Q: An ideal gas (0.5 mol), initially at 2 bar pressure, is compressed at a constant temperature of 600 K in two steps: first, against a constant external pressure of P bar (2 < P < 8), and then against c

Let the initial, intermediate, and final volumes of the gas be V_1, V_2, and V_3 respectively. Using the ideal gas equation V = nRT P , we have: V_1 = nRT 2 V_2 = nRT P V_3 = nRT 8 The work done on the gas during the first step against a constant external pressure P is: W_1 = -P_ ext,1 (V_2 - V_1) = -P ( nRT P - nRT 2 ) = nRT ( P 2 - 1 ) The work done on the gas during the second step against a constant external pressure of 8 bar is: W_2 = -P_ ext,2 (V_3 - V_2) = -8 ( nRT 8 - nRT P ) = nRT ( 8 P

Q: List-I contains various physical/chemical processes, and List-II contains combinations of changes in enthalpy ( H) and entropy ( S). Match each entry in List-I to the appropriate entry in List-II, and

(P) Physisorption is an exothermic process because attractive forces are formed between the adsorbate and the adsorbent, so H (Q) Diamond Graphite is an exothermic process because graphite is the thermodynamically more stable allotrope of carbon at standard conditions ( H 0). This matches (5). (R) Denaturation of protein involves the breaking of hydrogen bonds and the unfolding of its specific 3D structure into a more random coil. Breaking these interactions requires energy ( H > 0) and the unfo

Q: The entropy versus temperature plot for phases α and β at 1 bar pressure is given. S T and S 0 are entropies of the phases at temperatures T and 0 K , respectively. The tr

S = S 0 + ∫ C p d T T S α = S 0 + ∫ C p α d T T S β = S 0 + ∫ C p β d T T Now, S β - S α = S 0 + ∫ C p β - C p α d T T Given C p β - C p α = 1 S β - S α = lnT + C at any temperature of T S β - S α T 2 - S β - S α T 1 = lnT 2 - lnT 1 T 2 = 600 K , T 1 = 300 K and S β - S α = 1 at 600 K 1 - S β -

Q: The entropy versus temperature plot for phases α and β at 1 bar pressure is given. S T and S 0 are entropies of the phases at temperatures T and 0 K , respectively. The tr

According to Kirchhoff's law ΔH T 2 - ΔH T 1 = n × C p , β - C p , α ( T 2 - T 1 ) ΔH 600 - ΔH 300 = 1 × C p , β - C p , α ( 600 - 300 ) Now, at transition temperature, ∆ G = 0 ΔH 600 = TΔS 600 = 600 × ( 6 - 5 ) 600 - ΔH 300 = 1 × 1 × 300 ΔH 300 = 600 - 300 = 300 J mol - 1

Q: One mole of an ideal monoatomic gas undergoes two reversible processes ( A → B and B → C ) as shown in the given figure: A → B is an adiabatic process. If the total heat absorbed in

q A → C = RT 2 ln 10 q A → B = 0 ( ∵ adiabatic) q A → C = q A → B + q B → C q A → C = q B → C q A → c = nRT 2 ln V 3 V 2 ...(i) For B → C ΔE = q + w ΔE = 0 (since isothermic) q = - w = - - nRT 2 ln V 3 &#16

Q: In a one-litre flask, 6 moles of A undergoes the reaction A ( g ) ⇌ P ( g ) . The progress of product formation at two temperatures (in Kelvin), T 1 and T 2 , is shown in the figur

The equilibrium moles can be calculated as follows, A g ⇌ P g initial 6 &#

Q: The correct option(s) about entropy (S) is(are) [ R = gas constant, F = Faraday constant, T = Temperature]

(A) M s + 2 H + aq → H 2 g + M 2 + aq if dE cell dT = R F Δ S = nF dE dT = 2 F R F = 2 R (B) E cell = - 2 . 303 RT F log 0 . 01 0 . 1 = 2 . 303 RT F dE cell dT = 2 . 303 R F ∴ Δ S = nF dE dT > 0 It is an entropy driven process. (C) It is correct Racemisation is a thermodynamically favourable method, and it proceeds spontaneously if a suitable pathway is accessible for the interconversion of the enantiomers. During

Q: 2 mol of Hg g is combusted in a fixed volume bomb calorimeter with excess of O 2 at 298 K and 1 atm into HgO s . During the reaction, temperature increases from 298 . 0 K t

