QBManager

Thermodynamics

272764 For a process, entropy change of a system is expressed as

1

H - TS

2

\frac{q_{Tev}}{T}

3

\frac{T}{q_{rev}}

4

q_{rev} \times T

Explanation:

Entropy is a function of the state of the system , so the change in entropy of a system is determined by its initial and final states.
$Δ S_{5 y ytem} = \frac{q_{rz}}{T}$
Here, the system absorbs $q$ amount of heat from surrounding at temperature $T$ .

**[MHT CET-02.05.2019

Thermodynamics

272731 Among the following for spontaneity of chemical reaction there should be:

1 decrease in entropy and increase in free energy

2 decrease in entropy and free energy both

3 increase in entropy and decrease in free energy

4 increase in entropy and free energy both

Explanation:

From Gibbs - Helmholtz equation, $Δ G = Δ H - T Δ S$
For a spontaneous reaction at any temperature. $Δ G = -$ ve and it is possible when $Δ H = -$ ve and $Δ S = +$ ve Thus, for a spontaneous reaction, there must be increase in entropy and decrease in free energy.

MPPET - 2012

Thermodynamics

272732 The enthalpy of vaporization of a compound is $840 J / mole$ and its boiling point is $170 K$ . Its entropy of vaporization is

1

4.94 J / mol / K

2

12 J / mol / K

3

200 J / mol / K

4

49 J / mol / K

Explanation:

Given that $Δ H_{vap .} = 840 J / mol, T_{B . P} = 170 K$
We know that, $Δ S_{rap.} = \frac{Δ H_{rop.}}{T_{B . P}} = \frac{840}{170} = 4.94 J / mol / K$

Shift-I

Thermodynamics

272733 For an adiabatic change in a system, the condition which is applicable will be

1

w = 0

2

q = - w

3

q = w

4

q = 0

Explanation:

In adiabatic system there is neither heat exchange nor leaves the system (i.e. heat remains constant)
$Δ Q = Δ U + Δ W$
$Δ Q = Constant or Δ Q = 0$

COMEDK-2015

Thermodynamics

272735 A process is spontaneous at a given temperature if

1

Δ H = 0, Δ S < 0

2

Δ H > 0, Δ S < 0

3

Δ H < 0, Δ S > 0

4

Δ H > 0, Δ S = 0

Explanation:

$Δ G = Δ H - T Δ S$
A process is spontaneous if $Δ G$ is -ve, $Δ H$ is -ve and $Δ S$ is positive.

COMEDK-2017

Thermodynamics

272764 For a process, entropy change of a system is expressed as

1

H - TS

2

\frac{q_{Tev}}{T}

3

\frac{T}{q_{rev}}

4

q_{rev} \times T

Explanation:

Entropy is a function of the state of the system , so the change in entropy of a system is determined by its initial and final states.
$Δ S_{5 y ytem} = \frac{q_{rz}}{T}$
Here, the system absorbs $q$ amount of heat from surrounding at temperature $T$ .

**[MHT CET-02.05.2019

Thermodynamics

272731 Among the following for spontaneity of chemical reaction there should be:

1 decrease in entropy and increase in free energy

2 decrease in entropy and free energy both

3 increase in entropy and decrease in free energy

4 increase in entropy and free energy both

Explanation:

From Gibbs - Helmholtz equation, $Δ G = Δ H - T Δ S$
For a spontaneous reaction at any temperature. $Δ G = -$ ve and it is possible when $Δ H = -$ ve and $Δ S = +$ ve Thus, for a spontaneous reaction, there must be increase in entropy and decrease in free energy.

