272764
For a process, entropy change of a system is expressed as
1 $\mathrm{H}-\mathrm{TS}$
2 $\frac{\mathrm{q}_{\mathrm{Tev}}}{\mathrm{T}}$
3 $\frac{\mathrm{T}}{\mathrm{q}_{\mathrm{rev}}}$
4 $\mathrm{q}_{\mathrm{rev}} \times \mathrm{T}$
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. $\Delta \mathrm{S}_{5 y \text { ytem }}=\frac{\mathrm{q}_{\mathrm{rz}}}{\mathrm{T}}$ 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, $\Delta \mathrm{G}=\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S}$ For a spontaneous reaction at any temperature. $\Delta \mathrm{G}=-$ ve and it is possible when $\Delta H=-$ ve and $\Delta S=+$ 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 $840 \mathrm{~J} / \mathrm{mole}$ and its boiling point is $170 \mathrm{~K}$. Its entropy of vaporization is
1 $4.94 \mathrm{~J} / \mathrm{mol} / \mathrm{K}$
2 $12 \mathrm{~J} / \mathrm{mol} / \mathrm{K}$
3 $200 \mathrm{~J} / \mathrm{mol} / \mathrm{K}$
4 $49 \mathrm{~J} / \mathrm{mol} / \mathrm{K}$
Explanation:
Given that $\Delta \mathrm{H}_{\mathrm{vap.}}=840 \mathrm{~J} / \mathrm{mol}, \mathrm{T}_{\mathrm{B} . \mathrm{P}}=170 \mathrm{~K}$ We know that, $\Delta \mathrm{S}_{\text {rap. }}=\frac{\Delta \mathrm{H}_{\text {rop. }}}{\mathrm{T}_{\mathrm{B} . \mathrm{P}}}=\frac{840}{170}=4.94 \mathrm{~J} / \mathrm{mol} / \mathrm{K}$
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 $\mathrm{q}=\mathrm{w}$
4 $q=0$
Explanation:
In adiabatic system there is neither heat exchange nor leaves the system (i.e. heat remains constant) $\Delta \mathrm{Q}=\Delta \mathrm{U}+\Delta \mathrm{W}$ $\Delta \mathrm{Q}=\text { Constant or } \Delta \mathrm{Q}=0$
COMEDK-2015
Thermodynamics
272735
A process is spontaneous at a given temperature if
1 $\Delta \mathrm{H}=0, \Delta \mathrm{S}<0$
2 $\Delta \mathrm{H}>0, \Delta \mathrm{S}<0$
3 $\Delta \mathrm{H}<0, \Delta \mathrm{S}>0$
4 $\Delta \mathrm{H}>0, \Delta \mathrm{S}=0$
Explanation:
$\Delta \mathrm{G}=\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S}$ A process is spontaneous if $\Delta \mathrm{G}$ is -ve, $\Delta \mathrm{H}$ is -ve and $\Delta S$ is positive.
272764
For a process, entropy change of a system is expressed as
1 $\mathrm{H}-\mathrm{TS}$
2 $\frac{\mathrm{q}_{\mathrm{Tev}}}{\mathrm{T}}$
3 $\frac{\mathrm{T}}{\mathrm{q}_{\mathrm{rev}}}$
4 $\mathrm{q}_{\mathrm{rev}} \times \mathrm{T}$
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. $\Delta \mathrm{S}_{5 y \text { ytem }}=\frac{\mathrm{q}_{\mathrm{rz}}}{\mathrm{T}}$ 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, $\Delta \mathrm{G}=\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S}$ For a spontaneous reaction at any temperature. $\Delta \mathrm{G}=-$ ve and it is possible when $\Delta H=-$ ve and $\Delta S=+$ 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 $840 \mathrm{~J} / \mathrm{mole}$ and its boiling point is $170 \mathrm{~K}$. Its entropy of vaporization is
1 $4.94 \mathrm{~J} / \mathrm{mol} / \mathrm{K}$
2 $12 \mathrm{~J} / \mathrm{mol} / \mathrm{K}$
3 $200 \mathrm{~J} / \mathrm{mol} / \mathrm{K}$
4 $49 \mathrm{~J} / \mathrm{mol} / \mathrm{K}$
Explanation:
Given that $\Delta \mathrm{H}_{\mathrm{vap.}}=840 \mathrm{~J} / \mathrm{mol}, \mathrm{T}_{\mathrm{B} . \mathrm{P}}=170 \mathrm{~K}$ We know that, $\Delta \mathrm{S}_{\text {rap. }}=\frac{\Delta \mathrm{H}_{\text {rop. }}}{\mathrm{T}_{\mathrm{B} . \mathrm{P}}}=\frac{840}{170}=4.94 \mathrm{~J} / \mathrm{mol} / \mathrm{K}$
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 $\mathrm{q}=\mathrm{w}$
4 $q=0$
Explanation:
In adiabatic system there is neither heat exchange nor leaves the system (i.e. heat remains constant) $\Delta \mathrm{Q}=\Delta \mathrm{U}+\Delta \mathrm{W}$ $\Delta \mathrm{Q}=\text { Constant or } \Delta \mathrm{Q}=0$
COMEDK-2015
Thermodynamics
272735
A process is spontaneous at a given temperature if
1 $\Delta \mathrm{H}=0, \Delta \mathrm{S}<0$
2 $\Delta \mathrm{H}>0, \Delta \mathrm{S}<0$
3 $\Delta \mathrm{H}<0, \Delta \mathrm{S}>0$
4 $\Delta \mathrm{H}>0, \Delta \mathrm{S}=0$
Explanation:
$\Delta \mathrm{G}=\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S}$ A process is spontaneous if $\Delta \mathrm{G}$ is -ve, $\Delta \mathrm{H}$ is -ve and $\Delta S$ is positive.
