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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.
| 2501. |
When 1 mole of anhydrous `CuSO_(4)` is dissolved in excess of water, -66.4 kJ heat is evolved. When one mole of `CuSO_(4).5H_(2)O` is dissolved in water, the heat change is +11.7 kJ. Calculate enthalpy of hydration of `CuSO_(4)` (anhydrous). |
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Answer» We have to calculate enthalpy of reaction for `CuSO_(4) + 5H_(2)O rarr CuSO_(4).5H_(2)O, Delta H =`? Given that `CuSO_(4) ("anhyd.") + aq rarr CuSO_(4)(aq.), Delta H_(1)` = -66.4 kJ `CuSO_(4).5H_(2)O + aq rarr CuSO_(4)aq., Delta H_(2)` = +11.7 kJ On the basis of two equations `Delta H = Delta H_(1) - Delta H_(2)` = -66.4 kJ - (+11.7 kJ) = -78.1 kJ The enthalpy of hydration of `CuSO_(4)` = -78.41 kJ |
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| 2502. |
Calculate the heat of combustion of eithene `CH_(2) = CH_(2)(g) + 3O_(2)(g) rarr 2CO_(2)(g) + 2H_(2)O(I)` The bond energy data are given below C = C = 619 kJ `mol^(-1)` C - H = 414 kJ `mol^(-1)` O = O = 499 kJ `mol^(-1)` C = O = 724 kJ `mol^(-1)` O - H = 460 kJ `mol^(-1)` |
| Answer» Correct Answer - `-964 kJ" "mol^(-)` | |
| 2503. |
A metal chloride dissolves endothermally and when 7.45 g of its anhydrous form are dissolved in excess of water, the amount of heat absorbed is X kJ. Calculate the enthalpy of solution if molar mass of chloride is 74.5. |
| Answer» Correct Answer - 10X kJ | |
| 2504. |
An ice cube at `0.00^(@)C` is placed is 200g of distilled water at `25^(@)C`. The final temperature after ithe ice is completely metled is `5.00^(@)C`. What is the mass of the ice cube? `deltaH_(fus)=340J.g^(-1), C_(p)=4.18 J.g^(-1). .^(@)C^(-1)`A. `23.6g`B. `46.3g`C. `50.0g`D. `800g` |
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Answer» Correct Answer - B |
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| 2505. |
`DeltaG` is the net energy available to do useful work and is a measure of free energy. If a reaction has positive enthalpy change and positive entropy change, under what conditions will the reaction be spontaneous?A. `DeltaG` will be positive at low temperature hence reaction is spontaneous at low temperature.B. `DeltaG` is negative at high temperature hence reaction is spontaneous at high temperature.C. `DeltaG` is negative at low temperature hence.D. `DeltaG` is negative at all temperature hence reaction is spontaneous at all temperatures. |
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Answer» Correct Answer - B `DeltaG=DeltaH-T DeltaS` For a spontaneous reaction if `DeltaH` is +ve and `DeltaS=+ve,DeltaG` will be negative when `T DeltaS gt DeltaH`. |
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| 2506. |
`C_(2)H_(4) +Cl_(2) rarr C_(2)H_(4)Cl_(2) , Delta H = - 270.6 kJ mol^(-1), Delta S = - 139.0 J K^(-1) mol^(-1)` (i) Is the reaction favoured by entropy , enthalpy, both or none ? (ii) Find `Delta G` if `T = 300 K`. |
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Answer» (i) As `Delta H` is `-` ve , reaction isfavoured by enthalpy. As ` Delta S` is -ve, it is not favoured by entropy. (ii) `Delta G =Delta H - T DeltaS` `= - 270.6 k J mol^(-1) - 300 K ( -139 xx 10^(-3) kJ K^(-1) mol^(-1))` `= -270.6 + 41.7 kJ mol^(-1)` `= - 228.9kJ mol^(-1)` |
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| 2507. |
`C_(2)H_(4) +CI_(2) rarr C_(2)H_(4)CI_(2)` `DeltaH =- 270.6 kJ mol^(-1)K^(-1), DeltaS =- 139 J` a. Is the reaction favoured by entropy, enthalpy both or none? b. Find `DeltaG` if `T = 300K`. |
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Answer» a. Since `DeltaH =- ve`, exothermic process and is favoured, i.e., it will be spontaneous. b. `DeltaG = DeltaH - T DeltaS` `=- 270.6 xx 1000 - 300 xx (-139)` `=- 228900 J =- 228.9 kJ` |
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| 2508. |
Find out the value of equilibrium constant for the following reaction at 298 K. `2NH_(3)(g)+CO_(2)(g)hArrNH_(2)CONH_(2)(aq)+H_(2)O(1)` Standard Gibbs energy change, `Delta_(r)G^(Ө)` at the given temperature is `-13.6 kJ "mol"^(-1)` |