MPPET - 2012

Thermodynamics

272732 The enthalpy of vaporization of a compound is $840 J / mole$ and its boiling point is $170 K$ . Its entropy of vaporization is

1

4.94 J / mol / K

2

12 J / mol / K

3

200 J / mol / K

4

49 J / mol / K

Explanation:

Given that $Δ H_{vap .} = 840 J / mol, T_{B . P} = 170 K$
We know that, $Δ S_{rap.} = \frac{Δ H_{rop.}}{T_{B . P}} = \frac{840}{170} = 4.94 J / mol / K$

Shift-I

Thermodynamics

272733 For an adiabatic change in a system, the condition which is applicable will be

1

w = 0

2

q = - w

3

q = w

4

q = 0

Explanation:

In adiabatic system there is neither heat exchange nor leaves the system (i.e. heat remains constant)
$Δ Q = Δ U + Δ W$
$Δ Q = Constant or Δ Q = 0$

COMEDK-2015

Thermodynamics

272735 A process is spontaneous at a given temperature if

1

Δ H = 0, Δ S < 0

2

Δ H > 0, Δ S < 0

3

Δ H < 0, Δ S > 0

4

Δ H > 0, Δ S = 0

Explanation:

$Δ G = Δ H - T Δ S$
A process is spontaneous if $Δ G$ is -ve, $Δ H$ is -ve and $Δ S$ is positive.

COMEDK-2017

Thermodynamics

272764 For a process, entropy change of a system is expressed as

1

H - TS

2

\frac{q_{Tev}}{T}

3

\frac{T}{q_{rev}}

4

q_{rev} \times T

Explanation:

Entropy is a function of the state of the system , so the change in entropy of a system is determined by its initial and final states.
$Δ S_{5 y ytem} = \frac{q_{rz}}{T}$
Here, the system absorbs $q$ amount of heat from surrounding at temperature $T$ .

**[MHT CET-02.05.2019

Thermodynamics

272731 Among the following for spontaneity of chemical reaction there should be:

1 decrease in entropy and increase in free energy

2 decrease in entropy and free energy both

3 increase in entropy and decrease in free energy

4 increase in entropy and free energy both

Explanation:

From Gibbs - Helmholtz equation, $Δ G = Δ H - T Δ S$
For a spontaneous reaction at any temperature. $Δ G = -$ ve and it is possible when $Δ H = -$ ve and $Δ S = +$ ve Thus, for a spontaneous reaction, there must be increase in entropy and decrease in free energy.

MPPET - 2012

Thermodynamics

272732 The enthalpy of vaporization of a compound is $840 J / mole$ and its boiling point is $170 K$ . Its entropy of vaporization is

1

4.94 J / mol / K

2

12 J / mol / K

3

200 J / mol / K

4

49 J / mol / K

Explanation:

Given that $Δ H_{vap .} = 840 J / mol, T_{B . P} = 170 K$
We know that, $Δ S_{rap.} = \frac{Δ H_{rop.}}{T_{B . P}} = \frac{840}{170} = 4.94 J / mol / K$

Shift-I

Thermodynamics

272733 For an adiabatic change in a system, the condition which is applicable will be

1

w = 0

2

q = - w

3

q = w

4

q = 0

Explanation:

In adiabatic system there is neither heat exchange nor leaves the system (i.e. heat remains constant)
$Δ Q = Δ U + Δ W$
$Δ Q = Constant or Δ Q = 0$

COMEDK-2015

Thermodynamics

272735 A process is spontaneous at a given temperature if

1

Δ H = 0, Δ S < 0

2

Δ H > 0, Δ S < 0

3

Δ H < 0, Δ S > 0

4

Δ H > 0, Δ S = 0

Explanation:

$Δ G = Δ H - T Δ S$
A process is spontaneous if $Δ G$ is -ve, $Δ H$ is -ve and $Δ S$ is positive.

COMEDK-2017

Thermodynamics

272764 For a process, entropy change of a system is expressed as

1

H - TS

2

\frac{q_{Tev}}{T}

3

\frac{T}{q_{rev}}

4

q_{rev} \times T

Explanation:

Entropy is a function of the state of the system , so the change in entropy of a system is determined by its initial and final states.
$Δ S_{5 y ytem} = \frac{q_{rz}}{T}$
Here, the system absorbs $q$ amount of heat from surrounding at temperature $T$ .

**[MHT CET-02.05.2019

Thermodynamics

272731 Among the following for spontaneity of chemical reaction there should be:

1 decrease in entropy and increase in free energy

2 decrease in entropy and free energy both

3 increase in entropy and decrease in free energy

4 increase in entropy and free energy both

Explanation:

From Gibbs - Helmholtz equation, $Δ G = Δ H - T Δ S$
For a spontaneous reaction at any temperature. $Δ G = -$ ve and it is possible when $Δ H = -$ ve and $Δ S = +$ ve Thus, for a spontaneous reaction, there must be increase in entropy and decrease in free energy.

