4131 + Interview Questions in THERMODYNAMICS IN GENERAL AWARENESS Page 83 InterviewSolution

4101.	If a body A is in thermal equilibrium with three different bodies B, C and D thenA. B is in thermal equilibrium with CB. C is in thermal equilibrium with DC. D is in thermal equilibrium with BD. All of these
Answer» Correct Answer - D

Discussion

4102.	SI unit of heat isA. caloriesB. jouleC. watt hourD. kilojoule
Answer» Correct Answer - B

Discussion

4103.	What is the change in internal energy during Isothermal expansion of an ideal gas?
Answer» Zero, because ∆U α ∆T.

Discussion

4104.	An ideal gas is compressed in a closed container its UA. IncreasesB. DecreasesC. Remains sameD. Both (1) & (2)
Answer» Correct Answer - A

Discussion

4105.	if work is done by the system then W isA. (+)veB. (-)veC. ZeroD. Depends upon the thermodynamic process.
Answer» Correct Answer - A

Discussion

4106.	Statement-1.During isothermal expansion of an ideal gas, there is no change in the internal energy. Statement -2.During isothermal expansion of an ideal gas , as temperature remains constant, no heat enters or leaves the system.A. Statement -1 is True, Statement-2 is True, Statement-2 is a correct explanation of Statement-1B. Statement-1 is True, Statement-2 is True, Statement-2 is NOT a correct explanation for Statement -1.C. Statement-1 is True, Statement-2 is False.D. Statement-1 is False, Statement-2 is True
Answer» Correct Answer - c

Discussion

4107.	The SI unit of temperature is :A. KelvinB. celciusC. fahrenheitD. reaumur
Answer» Correct Answer - A

Discussion

4108.	During isothermal expansion of an ideal gas , the change in internal energy is `"…................"`.
Answer» Correct Answer - zero

Discussion

4109.	How does internal energy of a gas change in (i) isothermal expansion (ii) adiabatic expansion?
Answer» In isothermal expansion, `dT=0, :. dU=0` In adiabatic expansion, `dQ=0, dU= -dW` internal energy decrease.

Discussion

4110.	In an isothermal expansion change in internal energy and work done by the gas are respectivelyA. (-) ve, (-) veB. 0, (+) veC. 0, (-) veD. (+) ve, (-) ve
Answer» Correct Answer - B change in internal energy is 0 because internal energy depends on temperature. work done by gas can be + or- depending on whether the process is reversible or irreversible. work done by gas is equivalent to area under curve of p/v graph Correct option is: 0, (+) ve We know that, an isothermal expansion temperature is constant. So, change internal energy ∆u = nc_v ∆T ∵ ∆T = 0 ∆u = 0 work done will be positive because work done by the gas. ∆W = +ve

4110.

In an isothermal expansion change in internal energy and work done by the gas are respectivelyA. (-) ve, (-) veB. 0, (+) veC. 0, (-) veD. (+) ve, (-) ve

Answer» Correct Answer - B

change in internal energy is 0 because internal energy depends on temperature. work done by gas can be + or- depending on whether the process is reversible or irreversible. work done by gas is equivalent to area under curve of p/v graph

Correct option is: 0, (+) ve

We know that, an isothermal expansion temperature is constant.

So, change internal energy

∆u = nc_v ∆T

∵ ∆T = 0

∆u = 0

work done will be positive because work done by the gas.

∆W = +ve

Discussion

4111.	What makes a cycle tube burst when left in the sun ? (A) increase in volume (B) increase of pressure (C) increase of volume & pressure both. (D) Neither increase of pressure nor that of volume.
Answer» Answer is (B) increase of pressure

4111.

What makes a cycle tube burst when left in the sun ? (A) increase in volume (B) increase of pressure (C) increase of volume & pressure both. (D) Neither increase of pressure nor that of volume.

Answer»

Answer is (B) increase of pressure

Discussion

4112.	Work done in an isothermal expansion of gas depends upon : (A) temperature alone. (B) ratio of the initial and final volumes. (C) both temperature and the ratio of initial and final volumes. (D) neither temperature nor ratio of initial and final volumes.
Answer» Answer is (C) both temperature and the ratio of initial and final volumes. W_iso= RT log_e(V₂/V₁)

4112.

