only when the thermal capacities of both the bodies are equal. The coolant in a chemical or nuclear plant should have a high specific heat. This is because higher the specific heat of the coolant, higher is its heat-absorbing capacity and vice versa. Hence, a liquid having a high specific heat is the best coolant to be used in a nuclear or chemical plant. This would prevent different parts of the plant from getting too hot. When a car is in motion, the air temperature inside the car increases because of the motion of the air molecules. According to Charles’ law, temperature is directly proportional to pressure. Hence, if the temperature inside a tyre increases, then the air pressure in it will also increase. A harbour town has a more temperate climate (i.e., without the extremes of heat or cold) than a town located in a desert at the same latitude. This is because the relative humidity in a harbour town is more than it is in a desert town. |
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| 872. |
`100mL` of a liquid is contained in an insulated container at a pressure of 1bar. The pressure is steeply increased to `100`bar. The volume of the liquid is decreased by `1mL` at this constant pressure. Find `DeltaH` and `DeltaU`.A. 1 L atmB. 5 L atmC. 500 L atmD. 50 L atm |
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Answer» Correct Answer - A |
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| 873. |
Can a system be heated and its temperature remains constant? |
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Answer» If the system does work against the surroundings so that it compensates for the heat supplied, the temperature can remain constant |
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| 874. |
Consider a cycle followed by an engine (Fig. 12.6)1 to 2 is isothermal2 to 3 is adiabatic3 to 1 is adiabaticSuch a process does not exist because(a) heat is completely converted to mechanical energy in such a process, which is not possible.(b) mechanical energy is completely converted to heat in this process,which is not possible.(c) curves representing two adiabatic processes don’t intersect.(d) curves representing an adiabatic process and an isothermal process don’t intersect |
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Answer» (a) heat is completely converted to mechanical energy in such a process, which is not possible. (c) curves representing two adiabatic processes don’t intersect. |
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| 875. |
`100mL` of a liquid is contained in an insulated container at a pressure of 1bar. The pressure is steeply increased to `100`bar. The volume of the liquid is decreased by `1mL` at this constant pressure. Find `DeltaH` and `DeltaU`. |
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Answer» There are two process involved and both are under adiabatic condition. Process first: 1bar, `100mL rarr 100 "bar", 100 mL` Since change in volume `= 0` `:. W = 0` `dU = q +W` `dU = W` (Addiabatic condition `q = 0`) `:. qU = 0` Process second: `100 "bar", 100 mL rarr 100 "bar", 90 mL` `W =- P (V_(2) - V_(1))` `W =- 100 (99 - 100)` `:. W = + 100 mL-atm` Since container is still under adiabatic condition `dU = 100 mL-atm` Therefore, overall `dU = 0 + 100 = 100 mL-atm = 0.1 L-atm` For `DeltaH`: `DeltaH_(1) = DeltaU + PDeltaV + DeltaPV` `= DeltaU + 0 + 100 (99) = 9900 mL-atm` `DeltaH_(2) = DeltaH + P DeltaV + DeltaPV` `= 100 + P(V_(2) - V_(1)) +0 = 100 +100 (-1) = 0 mL-atm` `DeltaH = DeltaH_(1) + DeltaH_(2) = 9900 mL-atm = 9.9 L-atm` |
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| 876. |
For the reaction `N_(2)(g) + 3H_(2)(g) rarr 2NH_(2)(g)` Which of the following is correct?A. `DeltaH = DeltaU`B. `DeltaH gt DeltaU`C. `DeltaH lt DeltaU`D. `DeltaH = 2 DeltaU` |
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Answer» `Deltah = DeltaU + DeltanRT` `Deltan = (2-4) =- 2` `:. DeltaH = DeltaU - 2RT` or `DeltaU = DeltaH +2RT` |
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| 877. |
Consider a heat engine as shown in Fig. 12.7. Q1 and Q2 are heat added to heat bath T1 and heat taken from T2 in one cycle of engine. W is the mechanical work done on the engine.If W > 0, then possibilities are:(a) Q1 > Q2 > 0(b) Q2 > Q1 > 0(c) Q2 < Q1 < 0(d) Q1 < 0, Q2 > 0 |
