4131 + Interview Questions in THERMODYNAMICS IN GENERAL AWARENESS Page 78 InterviewSolution

3851.	`A` `0.50` kg ice cube at `-10^(@)C` is placed in `3.0` kg of coffce at `20^(@)C` . What will be the final temperature of mixture? Assume that specifec heat of tea is same as that of water.A. `5.1^(@)C`B. `8^(@)C`C. `10^(@)C`D. `6^(@)C`
Answer» Correct Answer - A In this situation there are three possibilities regarding the final state of the mixture: 1. All ice `2 .A` mixture of ice water at `0_(0)C` 3. All waer Energy release required to brong the `3.0` kg of water at `20^(@)` down to `0^(@)` , `Q_(1)=m_(w)C_(w)(20^(@)C-0^(@)C)` `=(3)(4186)(20)=251kJ` Energy required to change the ice from `-10^(@)C to 0^(@)C` , `Q_(2)=m_(ice)C_(ice)[0^(@)C-(-10^(@)C)]` `=(0.50)(2100)(10)=10.5kJ` Energy required to change the ice to water at `0^(@)C` `Q_(3)=m_(ice)L_(f)=(0.50)(333)=167kJ` `A` total of `10.5kJ+167kJ=177kJ` energy is required to bring ice at `-10^(@)C` to water at `20^(@)C` down to `0^(@)C` is `250kJ` . Thus the mixture must end up all water somewhere between `0^(@)C and 20^(@)C` . Let the final temperature be `T` Now applying energy conservation, `(("Heat to raise"),(0.50 "kg of ice"),("from" -10^(0) C "to"),(0^(0)C))+(("Heat to change"),(0.50 kg),("of ice"),("to water"))+(("Heat to raise"),(0.50 kg),("of water"),("from" 0^(0) C "to" T))=(("Heat lost by"),(3.0 "kg of"),("water colling"),("from" 20^(0) C "to" T))` Hence, `10.5+167+0.50(4184)T=(3)(4186)(20^(@)C-T)` Which on solving yilds `T=5.1^(@)C`

Discussion

3852.	Define spontaneous process.
Answer» A spontaneous process is an irreversible process and may only be reversed by some external agency.

Discussion

3853.	Define non-spontaneres process.
Answer» A process is said to be non-spontaneous if it does not occur of its own under given condition and occur only when an external force is continuously applied.

Discussion

3854.	What is the sign of enthalpy of formation of a highly stable compound?
Answer» Negative is the sign of enthalpy of formation of a highly stable compound.

Discussion

3855.	Two ideal gases under same pressure and temperature are allowed to mix in an isolated system – what will be sign of entropy change?
Answer» Entropy change is positive. It is because disorder or degree of freedom increase on mixing.

Discussion

3856.	Define extensive property/intensive property.
Answer» A property which depends on the quantity of matter present in the system is called extensive property. Examples :Mass, Volume, internal energy, heat, free energy, entropy etc. A property which is independent on the quantity of matter present in the system is called intensive property. Examples :Density, surface tension, viscosity, specific heat, thermal conductivity, refractive index, pressure, temperature, boiling point, freezing point etc.

3856.

Define extensive property/intensive property.

Answer»

A property which depends on the quantity of matter present in the system is called extensive property.

Examples :Mass, Volume, internal energy, heat, free energy, entropy etc. A property which is independent on the quantity of matter present in the system is called intensive property.

Examples :Density, surface tension, viscosity, specific heat, thermal conductivity, refractive index, pressure, temperature, boiling point, freezing point etc.

Discussion

3857.	Predict the sign of ΔS for the following reaction
Answer» ΔS is positive.

Discussion

3858.	Define heat capacity/ molar heat capacity/specific heat capacity.
Answer» Heat capacity (C) is the measurable physical quantity that characterizes the amount of heat required to change the temperature of substance by a given amount. The amount of heat required to raise the temperature of one mole of a substance by one degree Celsius or Kelvin is called molar heat capacity(C_m) The amount of heat required to raise the temperature of unit mass of a substance by one degree Celsius or one Kelvin is called specific heat capacity.

3858.

Define heat capacity/ molar heat capacity/specific heat capacity.

Answer»

Heat capacity (C) is the measurable physical quantity that characterizes the amount of heat required to change the temperature of substance by a given amount.