272764
For a process, entropy change of a system is expressed as
1 $\mathrm{H}-\mathrm{TS}$
2 $\frac{\mathrm{q}_{\mathrm{Tev}}}{\mathrm{T}}$
3 $\frac{\mathrm{T}}{\mathrm{q}_{\mathrm{rev}}}$
4 $\mathrm{q}_{\mathrm{rev}} \times \mathrm{T}$
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. $\Delta \mathrm{S}_{5 y \text { ytem }}=\frac{\mathrm{q}_{\mathrm{rz}}}{\mathrm{T}}$ 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, $\Delta \mathrm{G}=\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S}$ For a spontaneous reaction at any temperature. $\Delta \mathrm{G}=-$ ve and it is possible when $\Delta H=-$ ve and $\Delta S=+$ 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 $840 \mathrm{~J} / \mathrm{mole}$ and its boiling point is $170 \mathrm{~K}$. Its entropy of vaporization is
1 $4.94 \mathrm{~J} / \mathrm{mol} / \mathrm{K}$
2 $12 \mathrm{~J} / \mathrm{mol} / \mathrm{K}$
3 $200 \mathrm{~J} / \mathrm{mol} / \mathrm{K}$
4 $49 \mathrm{~J} / \mathrm{mol} / \mathrm{K}$
Explanation:
Given that $\Delta \mathrm{H}_{\mathrm{vap.}}=840 \mathrm{~J} / \mathrm{mol}, \mathrm{T}_{\mathrm{B} . \mathrm{P}}=170 \mathrm{~K}$ We know that, $\Delta \mathrm{S}_{\text {rap. }}=\frac{\Delta \mathrm{H}_{\text {rop. }}}{\mathrm{T}_{\mathrm{B} . \mathrm{P}}}=\frac{840}{170}=4.94 \mathrm{~J} / \mathrm{mol} / \mathrm{K}$
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 $\mathrm{q}=\mathrm{w}$
4 $q=0$
Explanation:
In adiabatic system there is neither heat exchange nor leaves the system (i.e. heat remains constant) $\Delta \mathrm{Q}=\Delta \mathrm{U}+\Delta \mathrm{W}$ $\Delta \mathrm{Q}=\text { Constant or } \Delta \mathrm{Q}=0$
COMEDK-2015
Thermodynamics
272735
A process is spontaneous at a given temperature if
1 $\Delta \mathrm{H}=0, \Delta \mathrm{S}<0$
2 $\Delta \mathrm{H}>0, \Delta \mathrm{S}<0$
3 $\Delta \mathrm{H}<0, \Delta \mathrm{S}>0$
4 $\Delta \mathrm{H}>0, \Delta \mathrm{S}=0$
Explanation:
$\Delta \mathrm{G}=\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S}$ A process is spontaneous if $\Delta \mathrm{G}$ is -ve, $\Delta \mathrm{H}$ is -ve and $\Delta S$ is positive.
