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An ideal gas is allowed to expand from 1L to 10L against a constant external pressure of 1 bar. The work done in KJ is :

• Option 1)

$+10.0$

• Option 2)

$-9.0$

• Option 3)

$-2.0$

• Option 4)

$-0.9$

Work (W) - Any type of energy transfer between system & surrounding which is not due to temperature difference is known as work. - wherein It is considered as ordered form of energy.       First law of Thermodynamics - Energy of universe is always conserved or total energy of an isolated system is always conserved   - wherein Internal Energy Heat work     We know that, option (4)...

Enthalpy of sublimation of iodine is $24calg^{-1}$ at $200^{\circ}C$. If specific heat of $I_{2}\left ( s \right )$ and $I_{2}\left ( vap \right )$ are $0.055$ and $0.031calg^{-1}K^{-1}$ respectively, then enthalpy of sublimation of iodine at $250^{\circ}C$ in $cal\: g^{-1}$ is :

• Option 1)

$2.85$

• Option 2)

$22.8$

• Option 3)

$5.7$

• Option 4)

$11.4$

Enthalpy of Sublimation - Amount of enthalpy change to sublimise 1 mole solid into 1 mole vapour at a temperature below its melting point - wherein       Specific heat of  &  are respectively,  &   Heat capacity of Product - Reactant                Applying Kirchoff's equation, we get:               Enthalpy of sublimation of iodine at   Option 1)Option 2)Option 3)Option 4)

For the reaction ,

$2SO_{2}\left ( g \right )+O_{2}\left ( g \right )\rightleftharpoons 2SO_{3}\left ( g \right ),$

$\Delta H=-57.2\; kJ\; mol^{-1}\; and$

$K_{c}=1.7\times 10^{16}.$

Which of the following statement is INCORRECT ?

• Option 1)

The equilibrium constant is large suggestive of reaction going to completion and so no catalyst is required .

• Option 2)

The equilibrium will shift in forward direction as the pressure increases.

• Option 3)

The equilibrium constant decreases as the temperature increases.

• Option 4)

The addition of inert gas at constant volume will not affect the equilibrium constant.

Option 2   so, increase in pressure will bring reaction in forward direction , it is correct. Option 3  means reaction is exothermic so, equlibrium constant decreases as the temperature increases. Option 4 Equilibrium constant depend or change only with temperature .  Option2,3,4 is correct so, Option (1) is incorrect.Option 1)The equilibrium constant is large suggestive of reaction going to...

The difference between $\Delta H$ and $\Delta U$ $\left ( \Delta H-\Delta U \right )$ , when the combustion of one mole of heptane $\left ( 1\right )$ is carried out at a temperature T, is equal to :

• Option 1)

$-4\: RT$

• Option 2)

$-3\: RT$

• Option 3)

$4\: RT$

• Option 4)

$3\: RT$

Combution of Heptane is -   Change in gaseous moles =                              Option 1)   Option 2) Option 3)Option 4)

A process will be spontaneous at all temperatures if :

• Option 1)

$\triangle H<0\: \: and\: \: \triangle S<0$

• Option 2)

$\triangle H>0\: \: and\: \: \triangle S<0$

• Option 3)

$\triangle H<0\: \: and\: \: \triangle S>0$

• Option 4)

$\triangle H>0\: \: and\: \: \triangle S>0$

Any process is spontaneous, when the GIbbs Free Energy () is negative; ie. Now,   or       or   Option (3) is correct.  Option 1)Option 2)Option 3)Option 4)

Which one of the following equations does not correctly represent the first law of thermodynamics for the given processes involving an ideal gas ? (Assume non-expansion work is zero)

• Option 1)

Isochoric process : $\Delta U=q$

• Option 2)

Adiabatic process : $\Delta U=-w$

• Option 3)

Isothermal process: $q=-w$

• Option 4)

Cyclic process : $q=-w$

Adiabatic process :- According to first law of thermodynamics So, Hence option 2 is incorrect.  Option 1)Isochoric process : Option 2)Adiabatic process : Option 3)Isothermal process:   Option 4)Cyclic process :

5 moles of an ideal gas at 100 K are allowed to undergo reversible compression till its temperature becomes 200 K. If $C_{V}=28JK^{-1}mol^{-1}$, calculate $\Delta U$ and $\Delta pV$ for this process. (R=8.0 JK-1mol-1)

• Option 1)

$\Delta U=14 kJ; \Delta (pV)=18kJ$

• Option 2)

$\Delta U=14J; \Delta (pV)=0.8J$

• Option 3)

$\Delta U=14 kJ; \Delta (pV)=4kJ$

• Option 4)

$\Delta U=2.8 kJ; \Delta (pV)=0.8kJ$

ans is  Option 1)Option 2)Option 3)Option 4)