The amount of heat required to raise the temperature of one mole of a substance by one degree Celsius or Kelvin is called molar heat capacity(C_m)

The amount of heat required to raise the temperature of unit mass of a substance by one degree Celsius or one Kelvin is called specific heat capacity.

Discussion

3859.	`2.8g` of `N_(2)` gas at `300K` and `20atm` was allowed to expand isothermally against a constant external pressure of `1atm`. Calculate `DeltaU, q,` and `W` for the gas.
Answer» `DeltaU = 0, DetaU = q +w or w =- q` `w =- p DeltaV =- p (V_(2)-V_(1))` `V_(2)` and `V_(1)` can be calculated from `V = (nRT)/(p)` `V_(1) = (0.1 xx 0.082 xx 300)/(20) = 0.123L` `V_(2) = (0.1 xx 0.082 xx 300)/(1) = 2.46 L` `w =- (1atm) xx (2.46 - 0.123) L =- 2.337 L atm` or `w =- 2.337 xx 101.3 = 236.7 J`

Discussion

3860.	State the second law of thermodynamics.
Answer» Naturally occurring processes are accompanied by increase in entropy. Hence, entropy of the universe is continuously increasing.

Discussion

3861.	State Hess’s law of heat summation.
Answer» Hess‟s law states that “the enthalpy change is same whether a reaction is carried out in one step or several steps”.

Discussion

3862.	Consider the following statements. (i) Thermodynamics is independent of atomic and molecular structure. (ii) It includes whether a particular reaction is feasible or not under a given set of temperature and concentration of reactants and products. (iii) It can determine the rate at which the reaction take place. Which of the above statements is/are not correct’? (a) (i) only (b) (ii) only (c) (iii) only (d) (i) and (ii) only
Answer» Answer: (c) (iii) only

Discussion

3863.	Find the relatio between volume and temperature of a gas in a process, in which the molar heat capacity C varies with temperature T as `C=C_(V)+alphaT`. [`alpha` is a constant] .
Answer» By first law of thermodymanics, molar specific heat is given as `C=C_(V)+R((dV//V)/(dT//T))` . . . . (i) Compare (i) with given equation `C+C_(V)+alphaT=C_(V)+R((dV//V)/(dT//T))` `impliesR(dV//V)=alphaT((dT)/(T))` `implies(dV)/(V)=(alphadT)/(R)` `implieslog" "V=(alphaT)/(R)+logC` `implieslogV=log_(e)e(alphaT)/(R)+logC` `implieslog" "V=log[C.e^(alphaT//R)]` `impliesVe^(-alphaT//R)=C`.

Discussion

3864.	A system which can exchange energy with the surroundings but not matter is called `"…......................."`
Answer» Correct Answer - closed

Discussion

3865.	A mixture contains `8g He` and `14 g N_(2)` in a vessel at `300K`. How much heat is required to increase the rms speed of these molecules to double their value. Also calculate the final temperatures.
Answer» Mole of `He = (8)/(4) = 2` Mole of `N_(2) = (14)/(28) = 0.5` `KE` of `He` at `300 K = (3)/(2) xx2xx 8.314 xx 300 = 7.4833 xx 10^(3)J` `KE` of `N_(2)` at `300K = (3)/(2) xx 0.5 xx 8.314 xx 300 = 1.873 xx 10^(3)J` Now, since `KE = (1)/(2) M (U_(rms))^(2)` On doubling `U_(rms)`, kinetic enegry will becomes four times of initial value. `:.` New `KE of He = 4 xx 7.483 xx 10^(3) J = 2.99 xx 10^(4)J` New `KE of N_(2) = 4 xx 1.871 xx 10^(3)J = 7.484 xx 10^(3)J` Thus, for `He, 4 xx 7.483 xx 10^(3) = (3)/(2)RT xxn (T` is new temperature) ltbRgt `= (3)/(2) xx2xx8.314 xxT` `T =1200K` Now, heat given at constant volume to gases to heat them from `300K` to `1200K` is `DeltaH_(He) = n xx C_(V) xxDeltaT` `= nxx (3)/(2) R xx DeltaT` `= 2 xx(3)/(2) xx 8.314 xx (1200 - 300) [C_(V)` for `He = (3)/(2)R]` `= 22447.8 J` `DeltaH_(N_(2)) = 0.5 xx (5)/(2) xx 8.314 xx (1200 - 300) = 9353.25J` `[C_(V)` for `N_(2) = (5)/(2)R]` `:.` Heat provded `= 22447.8 + 9353.25 = 31801.05J = 3.18 xx 10^(4)J`

Discussion

3866.	What is a spontaneous process/entropy?
Answer» A process which occurs on its own with or without proper initiation under the given set of conditions is called spontaneous process. Entropy is a measure of disorder.