272764
For a process, entropy change of a system is expressed as
1 $\mathrm{H}-\mathrm{TS}$
2 $\frac{\mathrm{q}_{\mathrm{Tev}}}{\mathrm{T}}$
3 $\frac{\mathrm{T}}{\mathrm{q}_{\mathrm{rev}}}$
4 $\mathrm{q}_{\mathrm{rev}} \times \mathrm{T}$
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. $\Delta \mathrm{S}_{5 y \text { ytem }}=\frac{\mathrm{q}_{\mathrm{rz}}}{\mathrm{T}}$ 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, $\Delta \mathrm{G}=\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S}$ For a spontaneous reaction at any temperature. $\Delta \mathrm{G}=-$ ve and it is possible when $\Delta H=-$ ve and $\Delta S=+$ 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 $840 \mathrm{~J} / \mathrm{mole}$ and its boiling point is $170 \mathrm{~K}$. Its entropy of vaporization is
1 $4.94 \mathrm{~J} / \mathrm{mol} / \mathrm{K}$
2 $12 \mathrm{~J} / \mathrm{mol} / \mathrm{K}$
3 $200 \mathrm{~J} / \mathrm{mol} / \mathrm{K}$
4 $49 \mathrm{~J} / \mathrm{mol} / \mathrm{K}$
Explanation:
Given that $\Delta \mathrm{H}_{\mathrm{vap.}}=840 \mathrm{~J} / \mathrm{mol}, \mathrm{T}_{\mathrm{B} . \mathrm{P}}=170 \mathrm{~K}$ We know that, $\Delta \mathrm{S}_{\text {rap. }}=\frac{\Delta \mathrm{H}_{\text {rop. }}}{\mathrm{T}_{\mathrm{B} . \mathrm{P}}}=\frac{840}{170}=4.94 \mathrm{~J} / \mathrm{mol} / \mathrm{K}$
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 $\mathrm{q}=\mathrm{w}$
4 $q=0$
Explanation:
In adiabatic system there is neither heat exchange nor leaves the system (i.e. heat remains constant) $\Delta \mathrm{Q}=\Delta \mathrm{U}+\Delta \mathrm{W}$ $\Delta \mathrm{Q}=\text { Constant or } \Delta \mathrm{Q}=0$
COMEDK-2015
Thermodynamics
272735
A process is spontaneous at a given temperature if
1 $\Delta \mathrm{H}=0, \Delta \mathrm{S}<0$
2 $\Delta \mathrm{H}>0, \Delta \mathrm{S}<0$
3 $\Delta \mathrm{H}<0, \Delta \mathrm{S}>0$
4 $\Delta \mathrm{H}>0, \Delta \mathrm{S}=0$
Explanation:
$\Delta \mathrm{G}=\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S}$ A process is spontaneous if $\Delta \mathrm{G}$ is -ve, $\Delta \mathrm{H}$ is -ve and $\Delta S$ is positive.
272764
For a process, entropy change of a system is expressed as
1 $\mathrm{H}-\mathrm{TS}$
2 $\frac{\mathrm{q}_{\mathrm{Tev}}}{\mathrm{T}}$
3 $\frac{\mathrm{T}}{\mathrm{q}_{\mathrm{rev}}}$
4 $\mathrm{q}_{\mathrm{rev}} \times \mathrm{T}$
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. $\Delta \mathrm{S}_{5 y \text { ytem }}=\frac{\mathrm{q}_{\mathrm{rz}}}{\mathrm{T}}$ 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, $\Delta \mathrm{G}=\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S}$ For a spontaneous reaction at any temperature. $\Delta \mathrm{G}=-$ ve and it is possible when $\Delta H=-$ ve and $\Delta S=+$ 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 $840 \mathrm{~J} / \mathrm{mole}$ and its boiling point is $170 \mathrm{~K}$. Its entropy of vaporization is
1 $4.94 \mathrm{~J} / \mathrm{mol} / \mathrm{K}$
2 $12 \mathrm{~J} / \mathrm{mol} / \mathrm{K}$
3 $200 \mathrm{~J} / \mathrm{mol} / \mathrm{K}$
4 $49 \mathrm{~J} / \mathrm{mol} / \mathrm{K}$
Explanation:
Given that $\Delta \mathrm{H}_{\mathrm{vap.}}=840 \mathrm{~J} / \mathrm{mol}, \mathrm{T}_{\mathrm{B} . \mathrm{P}}=170 \mathrm{~K}$ We know that, $\Delta \mathrm{S}_{\text {rap. }}=\frac{\Delta \mathrm{H}_{\text {rop. }}}{\mathrm{T}_{\mathrm{B} . \mathrm{P}}}=\frac{840}{170}=4.94 \mathrm{~J} / \mathrm{mol} / \mathrm{K}$
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 $\mathrm{q}=\mathrm{w}$
4 $q=0$
Explanation:
In adiabatic system there is neither heat exchange nor leaves the system (i.e. heat remains constant) $\Delta \mathrm{Q}=\Delta \mathrm{U}+\Delta \mathrm{W}$ $\Delta \mathrm{Q}=\text { Constant or } \Delta \mathrm{Q}=0$
COMEDK-2015
Thermodynamics
272735
A process is spontaneous at a given temperature if
1 $\Delta \mathrm{H}=0, \Delta \mathrm{S}<0$
2 $\Delta \mathrm{H}>0, \Delta \mathrm{S}<0$
3 $\Delta \mathrm{H}<0, \Delta \mathrm{S}>0$
4 $\Delta \mathrm{H}>0, \Delta \mathrm{S}=0$
Explanation:
$\Delta \mathrm{G}=\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S}$ A process is spontaneous if $\Delta \mathrm{G}$ is -ve, $\Delta \mathrm{H}$ is -ve and $\Delta S$ is positive.