2 Hg g + O 2 g ⟶ 2 HgO s Heat capacity of calorimeter = 20 kJ K - 1 Rise in temperature = 14 . 8 K Heat evolved = 20 × 14 . 8 = 296 kJ ΔH ° = ΔU ° + Δn g RT = - 296 - 3 × 8 . 3 × 298 × 10 - 3 = - 303 . 42 kJ ΔH ° = 2 ΔH f ° HgO s - 2 ΔH f ° Hg g - 303 . 42 = 2 ΔH f ° HgO s - 2 × 61 . 32 2 ΔH f ° HgO s = - 180 . 78 kJ ΔH f &#17

Question 1

An ideal gas (0.5 mol), initially at 2 bar pressure, is compressed at a constant temperature of 600 K in two steps: first, against a constant external pressure of P bar (2 < P < 8), and then against c

Accepted Answer

Let the initial, intermediate, and final volumes of the gas be V_1, V_2, and V_3 respectively. Using the ideal gas equation V = nRT P , we have: V_1 = nRT 2 V_2 = nRT P V_3 = nRT 8 The work done on the gas during the first step against a constant external pressure P is: W_1 = -P_ ext,1 (V_2 - V_1) = -P ( nRT P - nRT 2 ) = nRT ( P 2 - 1 ) The work done on the gas during the second step against a constant external pressure of 8 bar is: W_2 = -P_ ext,2 (V_3 - V_2) = -8 ( nRT 8 - nRT P ) = nRT ( 8 P

Question 2

List-I contains various physical/chemical processes, and List-II contains combinations of changes in enthalpy ( H) and entropy ( S). Match each entry in List-I to the appropriate entry in List-II, and

Accepted Answer

(P) Physisorption is an exothermic process because attractive forces are formed between the adsorbate and the adsorbent, so H (Q) Diamond Graphite is an exothermic process because graphite is the thermodynamically more stable allotrope of carbon at standard conditions ( H 0). This matches (5). (R) Denaturation of protein involves the breaking of hydrogen bonds and the unfolding of its specific 3D structure into a more random coil. Breaking these interactions requires energy ( H > 0) and the unfo

Question 3

Considering ideal gas behavior, the expansion work done (in kJ) when 144 g of water is electrolyzed completely under constant pressure at 300 K is _____ Use: Universal gas constant ( R )=8.3 ~J ~K ^ -

Accepted Answer

in official answer key answer was [-29.95 to -29.8] OR [29.8 to 29.95], we have updated the question to make it valid. array lcc H _2 O (l) & H _2( ~g )+ & 1 2 O _2( ~g ) \ 144 ~g & - & - \ 8 ~mol & 8 ~mol & 4 ~mol \ ~W =- P ~V = & -( n ) RT array aligned & Change in gaseous moles =12\ & W =- 12 8.3 300 1000 ~kJ =-29.88 ~kJ . aligned

Question 4

Consider the following volume-temperature ( V - T ) diagram for the expansion of 5 moles of an ideal monoatomic gas. Considering only P-V work is involved, the total change in enthalpy (in Joule) for

Accepted Answer

For ideal gas H = nC _ P T C _ P = C _ V + R =12+8.3=20.3 ~J / K -mole H=5 20.3 (415-335) H=8120 Joule

Question 5

The entropy versus temperature plot for phases &#945;  and &#946; at 1 bar pressure is given. S T and S 0  are entropies of the phases at temperatures T and 0 &#160; K , respectively. The tr

Accepted Answer

S = S 0 + &#8747; C p d T T S &#945; = S 0 + &#8747; C p &#945; d T T S &#946; = S 0 + &#8747; C p &#946; d T T Now,  S &#946; - S &#945; = S 0 + &#8747; C p &#946; - C p &#945; d T T Given   C p &#946; - C p &#945; = 1 S &#946; - S &#945; = lnT + C  at any temperature of  T S &#946; - S &#945; T 2 - S &#946; - S &#945; T 1 = lnT 2 - lnT 1 T 2 = 600 K &#160; , &#160; T 1 = 300 &#160; K &#160; and &#160; S &#946; - S &#945; = 1 &#160; at &#160; 600 &#160; K 1 - S &#946; -

Question 6

The entropy versus temperature plot for phases &#945;  and &#946; at 1 bar pressure is given. S T and S 0  are entropies of the phases at temperatures T and 0 &#160; K , respectively. The tr