Discussion

3867.	At what temperature entropy of a substance is zero?
Answer» At absolute zero (0 K).

Discussion

3868.	One mole of an ideal gas at `25^(@)C` is subjected to expand reversible ten times of its intial volume. The change in entropy of expansion is
Answer» We have, `DeltaS = (q_(1))/(T) = (2.303nRT"log"_(10)(V_(2))/(V_(1)))/(T)` Given, `n=1, R = 8.314J, T = 298K, V_(1) = V, V_(2) = 10V` `:. DeltaS = 2.303 xx1 xx 8.314 "log"_(10)(10V)/(V)` `= 19.15 J K^(-1) mol^(-1)`

Discussion

3869.	Why does entropy of a solid increase on fusion?
Answer» On fusion, more ordered solid form of the substance changes to less ordered liquid form. So, entropy (or randomness) increases.

Discussion

3870.	If `K lt1.0`, what will be the value of `DeltaG^(@)` out of the following `?`A. `1.0`B. ZeroC. NegativeD. Positive
Answer» Correct Answer - d `DeltaG^(@) -RT ln K . ` When `K lt 1 , DeltaG^(@) = + ve`

Discussion

3871.	At very low temperatures, heat capacity of a solid is proporional to `T^(3)` and can be written as: `C_(P) = alphaT^(3)` where `alpha = 3xx10^(8)J mol^(-1) K^(-1)`. What is the change in enthalpy when a solid is heated from `0K` to `300K`?
Answer» Using `DeltaH = int_(T_(1))^(T_(2)) C_(P) dT = int_(T_(1))^(T_(2)) alpha T^(3) dT` `= (alpha)/(4) [T_(2)^(4) - T_(1)^(4)]` `= (alpha)/(4) [(300)^(4) - 0] = 60.75 J mol^(-1)`

Discussion

3872.	A sample of `10g H_(2)O` is slowely heated from `27^(@)C to 87^(@)C`. Calculate the change in entropy during heating. (specific heat of water `=4200 J kg^(-1)K^(-1))`.
Answer» `DeltaS = 2.303 xx m xx C_(P) xx "log" (T_(2))/(T_(1))` `= 2.303 xx (10)/(1000) xx 4200"log" (360)/(300) = 7.65J`

Discussion

3873.	Why does entropy of a solid increase on fusion?
Answer» In a solid, the constituent particlesarefixed . On melting, they fall apart and are free to move,i.e., their randomness increases.

Discussion

3874.	The entropy change during heating of 50 g water (sp. Heat `=4200 J kg^(-1) K^(-1)`) from `20^(@)C` to `40^(@)C` will beA. `13.83 JK^(-1)`B. `-13.83 JK^(-1)`C. `27.66 JK^(-1)`D. `55.32 JK^(-1)`
Answer» Correct Answer - A `DeltaS=2.303 nC_(p)"log" T_(2)/T_(1)`

Discussion

3875.	The standard free energy change, `Delta G^(@)` is related to equilibrium constant `K_(p)` asA. `K_(p) = - RT ln Delta G^(@)`B. `K= ((e)/( RT))^(DeltaG^(@))`C. `K_(p) = (DeltaG^(@))/( RT)`D. `K_(p) = e^(- DeltaG^(@) //RT)`
Answer» Correct Answer - d `DeltaG^(@) = - RT lnK_(p) ` or `ln K_(p)=- DeltaG^(@) // RT ` or `K_(p) = e^(-DeltaG^(@) //RT )`

Discussion

3876.	Assertion (A): Enthalpy of graphite is lower than that of diamond. Reason (R ) : Entropy of graphite is lower than that of diamond.A. If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion.B. If both Assertion & Reason are True but Reason is not a correct explanation of the Assertion.C. If Assertion is True but the Reason is False.D. If both Assertion & Reason are false.
Answer» Correct Answer - B

Discussion

3877.	A process is taking place at constant temperature and pressure. ThenA. `Delta H = Delta E`B. `Delta H = T DeltaS`C. `Delta H = 0`D. `Delta S = 0`
Answer» Correct Answer - b At const.T & P,process is in equilibrium . `DeltaG=0`.Hence,`DeltaH = T DeltaS`

Discussion

3878.	Entropy of diamond is less than that of graphite.What conclusion do you draw from this ?
Answer» Less entropy of diamond meansless disorder,i.e., allthe C-atoms are linked to form a network structure. Greater entropy of graphite implies some disorder which is due to presence of free electrons and slipping of layers over each other.