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Answer» (a) Q1 > Q2 > 0 (c) Q2 < Q1 < 0 |
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| 878. |
The direct conversion of `A` to `B` is difficult, hence it is carried out as `A rarr C rarrD rarr B` Given, `DeltaS_((A^rarrC)) = 50 eU,DeltaS_((C^rarrD))=30eU,DeltaS_((BrarrD))=20 eU`, where `eU` is entropy unit. Thus the change in entropy in `(A rarrB)` is:A. `+ 100 eu`B. `+60 eu`C. `-100eu`D. `-60eu` |
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Answer» Correct Answer - b |
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| 879. |
The direct conversion of `A` to `B` is difficult, hence it is carried out as `A rarr C rarrD rarr B` Given, `DeltaS_((A^rarrC)) = 50 eU,DeltaS_((C^rarrD))=30eU,DeltaS_((BrarrD))=20 eU`, where `eU` is entropy unit. Thus the change in entropy in `(A rarrB)` is:A. `100 eU`B. `60 eU`C. `-100 eU`D. `-60 eU` |
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Answer» Correct Answer - b `DeltaS_(AtoB)=DeltaS_(AtoC)+DeltaS_(CtoD)+DeltaS_(BtoD)` `50+30-20=60 eU` `(DeltaS_(BtoD)=20, :. DeltaS_(DtoB)=-20 eU)` |
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| 880. |
The direct conversion of `A` to `B` is difficult, hence it is carried out as `A rarr C rarrD rarr B` Given, `DeltaS_((A^rarrC)) = 50 eU,DeltaS_((C^rarrD))=30eU,DeltaS_((BrarrD))=20 eU`, where `eU` is entropy unit. Thus the change in entropy in `(A rarrB)` is:A. `+100 eu`B. `+60 eu`C. `-100eu`D. `-60 eu` |
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Answer» `DeltaS_((A rarr B)) = DeltaS_((A rarr C)) +DeltaS_((C rarr D)) +DeltaS_((D rarr B))` `= 50 +30 - 20 = 60 eu` |
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| 881. |
The direct conversion of A to B is difficult. Hence, it is carried out by the following shown path `:` `DeltaS_((A rarr C)) =50 e.u., ` `DeltaS _((C rarr D)) = 30 e.u. ` `DeltaS_((BrarrD)) = 20 e.u.` where e.u. is entropy unit, then `DeltaS _((A rarr B )) ` isA. `+ 100 e.u. `B. ``+60 e.u. `C. `- 100 e.u. `D. ` - 60 e.u. ` |
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Answer» Correct Answer - B `DeltaS _(ArarrC )=S _(C)- S_(A) = 50 e.u.` ....(i) `DeltaS_(CrarrD) = S_(D) -S_(C) = 30e.u.` ....(ii) `DeltaS_(B rarrD) = S_(D) - S_(B) = 20 e.u. ` .....(iii) `DeltaS _(A rarrB)= S_(B) -S_(A) =`Eqn. (i) `+` Eqn. (ii)-Eqn.(iii) `=50 + 30-20= 60` e.u. |
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| 882. |
A thermodynamic system is taken from an initial state I with internal energy `U_i=-100J` to the final state f along two different paths iaf and ibf, as schematically shown in the figure. The work done by the system along the pat af, ib and bf are `W_(af)=200J, W_(ib)=50J and W_(bf)=100J` respectively. The heat supplied to the system along the path iaf, ib and bf are `Q_(iaf), Q_(ib),Q_(bf)` respectively. If the internal energy of the system in the state b is `U_b=200J and Q_(iaf)=500J`, The ratio `(Q_(bf))/(Q_(ib))` is |
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Answer» Correct Answer - B Here, `Q_(iaf)= 500 Jand W_(af)= W_(iaf)= 200J` `:. U_(iaf)= U_(f)-U_(i)= Q_(iaf)-W_(iaf)` `= 500-200= 300J` `:. U_(f)= 300+U_(i)= 300+100= 400 J` Now, `Q_(ib)= U_(ib)+W_(ib)= (U_(b)-U_(i))+W_(ib)` `=(200-100)+50= 150J` and `Q_(bf)=U_(bf)+W_(bf)=(U_(f)-U_(b))+W_(bf)` `=(400-200)+100= 300 J` `:. (Q_(bf))/(Q_(ib))=(300)/(150)=2` |
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| 883. |
Reactions involving gold have been of particular intrests to alchemists. Consider the following reactions , `Au(OH)_(3)+4HCl toHAuCl_(4)+3H_(2)O`, `" "DeltaH=-28kcal` `Au(OH)_(3)+4HBr toHAuBr_(4)+3H_(2)O` `" "DeltaH=-36.8kcal` In an experiment there was an absorption of 0.44 kcal when one mole of `HAuBr_(4)` was mixed with 4 moles of HCl. Then the fraction `HAuBr_(4)` converted into `HAuCl_(4)` : (percentage conversion)A. `5%`B. `6%`C. `7%`D. `8%` |
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Answer» Correct Answer - A |
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| 884. |
When a system is taken from state i to state f along the path iaf, it is found that `Q=50 cal and W=20 cal`. Along the path ibf `Q=36cal`. W along the path ibf is A. `14 cal.`B. `6 cal.`C. `16 cal.`D. `66 cal.` |