For silver $C_{p}(JK^{-1}mol^{-1})=23+0.01T.$ If the temperature $(T)$ of 3 moles of silver is raised from $300 \; K to\; 1000 \; K$ at 1 atm pressure, the value of $\Delta H$ will be close to :

• Option 1)

$21\; kJ$

• Option 2)

$13\; kJ$

• Option 3)

$16\; kJ$

• Option 4)

$62\; kJ$

Option 1)Option 2)Option 3)  Option 4)

The enthalpy change on freezing of 1 mol of water at 50C to ice at −50C is :

(Given fus H = 6 kJ mol−1 at 00C,

Cp (H 2O, ) = 75.3 J mol−1 K−1,

Cp (H2O, s) = 36.8 J mol−1 K−1)

• Option 1)

5.44 kJ mol−1

• Option 2)

5.81 kJ mol−1

• Option 3)

6.56 kJ mol−1

• Option 4)

6.00 kJ mol−1

Option 3 :

6.56 KJ.

For the chemical reaction $X\: \rightleftharpoons \: Y$ , the standered reaction Gibbs energy depend on temperature T (in K)  as

$\Delta _{r}G^{0} \left ( in \: k\: mol^{-1} \right )\: = \: 120\: -\frac{3}{8}T$

The major component of the reaction mixture at T is :

• Option 1)

Y if T =  300 K

• Option 2)

Y if T = 280 K

• Option 3)

X if T = 350 K

• Option 4)

X if T = 315 K

Δ G for reversible reaction -   - wherein Gibb's free energy at standard condition Universe gas constant Temperature Reaction Quotient     Δ G of equilibrium -   - wherein At Equilibrum     and       Spontanous process -   - wherein For spontanouse process must be negative. As we learnd At will be major component amount of will be greater at this temperature at...

Two blocks of the same metal having same mass and at temperature $T_{1}$ and $T_{2}$ respectively , are brough in contact with each other and allowed to attain thermal equilibrium at constant pressure . The change in entropy ,$\Delta S$ for this process is:

• Option 1)

$2C_{p} In \left ( \frac{T_{1}+T_{2}}{4T_ {1}T_{2} }\right )$

• Option 2)

$C_{p}\: In\: \left [ \frac{\left (T_{1}+T_{2} \right )^{2}}{4T_{1}T_{2}} \right ]$

• Option 3)

$2C_{p}\: In\: \left [ \frac{T_{1}+T_{2}}{2T_{1}T_{2}} \right ]$

• Option 4)

$2C_{p}\: In\: \left [ \frac{(T_{1}+T_{2})^{\frac{1}{2}}}{T_{1}T_{2}} \right ]$

Entropy for isobaric process - - wherein As we know    Option 1)Option 2)Option 3)    Option 4)

For the equilibrium,

$2H_{2}O \rightleftharpoons H_{3}O^{+}+OH^{-}$, the value of $\Delta G^{0}$ at 298 K is approximately :

• Option 1)

80 kJ mol -1

• Option 2)

-100 kJ mol -1

• Option 3)

-80 kJ mol -1

• Option 4)

100 kJ mol-1

Δ G of equilibrium -   - wherein At Equilibrum     and       ΔG{reaction} -   - wherein Sum of   of all product Sum of   of all reactant    Option 1)80 kJ mol -1Option 2)-100 kJ mol -1Option 3)-80 kJ mol -1Option 4)100 kJ mol-1

The standard reaction Gibbs energy for a chemical reaction at an absolute temperature T is given by

$\Delta _{r}G^{o}=A-BT$

Where A and B are non -zero constants. Which of the following is TRUE about this reaction ?

• Option 1)

Endothermic if A < 0 and B > 0

• Option 2)

Exothermic if B < 0

• Option 3)

Exothermic if A > 0 and B < 0

• Option 4)

Endothermic if A > 0

Gibb's free energy (Δ G) -   - wherein Gibb's free energy enthalpy of reaction entropy temperature     Δ G of equilibrium -   - wherein At Equilibrum     and     Spontanous process -   - wherein For spontanouse process must be negative. Formula   Option 1)Endothermic if A < 0 and B > 0 Option 2) Exothermic if B < 0Option 3)Exothermic if A > 0 and B < 0Option 4)Endothermic if A > 0

The reaction,

$MgO(s)+ C(s)\rightarrow Mg(s)+CO(g)$, for  which $\Delta _{r}H^{o}= + 491.1$ kJ mol-1 and $\Delta _{r}S^{o}= 198.0 \: JK^{-1}$ mol-1 , is not feasible at 298 K. Temperature above which reaction will be feasible is :

• Option 1)