Discussion

3879.	In an irreversible process taking place at constant T and P and in which only pressure volume work is being done, the change in Gibbs free energy (dG) and the change in entropy ( dS), satisfy the criterisA. `( dS) _(V,E) gt 0 , (dG) _(T,P) lt 0`B. `( dS) _(V,E) = 0 , (dG) _(T,P) = 0`C. `( dS) _(V,E) = 0 , (dG) _(T,P) gt 0`D. `( dS) _(V,E) lt 0 , (dG) _(T,P) lt 0`
Answer» Correct Answer - a For irrev., i.e.,spontaneous process , `d S gt 0, dG lt0`.

Discussion

3880.	In an irreversible process taking place at constant `T` and `P` and in which only pressure-volume work is being done, the change in Gibbs free energy `(dG)` and the change in entropy `(dS)` satisfy the criteriaA. `(dS)_(V,U)=0,(dG)_(T,P)=0`B. `(dS)_(V,U)=0,(dG)_(T,P)=+ve`C. `(dS)_(V,U)=-ve,(dG)_(T,P)=-ve`D. `(dS)_(V,U)=+ve,(dG)_(T,P)=-ve`
Answer» Correct Answer - d For a spontaneous change (work is beign done by the system) `(dG)_(T,P)=-ve` and `(dS)_(V,U)=+ve`

Discussion

3881.	In an irreversible process taking place at constant `T` and `P` and in which only pressure-volume work is being done, the change in Gibbs free energy `(dG)` and the change in entropy `(dS)` satisfy the criteriaA. `(dS)_(V , E) lt 0, (dG)_(T,P) lt 0`B. `(dS)_(V , E) = 0, (dG)_(T,P) gt 0`C. `(dS)_(V , E) = 0, (dG)_(T,P) = 0`D. `(dS)_(V , E) gt 0, (dG)_(T,P) lt 0`
Answer» Correct Answer - D According to the second law of thermodynamics, for an irreversible process (spontaneous process), `Delta S_(total) gt 0`. Since `- T Delta S_(Total) = Delta G`, we have `Delta F lt o` for a spontaneous process. Thus, `(dS)_(V, E) gt 0` and `(dG)_(T,P) lt 0`

Discussion

3882.	The standard molar enthalpies of formation of `IF_(3)(g)" and " IF_(5)(g)` are -470 kJ and -847 kJ, respectively. Valence shell electron-pair repulsion theory predicts that `IF_(3)(g)` is square pyramidal in shape in which all I-F bonds are equivalent while `IF_(3)(g)` is T-shaped (based on trigonal-bipyramial geometry) in which I-F bonds are of different lengths. It is observed that the axial I-F bonds in `IF_(3)` are equivalent to the I-F bonds in `IF_(5)`. Calculate the equitorial I-F bond strength `("in" "kJ"//"mol")` in `IF_(3)`. Some other informations given are : `I_(2)(s)toI_(2)(g)," "Delta H=62 kJ` `F_(2)(g)to 2F(g)," "Delta H =155 kJ` `I_(2)(g)to 2I(g)," "Delta H =149 kJ`
Answer» Correct Answer - 272

Discussion

3883.	One mole of an ideal monoatomic gas is caused to go through the cycle shown in figure. Then, the change in the internal energy is expanding the gas from a to `c` along the path `abc` is:A. `3P_(0)V_(0)`B. `6 RT_(0)`C. `4.5 RT_(0)`D. `10.5 RT_(0)`
Answer» Correct Answer - D `Pv= nRT` at point `C` `2P_(0)xx4V_(0)=1xxRT_(c)` `T_(c )=[(8P_(0)V_(0))/(R )]` at point a `P_(0)V_(0)=1xxRT_(0)` `T_(0)=(P_(0)V_(0))/(R ) , T_(c )=8T_(0)` change in internal energy `=[ nC_(v)dT]` For path `a` to `b=1xx(3)/(2)Rxx[3T_(0))=(9)/(2)RT_(0)` For path `b` to `c=1xx(3)/(2)Rxx[4T_(0)]=6T_(0)R` Total change `=(9)/(2)RT_(0)+6RT_(0)=(21RT_(0))/(2)=10.5 RT_(0)` So total change in internal energy `Delta U =10.5 RT_(0)`