Work done in an isothermal expansion of gas depends upon : (A) temperature alone. (B) ratio of the initial and final volumes. (C) both temperature and the ratio of initial and final volumes. (D) neither temperature nor ratio of initial and final volumes.

Answer»

Answer is (C) both temperature and the ratio of initial and final volumes.

W_iso= RT log_e(V₂/V₁)

Discussion

4113.	Which of the following is NOT a reversible process ?(A) Melting of ice.(B) Conduction of heat(C) Isothermal expansion of gas. (D) Adiabatic expansion of gas
Answer» Answer is (B) Conduction of heat

Discussion

4114.	A refrigerator is to maintain eatables kept inside at 9°C. If room temperature is 36° C. Calculate the coefficient of performance.
Answer» Given: T₁ = 36°C = (36 + 273) = 309 K T₂ = 10°C = (10 = 273) = 283 K ⇒ Cop = T₂/T₁- T₁ = 283/309 - 283 = 283/26 = 10.9.

4114.

A refrigerator is to maintain eatables kept inside at 9°C. If room temperature is 36° C. Calculate the coefficient of performance.

Answer»

Given:

T₁ = 36°C = (36 + 273) = 309 K

T₂ = 10°C = (10 = 273) = 283 K

⇒ Cop = T₂/T₁- T₁ = 283/309 - 283 = 283/26 = 10.9.

Discussion

4115.	When an ideal gas in a cylinder was compreswsed isothermally by a piston, the work done on the gas found to be `1.5xx10^(4)` cal. During this process about
Answer» Correct Answer - `3.6xx10^(3)`

Discussion

4116.	When an ideal gas in a cylinder was compreswsed isothermally by a piston, the work done on the gas found to be `1.5xx10^(4)` cal. During this process aboutA. `1.5xx10^(3)` cal of heat flowed out from the gasB. `1.5xx10^(3)` cal of heat flowed into the gasC. `1.5xx10^(4)` cal of heat flowed out from the gasD. `1.5xx10^(4)` cal of heat flowed out from the gas
Answer» Correct Answer - A

Discussion

4117.	Given are the following standard free energies of formation at 293 K. `Delta_(f)G^(@) //kJ mol^(-1)" "underset(-137.17)(CO(g))" "underset(-394.36)(CO_(2)(g))" "underset(-228.57)(H_(2)O(g))" "underset(-237.13)(H_(2)O(l))" "underset("zero")(H_(2)(g))` (a) Find `Ddelta_(r)G^(@) ` and the standard equilibrium constant `K_(p)^(0)` at 298 K for the reaction `" "CO(g)+ H_(2)O(g) iff CO_(2)(g) + H_(2)(g)` (b) If `CO, CO_(2)` and `H_(2)` are mixed so that the partical pressure of each is 101.325 kPa and the mixture is brought into contact with excess of liquid water, what will be the partial pressure of gas when equilibrium is attained at 298 K. The volume available to the gases is contant .
Answer» Correct Answer - `P_(CO_(2)) = 202.65 kPa; p_(H_(2)O)=3.16 kPa; p_(CO)= 0.124 kPa` (i) `DeltaG^(@) = - 394.36 - [ -137.17 -228.57]` `= - 28.62` `= - 2.303 RT log K_(p)` `K_(p)= 1.047 xx 10^(5)` `By " "H_(2)O(l) hArr H_(2)O(g)` `DeltaG^(@) = - 228.57 + 237.13` ` = 8.56 = RT In K_(p)` `K_(p) = 0.0316 "bar"` `P_(H_(2)O) (g) = 3.16 pa` `underset("at eq. pressure X") underset("Initial pressure (101.325)")(CO) + underset(3.16 kpa)underset((3.16 kpa))(H_(2)O(g))" "hArr" "underset(202.65 kpa)underset((101.325))(CO_(2)) + underset(202 .65 kpa)underset((101.325))(H_(2))` `1.047 xx 10^(5) =(202.65)^(5)/(X xx3.16)` `X = 0.124 kPa`