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Answer» We know log, `K=(-Delta_(r)G^(Ө))/(2.303RT)` `=((-13.6xx10^(3)J"mol"^(-1)))/(2.303(8.314JK^(-1)"mol"^(-1))(298K))` `=2.38` Hence K = antilog `2.38 = 2.4xx10^(2)` |
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| 2509. |
The average degree of freedom per molecule for a gas is 6. The gas performs `25J` of work when it expands at constant pressure. Find the heat absorbed by the gas:A. `25J`B. `50J`C. `75J`D. `100J` |
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Answer» Correct Answer - D |
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| 2510. |
At what temperature liquid water will be in equilibrium with water vapour? `DeltaH_("vap")="40.73 kJ mol"^(-1),DeltaS_("vap")="0.109 kJ K"^(-1)"mol"^(-1)`A. 282.4 KB. 373.6 KC. 100 KD. 400 K |
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Answer» Correct Answer - B At equilibrium `DeltaG=0" for "DeltaG=DeltaH-TDeltaS` `"or T"=(DeltaH)/(DeltaS)=(40.73)/(0.109)="373.6 K"` |
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| 2511. |
For a substance with the value of `DeltaH_(vap)` and `DeltaS_(vap)` given below, what is its normal boiling point in `.^(@)C (DeltaH_(vap)=59.0 kJ mol^(-1), DeltaS_(vap)=93.65 Jmol^(-1))`A. 357B. 630C. 1314D. 1587 |
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Answer» Correct Answer - A |
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| 2512. |
Which reactions is spontaneous at all temperature at standard pressure and concentration?A. exothermic reaction with a decrease in entropyB. exothermic reaction with an increase in entropyC. endothermic reaction with a decrease in entropyD. endothermic reaction with a increase in entropy |
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Answer» Correct Answer - B |
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| 2513. |
which of the following is incorrect regarding gibbs free energyA. it is a state functionB. it is extensive propertyC. it is maenoscopic propertyD. it is intensive property |
| Answer» Correct Answer - D | |
| 2514. |
Use the bond energies in the table to estimate `DeltaH`for this reaction . `H_(2)C=CH_(2)+Cl_(2)toClH_(2)C-CH_(2)Cl` `{:overset("Bond energies")((C-C,347KJmol^(-1)),(C=C,612KJmol^(-1)),(c-Cl,331KJmol^(-1)),(C-H,414KJmol^(-1)),(Cl-Cl,243KJmol^(-1))):}`A. `DeltaH=-684KJ`B. `DeltaH=-154KJ`C. `DeltaH=+89KJ`D. `DeltaH=+177KJ` |
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Answer» Correct Answer - b |
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| 2515. |
`C_(2)H_(6)(g) + 3.5O_(2)(g) rarr 2CO_(2)(g) + 3H_(2)O(g)``DeltaS_("vap") (H_(2)O,l) = "x"_(1)calK^(-1)` (boiling point `=T_(1)`) `DeltaH_(f)(H_(2)O,l) = "x"_(2)` `DeltaH_(f)(CO_(2)) = "x"_(3)` `DeltaH_(f)(C_(2)H_(6)) = "x"_(4)` Hence , `DeltaH` for the reaction is-A. `2"x"_(3) + 3"x"_(2)-"x"_(4)`B. `2"x"_(3) + 3"x"_(2)-"x"_(4) + 3"x"_(1)T_(1)`C. `2"x"_(3) + 3"x"_(2)-"x"_(4) - 3"x"_(1)T_(1)`D. `"x"_(1)T_(1) + "X"_(2) + "X"_(3) - "x"_(4)` |
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Answer» Correct Answer - B `H_(2)O(I) rarr H_(2)O(g) DeltaH_("vap") = DeltaS_("vap") T_(B.P) = "x"_(1)T_(1)` `DeltaH_(f)(H_(2)O,g) = DeltaH_(f)(H_(2)O,I) + DeltaH_("vap") = "x"_(2) + "x"_(1)"T"_(1)` `DeltaH_("reaction") = 2DeltaH_(f) (CO_(2),g) + 3DeltaH_(f)(H_(2)O,g) - DeltaH_(f)(C_(2)H_(6),g)` `" "=2"x"_(3) + 3("x"_(2) + "x"_(1)"T"_(1)) - "x"_(4)=2"x"_(3) + 3"x"_(2) + 3"x"_(1)"T"_(1) - "x"_(4).` |
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| 2516. |
A chemical reaction is carried out twice with the same quantity of reactants to forms the same products but the pressure is different for the two experiments. Which value does not change?A. `k_(p)`B. Heat releasedC. `DeltaT_("surroundings")`D. Work done |
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Answer» Correct Answer - A |
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| 2517. |