Discussion

3884.	Ethalpy for the reaction `Ag^(+)(aq)+Br^(-)(aq)toAgBr(s) " is" -90 kJ`. Magnitude of enthalpy of formation of `Ag^(+)(aq) " and " Br^(-)(aq)` in the ratio 5:6 Formation of `Ag^(+)(aq)` is an endothermic process whereas formation of `Br^(-)` is an exothermic process. Enthalpy of formation of AgBr is `-110 "kJ"//"mole"`. Calculate the enthalpy of formation of `Ag^(+)(aq) " in" " kJ"//"mol"`.
Answer» Correct Answer - 100

Discussion

3885.	The ethalpy of formation of methane at constant pressure and `300 K` is `-75.83 kJ`. What will be the heat of formation at constant volume? `[R = 8.3 J K^(-1) mol^(-1)]`
Answer» The equation for the formation of methane is `{:(C(s)+,2H_(2)(g)rarr,CH_(4)(g),DeltaH^(Theta) =- 67.83kJ,,),(,2mol,1mol,,):}` `Deltan = (1-2) =- 1` Given `DeltaH^(Theta) =- 75.83 kJ R = 8.3 xx 10^(-3) kJ K^(-1) mol^(-1), T = 300 K` Applying `DeltaH^(Theta) = DeltaU^(Theta) + DeltanRT` `- 75.83 = DeltaU^(Theta) +(-1) (8.3 xx 10^(-3)) (300)` So, `DeltaU^(Theta) =- 75.83 + 2.49 =- 73.34 kJ`

Discussion

3886.	Consider the following liquid -vapour equilibrium Liquid `hArr`Vapour Which of the following relations is correct.A. `(dlnP)/( dT^(2))= - ( DeltaH _(v))/( T^(2))`B. `(dlnP)/( dT)= ( DeltaH _(v))/( RT^(2))`C. `(dlnS)/( dT^(2))= ( DeltaH _(v))/( RT^(2))`D. `(dlnS)/( dT)= - ( DeltaH _(v))/( RT)`
Answer» Correct Answer - b It represents Clausius-Clapeyron euation.

Discussion

3887.	Consider the following liquid-vapour equilibrium. `"Liquid" hArr "Vapour"` Which of the following relations is correct?A. `("dInP")/(dT)=(-DeltaH_(v))/(RT)`B. `("dInP")/(dT^(2))=(-DeltaH_(v))/(T^(2))`C. `("dInP")/(dT)=(-DeltaH_(v))/(RT^(2))`D. `("dInG")/(dT^(2))=-(DeltaH_(v))/(RT^(2))`
Answer» Correct Answer - C The given phase equilibria is Liquid `hArr` Vapour This equilibrium states that, when liquid is heated, it converts into vapour , which is derived by Clausius Clapeyron and the relationship is written as, `(d"In" p)/(dT)=-(DeltaH_(v))/(RT^(2))` Where, `DeltaH_(v)` =Heat of vaporisation

Discussion

3888.	The correct thermodnamic conditions for the spontaneous reaction at all temperature is:A. `DeltaH gt 0 and DeltaS lt 0`B. `DeltaH lt 0 and DeltaS gt 0`C. `DeltaH lt 0 and DeltaS lt 0`D. `DeltaH lt 0 and DeltaS = 0`
Answer» Correct Answer - B::D We have the Gibbs Helmholtz reaction for spontaneity as `DeltaG=DeltaH-TDeltaS` For reaction to be spontaneous , `DeltaG` must be negative . For this , `DeltaH` should be negative and `DeltaS` should be positive. `therefore DeltaH lt 0 and DeltaS gt 0` and also `DeltaS=0` shows `DeltaG` a negative quantity.