MPPET - 2012

Thermodynamics

272732 The enthalpy of vaporization of a compound is $840 J / mole$ and its boiling point is $170 K$ . Its entropy of vaporization is

1

4.94 J / mol / K

2

12 J / mol / K

3

200 J / mol / K

4

49 J / mol / K

Explanation:

Given that $Δ H_{vap .} = 840 J / mol, T_{B . P} = 170 K$
We know that, $Δ S_{rap.} = \frac{Δ H_{rop.}}{T_{B . P}} = \frac{840}{170} = 4.94 J / mol / K$

Shift-I

Thermodynamics

272733 For an adiabatic change in a system, the condition which is applicable will be

1

w = 0

2

q = - w

3

q = w

4

q = 0

Explanation:

In adiabatic system there is neither heat exchange nor leaves the system (i.e. heat remains constant)
$Δ Q = Δ U + Δ W$
$Δ Q = Constant or Δ Q = 0$

COMEDK-2015

Thermodynamics

272735 A process is spontaneous at a given temperature if

1

Δ H = 0, Δ S < 0

2

Δ H > 0, Δ S < 0

3

Δ H < 0, Δ S > 0

4

Δ H > 0, Δ S = 0

Explanation:

$Δ G = Δ H - T Δ S$
A process is spontaneous if $Δ G$ is -ve, $Δ H$ is -ve and $Δ S$ is positive.

COMEDK-2017

Thermodynamics

272764 For a process, entropy change of a system is expressed as

1

H - TS

2

\frac{q_{Tev}}{T}

3

\frac{T}{q_{rev}}

4

q_{rev} \times T

Explanation:

Entropy is a function of the state of the system , so the change in entropy of a system is determined by its initial and final states.
$Δ S_{5 y ytem} = \frac{q_{rz}}{T}$
Here, the system absorbs $q$ amount of heat from surrounding at temperature $T$ .

**[MHT CET-02.05.2019

Thermodynamics

272731 Among the following for spontaneity of chemical reaction there should be:

1 decrease in entropy and increase in free energy

2 decrease in entropy and free energy both

3 increase in entropy and decrease in free energy

4 increase in entropy and free energy both

Explanation:

From Gibbs - Helmholtz equation, $Δ G = Δ H - T Δ S$
For a spontaneous reaction at any temperature. $Δ G = -$ ve and it is possible when $Δ H = -$ ve and $Δ S = +$ ve Thus, for a spontaneous reaction, there must be increase in entropy and decrease in free energy.

MPPET - 2012

Thermodynamics

272732 The enthalpy of vaporization of a compound is $840 J / mole$ and its boiling point is $170 K$ . Its entropy of vaporization is

1

4.94 J / mol / K

2

12 J / mol / K

3

200 J / mol / K

4

49 J / mol / K

Explanation:

Given that $Δ H_{vap .} = 840 J / mol, T_{B . P} = 170 K$
We know that, $Δ S_{rap.} = \frac{Δ H_{rop.}}{T_{B . P}} = \frac{840}{170} = 4.94 J / mol / K$

Shift-I

Thermodynamics

272733 For an adiabatic change in a system, the condition which is applicable will be

1

w = 0

2

q = - w

3

q = w

4

q = 0

Explanation:

In adiabatic system there is neither heat exchange nor leaves the system (i.e. heat remains constant)
$Δ Q = Δ U + Δ W$
$Δ Q = Constant or Δ Q = 0$

COMEDK-2015

Thermodynamics

272735 A process is spontaneous at a given temperature if

1

Δ H = 0, Δ S < 0

2

Δ H > 0, Δ S < 0

3

Δ H < 0, Δ S > 0

4

Δ H > 0, Δ S = 0

Explanation:

$Δ G = Δ H - T Δ S$
A process is spontaneous if $Δ G$ is -ve, $Δ H$ is -ve and $Δ S$ is positive.

COMEDK-2017