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Answer» Correct Answer - B Along the path `iaf`, Increase in internal energy, `U_(1)=Q_(1)-W_(1)= 50-20= 30Cal.` Along the path `ibf` Increase in inernal energy, `U_(2)=Q_(2)-W_(2)` `U_(2)= 36-W` As `U_(2)=U_(1) :. 36-W= 30` `W= 36-30= 6 Cal` |
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| 885. |
When a system is taken from state `i` to state `f` alone the path iaf, it is found that `Q=50` cal and `W=20` cal. Along the path ibf, `Q=36` cal (figure) (a) What is `W` along the path ibf ? (b) If `W=-13` cal for the curved return path `f i`, what is `Q` for this path ? (c) Take `U_(i)=10` cal. What is `U_(f)`? (d) If `U_(b)=22` cal, what is `Q` for the process in and for the process bf ? |
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Answer» Correct Answer - A::B::C::D (a) `DeltaU=Q-W=` same for both paths `W_(ibf)=36-(50-20)=6`cal (b) Q curved `-(-13)=-(50-20)` Q curved `=-43` cal (c) `U_(f)=U_(i)+30=40` cal (d) `Q_(ib)=W_(ib)+DeltaU_(ib)=6+(22-10)=18` cal `Q_(bf)=Q_(if)-Q_(ib)=36-18=18` cal |
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| 886. |
What is the relation between the enthalpy change of the system and heat exchanged with surroundings at constant pressure? |
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Answer» The enthalpy change of the system is equal to the heat exchanged between the system and surroundings at constant pressure that is, ΔH = qp. |
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| 887. |
A monoatomic gas is suddenly compressed to `1//8` of its original volume adiabatically. The pressure of gas will change to :A. `8` timesB. `16` timesC. `32` timesD. `24//5` times |
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Answer» Correct Answer - C `PV^(gamma)`=constant `P_(1)xx(V/8)^(gamma)=constant` `:. (P_(1))/Pxx(1/8)^(gamma)=1` or `P_(1)=P.8^(gamma)` or `=P.(8)^(5//3)=32 P` |
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| 888. |
Assetion : The temperature of a gas does not change when it undergoes on adiabatic process Reason: During adiabatic process , heat energy is exchanged between a system and surroundings.A. If both assetion and reason are true and reaasons is the correct expanation of assetionB. If both assetion and reason are tur but reason is not the correct explanation of assetionC. If assertion is true but reason is falseD. If both assertion and reason are false. |
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Answer» Correct Answer - D Temperature decreases during adiabatic expansion while it increases during adiabtic compression. |
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| 889. |
When ice melts at `1^(@)C` :A. an increase in entropyB. a decrease in enthalpyC. a decrease in free energyD. process is spontaneous |
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Answer» Correct Answer - A::C::D Entropy `uarr` intermolecular force `darr` `DeltaH gt 0` Spontaneous process `DeltaG lt 0` |
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| 890. |
For the reaction : `PCl_(5) (g) rarrPCl_(3) (g) +Cl_(2)(g)`:A. `DeltaH = DeltaU`B. `DeltaH gt DeltaU`C. `Delta lt DeltaU`D. None of the above |
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Answer» Correct Answer - B `DeltaH=DeltaU+Deltan_(g)RT, " "Deltan_(g)=1, " "DeltaH gt DeltaU` |
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| 891. |
36 mL of pure water takes 100 sec to evaporate from a vessel when a heater of 806 watt is used. The `DeltaH_("vaporisation")` of `H_(2)O` is (density of water `=1g//c c)`A. `40.3 kJ//mol`B. `43.2 kJ//mol`C. `4.03 kJ//mol`D. None of these |
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Answer» Correct Answer - A 1 watt = 1 J/sec Total heat supplied for 36 mL `H_(2)O` `" "=806 xx 100` `" "=80600 J` `" "DeltaH_("vap")=(80600)/(36)xx18` `" "=40300" J or 40.3 kJ"//"mol"` |
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| 892. |
Select the correct option(s).A. `Q=nC_(v) dt` is applicable to all substance during heating/cooling at constant volumeB. `gamma=(5)/(3)` for monomatomic ideal gas, at any temperatureC. `dU=nC_(v) dT` is applicable for real gas at constant volumeD. molar heat capacity , pressure and temperature are intensive properties |