Discussion

3889.	For a particular reverisble reaction at temperature T,`DeltaH` and `DeltaS` were found to be both `=ve`. If `T_(e)` is the temperature at equilibrium, the reaction would be spontaneous whenA. `T_(e)` is 5 times TB. `T = T_(e)`C. `T_(e) gt T`D. `T gt T_(e)`
Answer» Correct Answer - D `DeltaG =DeltaH - T DeltaS`. At equilibrium, `DeltaG = 0`. Hence, `T_(e)DeltaS = DeltaH`. As `DeltaH `and `DeltaS` are`+ve` , for reaction to be spontaneous `DeltaG` should be `-ve`. This can be so only if `T Delta S gt DeltaH `, i.e., `T Delta S gt T_(e) DeltaS ` or `T gt T_(e)`

Discussion

3890.	StatementI: Work done by a gas in isothermal expansion is more than the work done by the gas in the same expansion adaibatically. Statement II: Temperature remains constant in isothermal ecpansion but not in adiabatic expansion.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
Answer» Correct Answer - B

Discussion

3891.	A popular game is 'splatball', where people use pressurized CO2 cartridges to shoot paint-filled plastic balls at targets and at each other. A typical splatball gun takes a 1.00 L `CO_(2)` catridge filled to 300 PSI pressure. How many work is done as this cartridge is discharged (to atmospheric pressure) when the gun is fired ? (14.7 PSI = 1 atm)A. `-20.41 J`B. `+19.41 J`C. `-1970 J`D. `-2070 J`
Answer» Correct Answer - C `w = - P_(ext) Delta V = - (20.4 - 1.0) = - 19.4 L` atm `= -19.4 xx 101.4 = - 1967 J`

Discussion

3892.	The state of a gas can be described by quoting the relationship between `"…................"`.A. pressure, volume, temperatureB. temperature,amount,pressureC. amount, volume, temperatureD. pressure, volume, temperature, amount
Answer» Correct Answer - d `PV= nRT`. Thus , P,V,T and n describe the state of the system.

Discussion

3893.	The state of gas can be described by quoting the relationship betweenA. pressure, volume, temperatureB. temperature, amount, pressureC. amount, volume, temperatureD. pressure, volume, temperature, amount
Answer» Correct Answer - D

Discussion

3894.	Which is the correct relationship between `DeltaG^(@)` and equilibrium constant `K_(p)`?A. `K_(p)=-RT log DeltaG^(@)`B. `K_(p)=[E//RT]^(DeltaG^(@))`C. `K_(p)=-DeltaG^(@)//RT`D. `K_(p)=e^(-DeltaG^(@)//RT)`
Answer» Correct Answer - D

Discussion

3895.	Which is not correct relationship between `DeltaG^(Theta)` and equilibrium constant `K_(P)`.A. `K_(P) =- RT log DeltaG^(Theta)`B. `K_(P) =[e//RT]^(DeltaG^(Theta))`C. `K_(P) =- (DeltaG^(Theta))/(RT)`D. `K_(P) =e^(-DeltaG^(Theta)//RT)`
Answer» `DeltaG^(Theta) =- RT In K_(p) or K_(p) = e^(-DeltaG^(Theta)//RT)`

Discussion

3896.	Which is not correct relationship?A. `[(dH)/(dT)]_(P) - [(dU)/(dT)]_(V) = (+ve)`B. `[(dU)/(dV)]_(T) = 0`(for ideal gas)C. `[(dV)/(dT)]_(P) =(nR)/(P) =` (for ideal gas)D. All of these
Answer» `((delH)/(delT))_(P)-((delU)/(delT))_(V)=R`

Discussion

3897.	A gas expands from `3 dm^(3)` to `5dm^(3)` anainst a constant pressure of `3atm`. The work done during the expansion if used to heat `10mol` of water at temperature `290K`. Find the final temperature of water, if the specific heat of water `= 4.18g^(-1)K^(-1)`.
Answer» Since work is done against constant `P` and thus, irreversible `DeltaV=5-3=2 dm^(3)=2 litre, P=3 atm` `:. W=-P.DeltaV=-3xx2 litre atm` `=(6xx4.184xx1.987)/0.0821=-607.57` joule Now, this work is used up in heating water `:. W=nxxCxxDeltaT` `607.57=10xx4.184xx18xxDeltaT` `:. DeltaT=0.81` `:.` Final temperature `=T_(1)+DeltaT=290+0.81` `290.81 K`