A 500 gm sample of water is reacted with an equimoplar amount of CaO (both at an initial temperature of `25^(@)C`) . What I sthe final emperature of the product ? [Assume that the poduct absorbs all of the heat released in the reaction heat product per mol of `Ca(OH)_(2)` is 65.2 KJ and specific heat `Ca(OH)_(2)`is `1.2 J//g^(@)C`A. `~~735^(@)C`B. `~~760^(@)C`C. `~~746^(@)C`D. `~~789^(@)C` |
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Answer» Correct Answer - b |
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| 2518. |
For a reaction that is exothermic an non-spontaneous at `25^(@)C`, which quantity must be positive?A. `DeltaE^(@)`B. `DeltaG^(@)`C. `DeltaH^(@)`D. `DeltaS^(@)` |
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Answer» Correct Answer - B |
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| 2519. |
Enthalpy of atomiation of `C_(2)H_(6)(g)and C_(3)h_(8)(g)` are 620 and `880 KJ mol^(-1)` respectivelty. The C-C and C-H bond energies are respectively:A. `80and60KJMol^(-1)`B. `80 and 90KJ mol^(-1)`C. 70and 90KJ `mol^(-1)`D. `100and80 Kjmol^(-1)` |
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Answer» Correct Answer - b |
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| 2520. |
`DeltaS_("surroundings")=+959.1 JK^(-1) mol^(-1)` `DeltaS_("system")=-163.1 JK^(-1) mol^(-1)` Then the process isA. SpontaneousB. Non spontaneousC. At equilibriumD. Cannot be predicted from the information |
| Answer» Correct Answer - A | |
| 2521. |
What is a heat engine ? |
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Answer» A heat engine is a device in which a system is taken through cyclic processes that result in converting part of heat supplied by a hot reservoir into work (mechanical energy) and releasing the remaining part to a cold reservoir. At the end of every cycle involving thermodynamic changes, the system is returned to the initial state. [Note : Automobile engine is a heat engine.] |
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| 2522. |
Molten lead of mass `m = 5.0 g` at a temperature `t_2 = 327^@C` (the melting temperature of lead) was poured into a calorimeter packed with a large amount of ice at a temperature `t_1 = 0 ^@C`. Find the entropy increment of the system lead-ice by the moment the thermal equilibrium is reached. The specific latent heat of melting of lead is equal to `q = 22.5 J//g` and its specific heat capacity is equal to `c = 0.125 J//(g.K)`. |
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Answer» `Delta S = -(m q_1)/(T_2) - mc 1n (T_2)/(T_1) + (M q_(ice))/(T_1)` where `M q_(ice) = m(q_2 + c(T_2 - T_1)` =`mq_2 ((1)/(T_1) -(1)/(T_2)) + mc ((T_2)/(T_1) - 1 -(T_2)/(T_1))` =`0.2245 + 0.2564 ~~ 0.48 J//K`. |
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| 2523. |
Which amount the following is most soluble in water?A. `{:("Compound",Delta_("hydration")(KJ mol^(-1)),Delta H_("lattice")(K)),(A,-400,+500):}`B. `{:("Compound",Delta_("hydration")(KJ mol^(-1)),Delta H_("lattice")(K)),(B,-300,+650):}`C. `{:("Compound",Delta_("hydration")(KJ mol^(-1)),Delta H_("lattice")(K)),(C,-200,+150):}`D. `{:("Compound",Delta_("hydration")(KJ mol^(-1)),Delta H_("lattice")(K)),(D,-100,+250):}` |
| Answer» Correct Answer - C | |
| 2524. |
A hungry man weighing 80 kg take quickly 20 g lunck, and then climbs up a mountain making it to a height of 200 m. If 60% of food energy was wasted as heat and the rest was used as climbing work. The fuel intake could have been any one of the following with given enthalpy of combustion?A. Glucose 16 kJ/gB. Wheat bread 20 kJ/gC. Fructose syrup 13 kg/gD. Olive oil 35 kg/g |
| Answer» Correct Answer - B | |
| 2525. |
`A` gas is found to obey the law `P^(2)V`= constant. The initial temperature and volume are `T_(0)` and `V_(0^(@))` If the gas expands to a volume `3V_(0)`, thenA. final temperature become `sqrt3 T_(0)`B. internal energy of the gas will increaseC. final temperature becomes `(T_(0))/(sqrt3)`D. internal energy of the gas decrease. |
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Answer» Correct Answer - A::B `P^(2)V=` constant `(n^(2)R^(2)T^(2))/(V^(2)).V=` constant `P=(nRT)/(V)implies(T^(2))/(V)=` constant `(T_(o)^(2))/(V^(o))=(T^(2))/(3V_(o))impliesT=sqrt3T_(o)TalphasqrtV` here `V` increases `:. T` increases hence internal energy increases |