Discussion

4118.	Compute `Delta_(r)G` for the reaction `H_(2)O (l, 1, atm, 323 K ) to H_(2)O (g, 1 atm, 232K) ` Given that : `Delta_(vap) H at 373 K = 40.639 kJmol^(-1), C_(p)(H_(2)O, l)= 75.312 JK^(-1) mol^(-1)`, `" "C_(p)(H_(2)O, g) = 33.305 J K^(-1)mol^(-1)`.
Answer» Correct Answer - `Delta_(r)G = 5.59 KJ mol^(-1)` `DeltaH_(373) = DeltaH _(323) - DeltaC_(p)DeltaT` `40639 = DeltaH_(323)-(33.305 - 75.312) xx 50` `DeltaH_(323) = 42739. 35 J//mol`

Discussion

4119.	Find out the internal energy change for the reaction `A (l) rarr A (g)` at 373 K . Heat of vaporisation is 40.66 kJ `//` mol and `R = 8.3 J mol^(-1) K^(-1)`
Answer» `A (l) rarr A(g), Delta n_(g) = n_(p) - n_(r) = 1 -0 = 1` `Delta H = Delta U + Delta n_(g) RT` or `Delta U = Delta H - Delta n_(g) RT = 40660 J - 1 mol xx 8.314 J K^(-1) mol^(-1) xx 373 K ` `= 40660J - 3101 J = 37559 J mol^(-1) = 37.56 k J mol^(-1)`

Discussion

4120.	`I: H_(2)O_((s))" "II: H_(2)O_((l))" "III: H_(2)O_((g))` For the above compounds the correct of internal energy content isA. `I gt II gt III`B. `I = II = III`C. `I lt II lt III`D. `II gt II gt I`
Answer» Correct Answer - D Gas `gt` liquid `gt` solid

Discussion

4121.	A given mass of gas expands from state `A` to state `B` by three paths `1,2`, and `3` as shown in the figure below. If `w_(1),w_(2)` and `w_(3)`, respectively, be the work done by the gas along three paths, then A. `w_(1) gt w_(2) gt w_(3)`B. `w_(1) lt w_(2) lt w_(3)`C. `w_(1) = w_(2)=w_(3)`D. `w_(2) lt w_(3) lt w_(1)`
Answer» Correct Answer - B More than area under the curve, more is the work done

Discussion

4122.	A given mass of gas expands from state `A` to state `B` by three paths `1,2`, and `3` as shown in the figure below. If `w_(1),w_(2)` and `w_(3)`, respectively, be the work done by the gas along three paths, then A. `w_(1) gt w_(2) gt w_(3)`B. `w_(1) lt w_(2) lt w_(3)`C. `w_(1) = w_(2) = w_(3)`D. `w_(2) lt w_(3) lt w_(1)`
Answer» Correct Answer - B `w=-"Area of P-V curve"`

Discussion

4123.	The enthalpy of combustion of propance `(C_(3)H_(8))` gas in temes of given of geven data is , Bond energy (kJ/mol) `underset(+x_(1))(.^(epsi)C-H)underset(+x_(2))(.^(epsi)O-O)underset(+x_(3))(.^(epsi)C-O)underset(+x_(4))(.^(epsi)O-H)underset(+x_(5))(.^(epsi)C-C)` [Resonance energy of `CO_(2)` is -z KJ/mol and `DeltaH_("vaporization")[H_(2)O(l)" is y " KJ//mol]`A. `8x_(1)+2x_(5)+5x_(2)-6x_(3)-8x_(4)-4y-3z`B. `6x_(1)+x_(5)+5x_(2)-3x_(3)-4x_(4)-4y-3z`C. `8x_(1)+2x_(5)+5x_(2)-6x_(3)-8x_(4)-y-z`D. `8x_(1)+x_(5)+5x_(2)-6x_(3)-8x_(4)-4y+3z`
Answer» Correct Answer - A `C_(3)H_(8)(g)+5 O_(2)(g)rarr3 CO_(2)(g)+4H_(2)O(l)` `Delta_(C)H=[(8xxB.E.(C-H)),(+2xxB.E.(C-C)),(+5xxB.E.(O=O))]` `-[(6xxB.E. (C=O)),(+8xxB.E.(O-H)),(+3xx\|R.E.\|" of "CO_(2)),(+4xxDelta_("vap")H(H_(2)O))]`

Discussion

4124.	A sample of an ideal gas is expanded `1 m^(3)` to `3 m^(3)` in a reversible process for which `P=KV^(2)` , with `K=6 "bar"//m^(6).` What is work done by the gas (in kJ) ?A. 5200 JB. 15600 kJC. 52 kJD. 5267 .6 kJ
Answer» Correct Answer - A `intdw=-intP.dV` `implies w_("rev")=-int6.V^(2)dV` `=-6[(V_(2)^(3))/(3)-(V_(1)^(3))/(3)]"bar."m^(3),` `w=-5200 " kJ"` so, work done by the gas is 5200 kJ.