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Answer» Correct Answer - a,b,c,d |
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| 893. |
Glucose when dissolved in water leads to cooling of the solution . Suppose you take 250 mL Water at room temperature in an open container (such as a bowl) made of thermally insulated material and dissolve a sponnful of glucose in it .If you are able to accurately measure the heat absorbed by this (assuming negligible changes in the composition and the amount of solution during this process ), you will be measuring :A. the enthalpy of dissolution of the glucose in waterB. the Gibbs free energy of dissolution of the glucose in waterC. the work done by the atmosphere on the system during the dissolution procesD. the heat capacity of the solution |
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Answer» Correct Answer - a |
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| 894. |
If `Delta_(f)H^(@)(C_(2)H_(4))` and `Delta_(f)H^(@)(C_(2)H_(6))` are `x_(1)` and `x_(2)` kcal `mol^(-1)`, then heat of hydrogenation of `C_(2)H_(4)` is :A. `x_(1) + x_(2)`B. `x_(1) - x_(2)`C. `x_(2) - x_(1)`D. `x_(1) + 2x_(2)` |
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Answer» Correct Answer - C `C_(2)H_(4)(g)+H_(2)(g)rarrC_(2)H_(6)(g)` `Delta_(r )H=x_(2)-x_(1)` |
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| 895. |
For the ideal gas , the work of reversible expansion under isothermal conditions can be calculated by using the expression `w=-nRT"In"(v_(f))/(v_(I))` A sample containing 1.0 mol of an ideal gas is expanded isothermally and reversibly to ten times of its original is carried put at 300 K and at 600 K respectvely . Choose the correct option .A. Work done at 600 K is 20 times the work done at 300 K.B. Work done at 300 K is twice the work done at 600 K.C. Work done at 600 K is twice the work done at 300 K.D. `DeltaU=0 ` in both cases. |
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Answer» Correct Answer - c,d |
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| 896. |
The Standard enthalpy of formation of gaseous `H_(2)O` at 298K is `-242 mol^(-1)` . Calculate `DeltaH^(@)"at"373K` given the following values of the molar feat capacities at constant pressure . Molar heat capcity of `H_(2)O(g)=33.5JK^(-1)mol^(-1)` molar heat capacity of `H_(2)(g)=28.8JK^(-1)mol^(-1)` Molar heat capacity of `O_(2)(g)=29.4 JK^(-1)mol^(-1)` {Assume that the heat capacities are independent of temperature.}A. `508KJ mol^(-1)`B. `-242KJ mol^(-1)`C. `-242.75 KJ mol^(-1)`D. none of these |
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Answer» Correct Answer - c |
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| 897. |
The correct statement (s) is/are:A. All system maximze their entropy at equilibriumB. Al substance have zero entropy at absolute zero temperatureC. At constant P,T if `DeltaGlt 0 ` then the process must be work producingD. In all adiabatic processes, entropy of surroundings is constant |
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Answer» Correct Answer - c,d |
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| 898. |
Which of the following statement is /are correct ?A. `DeltaG_(f)^(@) of H_(2)(g)` at 298 K is zeroB. `DeltaG_(f)^(@) of D_(2)(g)` at 298 K is zeroC. `DeltaH_(f)^(@) of H_(2)(g)`at 298 K is zeroD. `S^(@)` of `H^(+)(aq.)` at 298 K is zero |
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Answer» Correct Answer - a,b,c,d |
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| 899. |
The enthalpies of vombustion of formaldehyde and paraformaldehyde (a ploymer of formaldehyde) arae -134 and -732 kcal/mol respectively .The enthalpy of polymerisation per mole of paraformaldehtde is -72 kcal the molecular formula of paraformaldehyde is:A. `CH_(2)O`B. `C_(6)H_(12)O_(6)`C. `c_(12)H_(22)O_(11)`D. `C_(3)H_(6)O_(3)` |
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Answer» Correct Answer - b |
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| 900. |
`1 mol` of `NH_(3)` gas at `27^(@)C` is expanded under adiabatic condition to make volume `8` times `(gamma=1.33)`. Final temperature and work done, respectively, areA. `1500 K, 900 cal`B. `150 K, 400 cal`C. `250 K, 1000 cal`D. `200 K, 800 cal` |
| Answer» Correct Answer - A | |