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| 2526. |
The figure shows the `P-V` plot of an ideal gas taken through a cycle `ABCDA`. The part `ABC` is a semi-circle and `CDA` is half of an ellipseThen A. The process during the path `ArarrB` is isthermalB. Heat flows out of the gas during then path `BrarrCrarrD`C. Work done during the path `ArarrBrarrC` is zeroD. Positive work is done by the gas in cycle `ABCDA` |
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Answer» Correct Answer - B::D Temperature at `Bgt` temperature at D `:. Delta U` is inegative (for `B rarr C rarr D`) Also `W` is negative (for `B rarr C rarr D`) Tracing is clockwise on `PV` diagram. `:.` Total `W` is positive `:. (B) and(D)` are Correct |
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| 2527. |
Define Thermodynamics. |
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Answer» Science which deals with study of different forms of energy and quantitative relationship is called Thermodynamics. |
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| 2528. |
Define System & Surroundings. |
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Answer» The part of universe for study is called system and remaining portion is surroundings. |
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| 2529. |
Explain how is enthalpy related to spontaneity of a reaction? |
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Answer» Majority of the exothermic reactions are spontaneous because there is decrease in energy. Burning of a substance is spontaneous reaction. C (s) + O2 (g) → CO2 (g); ΔH = -394 kJ mol-1 Neutralization of an acid with a base is a spontaneous reaction. HCl (aq) + NaOH (aq) → NaCl (aq) + H2O; ΔH = -57 kJ mol-1 However, many spontaneous reactions proceed with the absorption of heat. Conversion of liquid water into its vapour by absorption of heat is an endothermic spontaneous change. Therefore, change in enthalpy is not the only criterion for deciding the spontaneity of a reaction. |
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| 2530. |
The standard free energy of a reaction is found to be zero. What is the value of its equilibrium constant? |
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Answer» ΔG = -2.303 RT log K ΔG0 = 0, it means log K = 0 or K = 1 Value of equilibrium constant, K is unity. |
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| 2531. |
In the following equation ΔE = q + W State whether ΔE, q and W are state functions or not. |
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Answer» ΔE is a state function, while q and W are not. |
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| 2532. |
Why does the temperature of a gas decrease, when it is allowed to expand adiabatically? |
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Answer» For an adiabatic change, the first law of thermodynamics may be expressed as dU + PdV = 0 or, −PdV = dU …(i) During expansion, dV is + ve. Therefore the equation (i) will hold, if dU is −ve, i.e., temperature decreases. |
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| 2533. |
A gas does work during adiabatic change. What is the source of mechanical energy to produced? |
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Answer» For an adiabatic change, the first law of thermodynamics may be expressed as dU + PdV = 0 or, −PdV = dU …(i) When a gas does work adiabatically PdV is +ve. For this dU should ne negative. Hence, the source of mechanical energy produced during the adiabatic change is the decrease in internal energy of the gas. |
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| 2534. |
The heat capactiy of liquid water is `75.6 J//mol K`, while the enthalpy of fusion of ice is `6.0 kj//mol`. What is the smallest number of ice cubes at `0^(@)` C, each containing 9.0 g of the of water, needed to cool 500 g of liquid water from `20^(@)`C to `0^(@)`C ?A. 1B. 7C. 14D. None of these |
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Answer» Correct Answer - C |
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| 2535. |