2380.5 K

• Option 2)

2480.3 K

• Option 3)

2040.5 K

• Option 4)

1890.0 K

ΔG{reaction} -   - wherein Sum of   of all product Sum of   of all reactant     Gibb's free energy (Δ G) -   - wherein Gibb's free energy enthalpy of reaction entropy temperature     Spontanous process -   - wherein For spontanouse process must be negative.  Option 1)2380.5 K Option 2)2480.3 KOption 3)2040.5 K Option 4)1890.0 K

The combination of plots which does not represent isothermal expansion of an ideal gas is :

• Option 1)

(A) and (D)

• Option 2)

(A) and (C)

• Option 3)

(B) and (D)

• Option 4)

(B) and (C)

Work Done for Irreversible Isothermal Expansion of an ideal gas - - wherein may be equal to or may not be equal to , but work done is always calculated by     Work Done for Irreversible Isothermal Expansion of an ideal gas - - wherein may be equal to or may not be equal to , but work done is always calculated by As we have learned in isothermal expansion The correct graphs...

An ideal gasundergoes isothermal compression from 5 m3 against a constant external pressure of 4 Nm-2  Heat relesed in this process is used to increase the temprature of 1 mole of Al . If molar heat capacity of Al is 24 J mol-1 K-1,the temperture of Al increases by:

• Option 1)

$\frac{3}{2}K$

• Option 2)

$\frac{2}{3}K$

• Option 3)

1K

• Option 4)

2K

Work Done for Irreversible Isothermal Expansion of an ideal gas - - wherein may be equal to or may not be equal to , but work done is always calculated by   As we have learned in isothermal expansion. The work done on isothermal irreversible for ideal gas = -Pext   (V2 - V1) = -4N/m2  (1m3 - 5m3) =16 Nm For an isothermal process for ideal gas = -16 Nm = -16 J Heat used to increase...

The process with nrgative entropy change is

• Option 1)

sublimation of dry ice

• Option 2)

aadissolution of iodine in water

• Option 3)

Synthesis of ammonia from $N_{2}andH_{2}$

• Option 4)

Dissociation of$CaSO_{4\left ( s \right )}\: to\: CaO_{\left ( s \right )}\: and\: SO_{3\left ( g \right )}$

Second Law of Thermodynamics - The disorderness or entropy of universe increasing continuously.   - wherein    As we have learned in Enthalpy Change . When ammonia is treated with hydrogen .  Option 1)sublimation of dry iceOption 2)aadissolution of iodine in waterOption 3)Synthesis of ammonia from  Option 4)Dissociation of

A process has  $\Delta H=200\: Jmol^{-1}\: \: and\: \: \Delta S=40\: JK^{-1}mol^{-1}$. Out  of the values given below , choose the minimum temperature above which the process will be spontaneous :

• Option 1)

20 K

• Option 2)

12 K

• Option 3)

5 K

• Option 4)

4 K

Spontanous process -   - wherein For spontanouse process must be negative. As we have learned in Gibbs free energy  Option 1)20 KOption 2)12 KOption 3)5 K  Option 4)4 K

For a diatomic ideal gas in a closed system , which of the following plots does not correctly describe the relation between various thermodynamic quantities ?

• Option 1)

• Option 2)

• Option 3)

• Option 4)

Closed System - System which can exchange only energy but can not exchange matter with the surrounding - wherein Any matter in close container     Change in entropy for ideal gas in terms of C(p) - - wherein Molar heat capacity at constant pressure As ew have learned in closed system higher temperture,rotational degree of  freedom becomes active Variation of U  vs T is similar as Cv vs...

The entropy change associated with the conversion of 1kg of ice at 273 K to water vapours  at 383 K is:

( Specific heat of water liquid and water vapour are $4.2kJK^{-1}kg^{-1} and \: 2.0kJ K^{-1}kg^{-1}$ ; heat of liquid fusion and vapourisation of water are $334kJkg^{-1}\: and\: \: 2491kJkg^{-1}$ , respectively ). ( log 273 = 2.436 , log 373=2.572 , log 383 = 2.583)

• Option 1)

$2.64kJkg^{-1}K^{-1}$

$\: \:$

• Option 2)

$7.90kJkg^{-1}K^{-1}$

• Option 3)

$8.49 kJ kg^{-1}K^{-1}$

• Option 4)

$9.26\: \: kJkg^{-1}K^{-1}$

Entropy for phase transition at constant pressure - - wherein Transition Fusion, Vaporisition, Sublimation Enthalpy Internal Energy Transitional temperature     Entropy for solid and liquid - or   - wherein no. of moles molar heat capacity mass specific heat capacity   Phase change path  273 K                    273 K                 373 K            373 K                   383 K   ...
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