For the process H 2 O l → H 2 O g at T = 100 o C and 1 atmosphere pressure, the correct choice is-

∆ S s y s t e m > 0 a n d ∆ S s u r r o u n d i n g < 0

∆ S s y s t e m > 0 a n d ∆ S s u r r o u n d i n g > 0

∆ S s y s t e m < 0 a n d ∆ S s u r r o u n d i n g > 0

∆ S s y s t e m < 0 ∆ S s u r r o u n d i n g < 0

A reversible cyclic process for an ideal gas is shown below. Here, P , V , a n d T are pressure, volume and temperature, respectively. The thermodynamic parameters q , w , H a n d U are heat, work, enthalpy and internal energy, respectively. The correct option(s) is (are)

∆ H C A < ∆ U C A a n d q A C = ∆ U B C

q A C = ∆ U B C a n d w A B = P 2 V 2 - V 1

q B C = ∆ H A C a n d ∆ H C A > ∆ U C A

HARD

JEE Advanced

IMPORTANT

Earn 100

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.

$298 K : Δ_{f} H^{0} ({SnO}_{2} (s)) = - 581.0 kJ {mol}^{- 1}, Δ_{f} H^{0} ({CO}_{2} (g)) = - 394.0 kJ {mol}^{- 1}$

$S^{0} ({SnO}_{2} (s)) = 56.0 {JK}^{- 1} {mol}^{- 1}, S^{0} (Sn (s)) = 52.0 {JK}^{- 1} {mol}^{- 1}$

$S^{0} (C (s)) = 6.0 {JK}^{- 1} {mol}^{- 1}, S^{0} ({CO}_{2} (g)) = 210.0 {JK}^{- 1} {mol}^{- 1}$

Assume that the enthalpies and the entropies are temperature independent.

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Important Questions on Thermodynamics

HARD

JEE Advanced

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR CHEMISTRY>Chapter 4 - Thermodynamics>JEE Advanced Paper 2 - 2018>Q 4

For a reaction,

A ⇌ P

, the plots of

[A] a n d [P]

with time at temperatures

T_{1} a n d T_{2}

are given below.

Question Image

T_{2} > T_{1}

, the correct statement(s) is (are)

(Assume

∆ H^{⊝} a n d ∆ S^{⊝}

are independent of temperature and ratio of

\ln K a t T_{1} t o \ln K a t T_{2}

is greater than

T_{2} / T_{1}

. Here H, S , G and K are enthalpy, entropy, Gibbs energy and equilibrium constant, respectively.)

HARD

JEE Advanced

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR CHEMISTRY>Chapter 4 - Thermodynamics>JEE Advanced Paper 2 - 2018>Q 5

The surface of copper gets tarnished by the formation of copper oxide. $N_{2}$ gas was passed to prevent the oxide formation during heating of copper at 1250 K. However, the $N_{2}$ gas contains 1 mole % of water vapour as impurity. The water vapour oxidises copper as per the reaction given below:

$2 Cu (s) + H_{2} O (g) \to {Cu}_{2} O (s) + H_{2} (g)$

$P_{H_{2}}$ Is the minimum partial pressure of $H_{2}$ (in bar) needed to prevent the oxidation at $1250 K$ . The magnitude of value of $\ln (P_{H_{2}})$ is ____.

(Given: total pressure = 1 bar, R (universal gas constant) $= 8 J K^{- 1} {mol}^{- 1}, \ln (10) = 2 .3 . Cu (s) and {Cu}_{2} O (s)$ are mutually immiscible.

At $1250 K : 2 Cu (s) + \frac{1}{2} O_{2} (g) \to {Cu}_{2} O (s); ∆ G^{⊝} = - 78,000 J {mol}^{- 1}$

$H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (g); ∆ G^{⊝} = - 1,78,000 J {mol}^{- 1};$ )

Round off the answer up to the nearest integer.

HARD

JEE Advanced

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR CHEMISTRY>Chapter 4 - Thermodynamics>JEE Advanced Paper 2 - 2017>Q 6

The standard state Gibb's free energies of formation of

C

(graphite) and

C

(diamond) at

T = 298 K

are

Δ_{f} G^{o} [C (graphite] = 0 kJ {mol}^{- 1}

Δ_{f} G^{o} [C (diamond)] = 2 .9 kJ {mol}^{- 1}

The standard state means that the pressure should be 1 bar, and substance should be pure at a given temperature. The conversion of graphite [

C

(graphite)] to diamond [

C

(diamond)] reduces its volume by

2 \times 10^{- 6} m^{3} {mol}^{- 1}

. If

C

(graphite) is converted to

C

(diamond) isothermally at

T = 298 K

, the pressure at which

C

(graphite) is in equilibrium with

C

(diamond), is

[Useful information:

1 J = 1 kg m^{2} s^{- 2}, 1 Pa = 1 kg m^{- 1} s^{- 2}; 1 bar = 10^{5} Pa

]

HARD

JEE Advanced

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR CHEMISTRY>Chapter 4 - Thermodynamics>JEE Advanced Paper 2 - 2017>Q 7

For a reaction taking place in a container in equilibrium with its surroundings, the effect of temperature on its equilibrium constant

K

in terms of change in entropy is described by:

HARD

JEE Advanced

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR CHEMISTRY>Chapter 4 - Thermodynamics>JEE Advanced Paper 2 - 2015>Q 8

When 100 mL of 1.0 M HCl was mixed with 100 mL of 1.0 M NaOH in an insulated beaker at constant pressure, a temperature increase of

{5.7}^{o} C

was measured for the beaker and its contents (Expt.1). Because the enthalpy of neutralization of a strong acid with a strong base is a constant

(- 57.0 k J m o l^{- 1})

, this experiment could be used to measure the calorimeter constant. In a second experiment (Expt. 2), 100 mL of 2.0 M acetic acid

(K_{a} = 2.0 \times 10^{- 5})

was mixed with 100 mL of 1.0 M NaOH (under identical conditions to Expt. 1) where a temperature rise of

{5.6}^{o} C

was measured. (Consider heat capacity of all solutions as

4.2 J g^{- 1} K^{- 1}

and density of all solutions as

1.0 g m L^{- 1}

)

Enthalpy of dissociation

(i n k J m o l^{- 1})

of acetic acid obtained from the Expt. 2 is

MEDIUM

JEE Advanced

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR CHEMISTRY>Chapter 4 - Thermodynamics>JEE Advanced Paper 2 - 2014>Q 9

For the process

H_{2} O (l) \to H_{2} O (g)

T = 100^{o} C

and

1

atmosphere pressure, the correct choice is-

HARD

JEE Advanced

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR CHEMISTRY>Chapter 4 - Thermodynamics>JEE Advanced Paper 1 - 2021>Q 11

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)

representing the changes from state

I

to state

III

is(are)

( $p$ : pressure, $V$ : volume, $T$ : temperature, $H$ : enthalpy, $S$ : entropy)

HARD

JEE Advanced

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR CHEMISTRY>Chapter 4 - Thermodynamics>JEE Advanced Paper 1 - 2019>Q 12

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

Important Questions on Thermodynamics

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