Write the conditions and meaning of the relation q = -W in the light of the first law of thermodynamics. |
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Answer» First law of thermodynamics is stated as ΔE = q + W, if there is no change in the internal energy of the system. ΔE = 0 and the first law of thermodynamics may be written as q = - W .... for ΔE = 0 But - W represents the work done by the system. Therefore, heat absorbed by the system is equal to the work done by the system. The relation q = - W may also be written as W = - q ..... for ΔE = 0 That is, when the energy of the system is kept constant and work is done on the system, then heat must flow from the system to the surroundings. |
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| 2536. |
By applying the first law of thermodynamics to isobaric process. Obtain relation between two specific heats of a gas. |
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Answer» In an isobaric process, pressure remains constant. If an amount of heat dQ is supplied to one mole of a gas at constant pressure and its temperature increases by dT, then dQ = CpdT Here Cp is molar specific heat of the gas at constant pressure. Therefore, for an isobaric process, the first law of thermodynamics becomes : Cpdt = dU + PdV …(i) From perfect gas equation it follows that PdV = RdT In the eqn. (i), substituting PdV and dU, we have CPdT = CvdT+ RdT Cp = Cv+ R. |
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| 2537. |
For a reaction `N_(2)(g)+3H_(2)(g)to2NH_(3)(g),DeltaH=-24` kcal at 700 K and 10 atm pressure, calculate magnitude of change in internal energy if 1.68 kg of `N_(2)(g)` and 0.3 kg of `H_(2)` are mixed and reaction undergoes `60%` completion :A. 21.2 kcalB. 636 kcalC. 1200 kcalD. 1090 kcal |
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Answer» Correct Answer - D |
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| 2538. |
What will be the value of maximum work one by the gas when pressure of 20 gm `H_(2)` is reduced from 20 to 2 atm at constant tempreature of 300 k, assuming gas to behave ideally?A. 57.44 kJB. 114.88 JC. 224.478 kJD. 22.4478 kJ |
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Answer» Correct Answer - A |
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| 2539. |
Out of pressure, mass, moles and volume, which is an intensive property? |
| Answer» Correct Answer - Pressure | |
| 2540. |
The heat of combustion of naphthalene `(C_(10)H_(8)(s))` at constant volume was found to be - 5133 k J `mol^(-1)` . Calculate the value of enthalpy change. |
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Answer» Correct Answer - `-5138kJ mol^(-1)` `C_(10)H_(g)(s) +12O_(2)(g) rarr 10CO_(2)(g)+4H_(2)O(l), DeltaU = - 5133 k J mol^(-1), Deltan_(g) =10-12=-2` `DeltaH = DeltaU + Deltan_(g) RT = - 5132 kJ mol^(-1) + ( - 2 mol) ( 8.314 xx 10^(-3)kJK^(-1) mol^(-1)) ( 298K)` `= -5133kJ mol^(-1) -5 kJ mol^(-1) = - 5138 kJ mol^(-1)` |
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| 2541. |
Whgen the door of a refrigerator is kept open then the room temperature startsA. coll downB. hot upC. first cool down thern hot upD. neither cool down nor hot up |
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Answer» Correct Answer - B Room may become hotter as in that event amout of heat removed would be less than the amount of heat released in the room. |
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| 2542. |
Study the following graph and fill in the blanks. |
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Answer» Correct Answer - c X represents melting of solid into liquid and P being the meling point. Y represents vaporisation of liquid into gas and Q represents boiling point of the liquid. |
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| 2543. |
Consider a heat engine as shown in (figure). `Q_(1) and Q_(2)` are heat added to heat bath `T_(1)` and heat taken from `T_(2)` one cycle of engine. `W` is the mechanical work done on the engine. If `Wgt0`, then possibillities are:A. (i) and (ii)B. (i) and (iii)C. (ii) and (iii)D. (ii) and (iv) |