Discussion

3898.	A gas expands from `3 dm^(3)` to `5dm^(3)` anainst a constant pressure of `3atm`. The work done during the expansion if used ti heat `10mol` of water at temperature `290K`. Find the final temperature of water, if the specific heat of water `= 4.18g^(-1)K^(-1)`.
Answer» Work done `P DeltaV = (3 xx2) L-atm` `DeltaV = 5- 3 = 2=3xx2xx101.3J = 607.8J` Heat lost `=` Heat gained by `H_(2)O (ms Deltat)` `Deltat = (607.8)/(10xx18xx4xx4.18) = 0.808K` `rArr T_(f) = 290 +0.808 = 290.808K`

Discussion

3899.	If a refrigerator’s door is kept open, will the room become cool or hot? Explain.
Answer» Here heat removed is less than the heat supplied and hence the room, including the refrigerator (which is not insulated from the room) becomes hotter.

Discussion

3900.	If a refrigerator’s door is kept open, will the room become cool or hot? Explain.
Answer» The room will becomes hot. Amount of heat removed is less than the heat supplied and hence the room, including the refrigerator (which is not insulated from the room) becomes hotter.

Discussion

Explore topic-wise InterviewSolutions in .

`A` `0.50` kg ice cube at `-10^(@)C` is placed in `3.0` kg of coffce at `20^(@)C` . What will be the final temperature of mixture? Assume that specifec heat of tea is same as that of water.A. `5.1^(@)C`B. `8^(@)C`C. `10^(@)C`D. `6^(@)C`

Define spontaneous process.

Define non-spontaneres process.

What is the sign of enthalpy of formation of a highly stable compound?

Two ideal gases under same pressure and temperature are allowed to mix in an isolated system – what will be sign of entropy change?

Define extensive property/intensive property.

Predict the sign of ΔS for the following reaction

Define heat capacity/ molar heat capacity/specific heat capacity.

`2.8g` of `N_(2)` gas at `300K` and `20atm` was allowed to expand isothermally against a constant external pressure of `1atm`. Calculate `DeltaU, q,` and `W` for the gas.

State the second law of thermodynamics.

State Hess’s law of heat summation.

Find the relatio between volume and temperature of a gas in a process, in which the molar heat capacity C varies with temperature T as `C=C_(V)+alphaT`. [`alpha` is a constant] .

A system which can exchange energy with the surroundings but not matter is called `"…......................."`

A mixture contains `8g He` and `14 g N_(2)` in a vessel at `300K`. How much heat is required to increase the rms speed of these molecules to double their value. Also calculate the final temperatures.

What is a spontaneous process/entropy?

At what temperature entropy of a substance is zero?

One mole of an ideal gas at `25^(@)C` is subjected to expand reversible ten times of its intial volume. The change in entropy of expansion is

Why does entropy of a solid increase on fusion?

If `K lt1.0`, what will be the value of `DeltaG^(@)` out of the following `?`A. `1.0`B. ZeroC. NegativeD. Positive

At very low temperatures, heat capacity of a solid is proporional to `T^(3)` and can be written as: `C_(P) = alphaT^(3)` where `alpha = 3xx10^(8)J mol^(-1) K^(-1)`. What is the change in enthalpy when a solid is heated from `0K` to `300K`?

A sample of `10g H_(2)O` is slowely heated from `27^(@)C to 87^(@)C`. Calculate the change in entropy during heating. (specific heat of water `=4200 J kg^(-1)K^(-1))`.

Why does entropy of a solid increase on fusion?

The entropy change during heating of 50 g water (sp. Heat `=4200 J kg^(-1) K^(-1)`) from `20^(@)C` to `40^(@)C` will beA. `13.83 JK^(-1)`B. `-13.83 JK^(-1)`C. `27.66 JK^(-1)`D. `55.32 JK^(-1)`

The standard free energy change, `Delta G^(@)` is related to equilibrium constant `K_(p)` asA. `K_(p) = - RT ln Delta G^(@)`B. `K= ((e)/( RT))^(DeltaG^(@))`C. `K_(p) = (DeltaG^(@))/( RT)`D. `K_(p) = e^(- DeltaG^(@) //RT)`

A process is taking place at constant temperature and pressure. ThenA. `Delta H = Delta E`B. `Delta H = T DeltaS`C. `Delta H = 0`D. `Delta S = 0`

Entropy of diamond is less than that of graphite.What conclusion do you draw from this ?