Accepted Answer

According to Kirchhoff's law  &#916;H T 2 - &#916;H T 1 = n &#215; C p , &#946; - C p , &#945; ( T 2 - T 1 ) &#916;H 600 - &#916;H 300 = 1 &#215; C p , &#946; - C p , &#945; ( 600 - 300 ) Now, at transition temperature,  &#8710; G = 0 &#916;H 600 = T&#916;S 600 = 600 &#215; ( 6 - 5 ) 600 - &#916;H 300 = 1 &#215; 1 &#215; 300 &#916;H 300 = 600 - 300 = 300 &#160; J &#160; mol - 1

Question 7

One mole of an ideal monoatomic gas undergoes two reversible processes ( A &#8594; B and B &#8594; C ) as shown in the given figure: A &#8594; B is an adiabatic process. If the total heat absorbed in

Accepted Answer

q A &#8594; C = RT 2 ln 10 q A &#8594; B = 0                                &#160;( &#8757; &#160;adiabatic)&#160; q A &#8594; C = q A &#8594; B + q B &#8594; C q A &#8594; C = q B &#8594; C q A &#8594; c = nRT 2 ln V 3 &#160; V 2      ...(i) For B &#8594; C &#916;E = q + w &#916;E = 0                   (since isothermic) q = - w = - - nRT 2 ln V 3 &#16

Question 8

In a one-litre flask, 6 moles of A undergoes the reaction A ( &#160; g ) &#8652; P ( &#160; g ) . The progress of product formation at two temperatures (in Kelvin), T 1 and T 2 , is shown in the figur

Accepted Answer

The equilibrium moles can be calculated as follows, &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; A g &#8652; P g initial &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; 6 &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#

Question 9

The correct option(s) about entropy (S) is(are) [ R = gas constant, F = Faraday constant, T = Temperature]

Accepted Answer

(A)  M s + 2 H + aq &#8594; H 2 g + M 2 + aq if dE cell &#160; dT = R F &#916; S = nF dE dT = 2 &#160; F R F = 2 R (B)  E cell &#160; = - 2 . 303 RT F log 0 . 01 0 . 1 = 2 . 303 RT F dE cell &#160; dT = 2 . 303 R F &#8756; &#916; S = nF dE dT &#62; 0 It is an entropy driven process. (C) It is correct Racemisation is a thermodynamically favourable method, and it proceeds spontaneously if a suitable pathway is accessible for the interconversion of the enantiomers.  During

Question 10

2 &#160; mol of Hg g is combusted in a fixed volume bomb calorimeter with excess of O 2 at 298 &#160; K and 1 &#160; atm into HgO s . During the reaction, temperature increases from 298 . 0 &#160; K t

Accepted Answer

2 Hg g + O 2 g &#10230; 2 HgO s Heat capacity of calorimeter  = 20 &#160; kJ &#160; K - 1 Rise in temperature  = 14 . 8 &#160; K Heat evolved  = 20 &#215; 14 . 8 = 296 &#160; kJ &#916;H &#176; = &#916;U &#176; + &#916;n g RT = - 296 - 3 &#215; 8 . 3 &#215; 298 &#215; 10 - 3 = - 303 . 42 &#160; kJ &#916;H &#176; = 2 &#916;H f &#176; HgO s - 2 &#916;H f &#176; Hg g - 303 . 42 = 2 &#916;H f &#176; HgO s - 2 &#215; 61 . 32 2 &#916;H f &#176; HgO s = - 180 . 78 &#160; kJ &#916;H f &#17

Question 11

One mole of an ideal gas at 900 &#160; K , undergoes two reversible processes, I followed by II , as shown below. If the work done by the gas in the two processes are same, the value of ln V 3 V 2 is

Accepted Answer

Process -I: Adiabatic reversible process. (Since entropy is constant) W I = &#916;U = 450 &#8722; 2250 R = - 1800 &#160; R Process II: Isothermal reversible process. (since internal energy is constant and entropy is increased) Work done:   W II = &#8722; nRTln V f V i W II = &#8722; nRTln V 3 V 2 W II = &#8722; 900 5 Rln V 3 V 2 W II = 900 5 Rln V 3 V 2 U = 5 2 nRT 450 &#160; R = 5 2 nRT nRT = 900 5 R Given W I = W II 1800 R = 900 5 R &#160; ln V 3 V 2 ln V 3 V 2 = 10