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Answer» Correct Answer - B Here the given figure reprsents the working of a refrigeratiro `Q_(1)=Q_(2)+W` If ,`Wgt 0`, then `Q_(1)gtQ_(2)gt0` If `Q_(2)` is negative `Q_(1)` is also negative `therefore Q_(2)ltQ_(1)lt0` Hence (iii) is also correct. |
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| 2544. |
Which of the following options consist of only intensive parametres?A. pH of solution, Temperature and volume.B. `Deltap`, Specific heat capacity, Molar internal energy, E.M.F.C. Resistence, Molar mass, Vapour densityD. Density, Mass and temperature. |
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Answer» Correct Answer - B |
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| 2545. |
In an adibatic process, no transfer of heat takes place between system and surroundings. Choose the correct option for free expansion of an ideal gas under adiabatic conditions from the following :A. `q=0,DeltaTne0, w=0`B. `qne0,DeltaT=0, w=0`C. `q=0,DeltaT=0, w=0`D. `q=0,DeltaTlt0, wne0` |
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Answer» Correct Answer - C |
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| 2546. |
Consider a heat engine as shown in (figure). `Q_(1) and Q_(2)` are heat added to heat bath `T_(1)` and heat taken from `T_(2)` one cycle of engine. `W` is the mechanical work done on the engine. If `Wgt0`, then possibillities are:A. `Q_(1) gt Q_(2) gt 0`B. `Q_(2) gt Q_(1) gt 0`C. `Q_(2) lt Q_(1) lt 0`D. `Q_(1) lt 0,Q_(2) gt 0` |
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Answer» Correct Answer - A::C (figure). Represents the working of a refrigerator, wherein `Q_(1)= Q_(2)+W` If `W gt 0, Q_(1) gt 0`. Both `Q_(1)` and `Q_(2)` are positive. If `Q_(2)` is negative, `Q_(1)` is also negative (but less negative as `W gt 0`). `:. Q_(2) lt Q_(1) lt 0` Choices (a) and (c ) are corrcet. |
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| 2547. |
Can the absolute value of internal energy be determined ? Why of why not ? |
| Answer» No, because it is the sum of different types of energies, some of which cannot be determmined accurately. | |
| 2548. |
The heat of combustion of napthalene `{C_(10)H_(8)(s)}` at constant volume was measured to be `-5133 kJ mol^(-1)` at 298 K. Calculate the value of enthalpy change (Given `R=8.314 JK^(-1)mol^(-1)`) |
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Answer» The combustion reaction of napthalene. `C_(10)H_(8)(s)+12O_(2)(g)rarr 10CO_(2)(g)+4H_(2)O(l), " " Delta E=-5133 KJ` `Delta n_(g)=10-12=-2` mol. Now applying the relation. `Delta H = Delta H+Delta n_(g)RT=-5133xx10^(3)+(-2)(8.314)(298)` `=-513000J-4955.14 J=-5137955.14` joule. |
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| 2549. |
In an adibatic process, no transfer of heat takes place between system and surroundings. Choose the correct option for free expansion of an ideal gas under adiabatic conditions from the following :A. `q = 0, Delta T != 0, w=0`B. `q 1= 0, Delta T = 0, w=0`C. `q = 0, Delta T = 0, w=0`D. `q = 0, Delta T lt 0, w != 0` |
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Answer» Correct Answer - C For free expansion w=0 For adiabatic process q=0 From first law of thermodynamics, ltbr. `Delta U = q+w=0` Since there is no change of internal energy, hence temperature will also remain constant, i.e., `Delta T=0` |
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| 2550. |
Calculate the internal energy change in each of the following cases : (i) A system absorbs 5 kJ of heat and does 1 kJ of work. (ii) 5 kJ of work is done on the system and 1 kJ of heat is given out by the system. |
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Answer» (i) Here q + 5 kJ and w = - 1 kJ `therefore` According to first law of thermodynamics, `Delta E=q+w=5+(-1)=4kJ` (ii) Here w = + 5kJ and q = - 1kJ `therefore` According to first law of thermodynamic `Delta E=q+w=-1+(+5)=4 kJ` i.e. the internal energy of the system increases by 4 by in each case. |
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