Discussion

4125.	A sample of an ideal gas is expanded `1 m^(3)` to `3 m^(3)` in a reversible process for which `P=KV^(2)` , with `K=6 "bar"//m^(6).` What is work done by the gas (in kJ) ?
Answer» Correct Answer - 5200

Discussion

4126.	For the given real gas reaction, `2A(g)+B(g)rarrD(g)` carried out in a `10` liter rigid vesses, the initial pressure is `50` bar which decreases to `20` bar, during the course of reaction. If heat liberated in the reaction is `400` kJ then what is the change in magnitude of internal energy of the reaction (in kJ) ?
Answer» Correct Answer - 400

Discussion

4127.	a thermodynamic process is carried out from an original state D to an intermediate state E by the linear process shown in figure. The total work is done bvy the gas from D to E to F is A. 100 JB. 800 JC. 300 JD. 250 J
Answer» Correct Answer - B Total work done by the gas from D to E to F is equal to the area of `DeltaDEF` `therefore` The area of `Delta DEF=(1)/(2)DFxxEF` Here DF change in pressure =600-200=400 N `m^(-2)` Also EF=change in volume = `7m^(3)-3m^(3)=4m^(3)` Area of `DeltaDEF =(1)/(2)xx400xx4=800J` Thus the total work done by the gas from D to E to F is 800 J.

Discussion

4128.	Bromine boils at `59^(@)C` with `DeltaH_(vap)^(@)=29.6kJ mol^(-1)` What is the value of `DeltaS_(vap)^(@)"in"29.6 J mol^(-1)K^(-1)`A. 11.2B. 89.2C. 50.1D. 1750
Answer» Correct Answer - B

Discussion

4129.	What is the melting point of benzene if `DeltaH_("fusion")= 9.95kJ//mol` and `DeltaS_("fusion")= 35.7J//K-"mol"`A. `278.7^(@)C`B. `278.7^(@)K`C. 300 KD. 298 K
Answer» Correct Answer - B at M.P. `DeltaS_("fusion")=(DeltaH_("fusion"))/(T)` `implies" "T=(DeltaH)/(DeltaS)=(9.95 xx1000)/(35.7)=278.7 K`

Discussion

4130.	Calculate `DeltaG^(Theta)` for the conversion of oxygen to ozone, `((3)/(2)) O_(2)(g) hArr O_(3)(g) at 298 K`, of `K_(p)` for this conversion is `2.47 xx 10^(-29)`.
Answer» We know `Delta_(r)G^(Ө)=-2.303RT` log `K_(p)` and R=8.31 `JK^(-1)"mol"^(-1)` Thererfore, `Delta_(r)G^(Ө)=-2.303(8.314JK^(-1)"mol"^(-1))xx(298K)("log"2.47xx10^(-29))` `=163000 J "mol"^(-1)` `=163 kJ "mol"^(-1)`

Discussion

4131.	Calculate the enthalpy change during the reaction `:` `H_(2(g))+Br_(2(g))rarr 2HBr_((g))` Given, `e_(H-H)=435kJ mol^(-1),e_(Br-Br)=192kJ mol^(-1)` and `e_(H-Br)=368kJ mol^(-1).`
Answer» `-109 kJ" "mol^(-1)`

Discussion

Explore topic-wise InterviewSolutions in .

If a body A is in thermal equilibrium with three different bodies B, C and D thenA. B is in thermal equilibrium with CB. C is in thermal equilibrium with DC. D is in thermal equilibrium with BD. All of these

SI unit of heat isA. caloriesB. jouleC. watt hourD. kilojoule

What is the change in internal energy during Isothermal expansion of an ideal gas?

An ideal gas is compressed in a closed container its UA. IncreasesB. DecreasesC. Remains sameD. Both (1) & (2)

if work is done by the system then W isA. (+)veB. (-)veC. ZeroD. Depends upon the thermodynamic process.