Question 12

For the reaction, X (s) Y (s)+ Z (g), the plot of p_ Z p^ versus 10^ 4 T is given below (in solid line), where p_ Z is the pressure (in bar) of the gas Z at temperature T and p^ =1 bar. (Given, d ( K)

Accepted Answer

Slop  g  the Given graph d &#160; ln p z p &#981; d &#160; 10 4 T = &#8722; 7 + 3 12 &#8722; 10 = &#8722; 2 d &#160; ln p z p &#981; d 1 T &#8722; 2 &#215; 10 4 = &#8722; &#916;H o R &#916;H o = 2 &#215; 10 4 &#215; R = 2 &#215; 8 .314 &#215; 10 &#160; kJ / mol = 166 . 28

Question 13

For the reaction, X (s) Y (s)+ Z (g), the plot of p_ Z p^ versus 10^ 4 T is given below (in solid line), where p_ Z is the pressure (in bar) of the gas Z at temperature T and p^ =1 bar. (Given, d ( K)

Accepted Answer

According to graph when  T = 1000 &#160; K  and  10 4 T = 10  then  ln p z p &#981; = &#8722; 3 Now,  &#916;G o = &#916;H o &#8722; T &#160; &#916;S o &#8722; RT &#160; lnK p = &#916;H o &#8722; T &#160; &#916;S o R &#160; ln &#160; K p = &#916;S o &#8722; &#916;H o T &#916;S o = RlnK p + &#916;H o T lnK p = ln p z p &#981; = &#8722; 3 &#916;S o = &#8722; 3 R + 166 .28 &#215; 10 3 J 1000 166 .28 &#8722; 3 &#215; 8 .314 &#916;S o = 141 .34

Question 14

An ideal gas undergoes a reversible isothermal expansion from state I to state II followed by a reversible adiabatic expansion from state II  to state III .  The correct plot  s represe

Accepted Answer

Reversible isothermal expansion from  I &#8594; II  during this process T =  constant &#916;H = &#916;U = 0 H =  constant. &#916;S &#8800; 0  Entropy will increase  PV =  constant. and during adiabatic reversible expansion. q = 0 &#916;U = w = &#8722; ve so  T  will decrease &#916;H = nc p &#916;T = - ve H  will decrease &#916;S = 0  (isoentropic process) S = constant. and  pv r =  constant. Keeping in mind correct graph are 1

Question 15

Tin is obtained from cassiterite by reduction with coke. Use the data given below to determine the minimum temperature (in K ) at which the reduction of cassiterite by coke would take place. 298 &#160

Accepted Answer

SnO 2 ( s ) + C ( s ) &#8594; Sn ( s ) + CO 2 ( g ) &#916;H rxn 0 = &#916;H f 0 CO 2 ( g ) - &#916;H f 0 SnO 2 ( s ) = - 394 . 0 - [ - 581 . 0 ] = 187 &#160; KJ &#916;S rxn 0 = S 0 ( Sn ( s ) ) + S 0 CO 2 ( g ) - S 0 SnO 2 , ( s ) - S 0 ( C ( s ) ) = 52 + 210 - 56 - 6 = 200 &#160; J &#916;G &#8728; = &#916;H &#8728; - T&#916;S &#8728; For reaction to be spontaneous &#916;G &#8728; &#60; 0 &#916;H &#8728; - T&#916;S &#8728; &#60; 0 187 &#215; 1000 - T &#215; 200 &#60; 0 T &#62; 935 &#160; K &nbsp

Question 16

In thermodynamics, the  P - V  work done is given by  w = - &#8747; dVP ext . For a system undergoing a particular process, the work done is, w = - &#8747; dV RT V - b - a V 2 .&nb

Accepted Answer

$W=- P_ ext d v$ From van der waal equation of state for one mole gas $p+ a v^ 2 (v-b)= RT $ $p= RT (v-b) - a v^ 2 $ For reversible process $p_ ext =p_ gas $ $W=- RT (v-b) - a v^ 2 $ process is only not applicable for irreversible process.

Question 17

Choose the reaction(s) from the following options, for which the standard enthalpy of reaction is equal to the standard enthalpy of formation.

Accepted Answer

Standard enthalpy formation of a compound is the amount of enthalpy charge involved when one mole of a compound is formed from the reactants present in their elementary form,

Thermodynamics (C) — Previous Year Questions & Solutions