One mole of an ideal monoatomic gas is caused to go through the cycle shown in figure. Then, the change in the internal energy is expanding the gas from a to `c` along the path `abc` is:A. `3P_(0)V_(0)`B. `6 RT_(0)`C. `4.5 RT_(0)`D. `10.5 RT_(0)`

The ethalpy of formation of methane at constant pressure and `300 K` is `-75.83 kJ`. What will be the heat of formation at constant volume? `[R = 8.3 J K^(-1) mol^(-1)]`

Consider the following liquid -vapour equilibrium Liquid `hArr`Vapour Which of the following relations is correct.A. `(dlnP)/( dT^(2))= - ( DeltaH _(v))/( T^(2))`B. `(dlnP)/( dT)= ( DeltaH _(v))/( RT^(2))`C. `(dlnS)/( dT^(2))= ( DeltaH _(v))/( RT^(2))`D. `(dlnS)/( dT)= - ( DeltaH _(v))/( RT)`

Consider the following liquid-vapour equilibrium. `"Liquid" hArr "Vapour"` Which of the following relations is correct?A. `("dInP")/(dT)=(-DeltaH_(v))/(RT)`B. `("dInP")/(dT^(2))=(-DeltaH_(v))/(T^(2))`C. `("dInP")/(dT)=(-DeltaH_(v))/(RT^(2))`D. `("dInG")/(dT^(2))=-(DeltaH_(v))/(RT^(2))`

The correct thermodnamic conditions for the spontaneous reaction at all temperature is:A. `DeltaH gt 0 and DeltaS lt 0`B. `DeltaH lt 0 and DeltaS gt 0`C. `DeltaH lt 0 and DeltaS lt 0`D. `DeltaH lt 0 and DeltaS = 0`

For a particular reverisble reaction at temperature T,`DeltaH` and `DeltaS` were found to be both `=ve`. If `T_(e)` is the temperature at equilibrium, the reaction would be spontaneous whenA. `T_(e)` is 5 times TB. `T = T_(e)`C. `T_(e) gt T`D. `T gt T_(e)`

The state of a gas can be described by quoting the relationship between `"…................"`.A. pressure, volume, temperatureB. temperature,amount,pressureC. amount, volume, temperatureD. pressure, volume, temperature, amount

The state of gas can be described by quoting the relationship betweenA. pressure, volume, temperatureB. temperature, amount, pressureC. amount, volume, temperatureD. pressure, volume, temperature, amount

Which is the correct relationship between `DeltaG^(@)` and equilibrium constant `K_(p)`?A. `K_(p)=-RT log DeltaG^(@)`B. `K_(p)=[E//RT]^(DeltaG^(@))`C. `K_(p)=-DeltaG^(@)//RT`D. `K_(p)=e^(-DeltaG^(@)//RT)`

Which is not correct relationship between `DeltaG^(Theta)` and equilibrium constant `K_(P)`.A. `K_(P) =- RT log DeltaG^(Theta)`B. `K_(P) =[e//RT]^(DeltaG^(Theta))`C. `K_(P) =- (DeltaG^(Theta))/(RT)`D. `K_(P) =e^(-DeltaG^(Theta)//RT)`

Which is not correct relationship?A. `[(dH)/(dT)]_(P) - [(dU)/(dT)]_(V) = (+ve)`B. `[(dU)/(dV)]_(T) = 0`(for ideal gas)C. `[(dV)/(dT)]_(P) =(nR)/(P) =` (for ideal gas)D. All of these

A gas expands from `3 dm^(3)` to `5dm^(3)` anainst a constant pressure of `3atm`. The work done during the expansion if used to heat `10mol` of water at temperature `290K`. Find the final temperature of water, if the specific heat of water `= 4.18g^(-1)K^(-1)`.

A gas expands from `3 dm^(3)` to `5dm^(3)` anainst a constant pressure of `3atm`. The work done during the expansion if used ti heat `10mol` of water at temperature `290K`. Find the final temperature of water, if the specific heat of water `= 4.18g^(-1)K^(-1)`.

If a refrigerator’s door is kept open, will the room become cool or hot? Explain.

If a refrigerator’s door is kept open, will the room become cool or hot? Explain.