The SI unit of temperature is :A. KelvinB. celciusC. fahrenheitD. reaumur

During isothermal expansion of an ideal gas , the change in internal energy is `"…................"`.

How does internal energy of a gas change in (i) isothermal expansion (ii) adiabatic expansion?

In an isothermal expansion change in internal energy and work done by the gas are respectivelyA. (-) ve, (-) veB. 0, (+) veC. 0, (-) veD. (+) ve, (-) ve

What makes a cycle tube burst when left in the sun ? (A) increase in volume (B) increase of pressure (C) increase of volume &amp; pressure both. (D) Neither increase of pressure nor that of volume.

Work done in an isothermal expansion of gas depends upon : (A) temperature alone. (B) ratio of the initial and final volumes. (C) both temperature and the ratio of initial and final volumes. (D) neither temperature nor ratio of initial and final volumes.

Which of the following is NOT a reversible process ?(A) Melting of ice.(B) Conduction of heat(C) Isothermal expansion of gas. (D) Adiabatic expansion of gas

A refrigerator is to maintain eatables kept inside at 9°C. If room temperature is 36° C. Calculate the coefficient of performance.

When an ideal gas in a cylinder was compreswsed isothermally by a piston, the work done on the gas found to be `1.5xx10^(4)` cal. During this process about

Compute `Delta_(r)G` for the reaction `H_(2)O (l, 1, atm, 323 K ) to H_(2)O (g, 1 atm, 232K) ` Given that : `Delta_(vap) H at 373 K = 40.639 kJmol^(-1), C_(p)(H_(2)O, l)= 75.312 JK^(-1) mol^(-1)`, `" "C_(p)(H_(2)O, g) = 33.305 J K^(-1)mol^(-1)`.

Find out the internal energy change for the reaction `A (l) rarr A (g)` at 373 K . Heat of vaporisation is 40.66 kJ `//` mol and `R = 8.3 J mol^(-1) K^(-1)`

`I: H_(2)O_((s))" "II: H_(2)O_((l))" "III: H_(2)O_((g))` For the above compounds the correct of internal energy content isA. `I gt II gt III`B. `I = II = III`C. `I lt II lt III`D. `II gt II gt I`

A sample of an ideal gas is expanded `1 m^(3)` to `3 m^(3)` in a reversible process for which `P=KV^(2)` , with `K=6 "bar"//m^(6).` What is work done by the gas (in kJ) ?A. 5200 JB. 15600 kJC. 52 kJD. 5267 .6 kJ

A sample of an ideal gas is expanded `1 m^(3)` to `3 m^(3)` in a reversible process for which `P=KV^(2)` , with `K=6 "bar"//m^(6).` What is work done by the gas (in kJ) ?

a thermodynamic process is carried out from an original state D to an intermediate state E by the linear process shown in figure. The total work is done bvy the gas from D to E to F is A. 100 JB. 800 JC. 300 JD. 250 J

Bromine boils at `59^(@)C` with `DeltaH_(vap)^(@)=29.6kJ mol^(-1)` What is the value of `DeltaS_(vap)^(@)"in"29.6 J mol^(-1)K^(-1)`A. 11.2B. 89.2C. 50.1D. 1750

What is the melting point of benzene if `DeltaH_("fusion")= 9.95kJ//mol` and `DeltaS_("fusion")= 35.7J//K-"mol"`A. `278.7^(@)C`B. `278.7^(@)K`C. 300 KD. 298 K

Calculate `DeltaG^(Theta)` for the conversion of oxygen to ozone, `((3)/(2)) O_(2)(g) hArr O_(3)(g) at 298 K`, of `K_(p)` for this conversion is `2.47 xx 10^(-29)`.

Calculate the enthalpy change during the reaction `:` `H_(2(g))+Br_(2(g))rarr 2HBr_((g))` Given, `e_(H-H)=435kJ mol^(-1),e_(Br-Br)=192kJ mol^(-1)` and `e_(H-Br)=368kJ mol^(-1).`

What makes a cycle tube burst when left in the sun ? (A) increase in volume (B) increase of pressure (C) increase of volume & pressure both. (D) Neither increase of pressure nor that of volume.