369261
The enthalpy of vaporisation of chloroform is \(\mathrm{29.4 \mathrm{~kJ} / \mathrm{mol}}\) at its normal boiling point of \(\mathrm{61.7^{\circ} \mathrm{C}}\). What is the entropy of condensation of chloroform at this temperature ?
1 \({\rm{ - 57}}{\rm{.3}}\,{\rm{R}}\)
2 \({\rm{ - 10}}{\rm{.6}}\,{\rm{R}}\)
3 \({\rm{ - 1}}{\rm{.18}}\,{\rm{R}}\)
4 \({\rm{10}}{\rm{.6}}\,{\rm{R}}\)
Explanation:
\(\mathrm{\Delta_{v a p} H^{\circ}=29.4 \mathrm{~kJ} / \mathrm{mol}}\) and \(\rm{T_{\text {boiling }}=334.7 \mathrm{~K}}\) \(\mathrm{\Rightarrow \Delta_{v a p} S^{\circ}=\dfrac{\Delta_{v a p} H^{\circ}}{T_{\text {boiling }}}=87.84 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}=10.6 \mathrm{R}}\) \(\rm{\left(R=8.314 J^{-1} \mathrm{~mol}^{-1}\right)}\) \(\mathrm{\Rightarrow \Delta_{\text {condensation }} S^{\circ}=-10.6 R}\)
CHXI06:THERMODYNAMICS
369262
One mole of which of the following has the lowest entropy?
1 Mercury
2 Diamond
3 Liquid nitrogen
4 Hydrogen gas
Explanation:
Solids possess least entropy than liquids and gases.
CHXI06:THERMODYNAMICS
369263
In which of the following process, a maximum increase in entropy is observed?
1 Melting of ice
2 Sublimation of naphthalene
3 Condensation of water
4 Dissolution of salt in water
Explanation:
As sublimation involves the conversion of solid into gaseous state directly. Therefore there is a large/maximum increase in entropy as the highly ordered solid goes to highly (most) disordered gaseous state. Therefore \(\mathrm{\Delta \mathrm{S}=}\) maximum for the sublimation of naphthalene.
KCET - 2007
CHXI06:THERMODYNAMICS
369264
For which of the following processes, \(\mathrm{\Delta S}\) is negative?
\(\mathrm{N_{2}(g, 1 \mathrm{~atm}) \rightarrow N_{2}(\mathrm{~g}, 5 \mathrm{~atm})}\) \(\rm{\Delta S=\left(n C_{p} \ln \dfrac{T_{2}}{T_{1}}\right)+n R \ln \dfrac{V_{2}}{V_{1}}}\) For isothermal process, \(\mathrm{T_{1}=T_{2}}\) and \(\mathrm{\dfrac{V_{2}}{V_{1}}=\dfrac{P_{1}}{P_{2}}}\) \({\rm{(\Delta )S = 0 + nRln}}\frac{{{{\rm{P}}_{\rm{1}}}}}{{{{\rm{P}}_{\rm{2}}}}}{\rm{ = nRln}}\frac{{\rm{1}}}{{\rm{5}}}{\rm{ = - nRln}}\,{\rm{5}}\) \(\therefore {\rm{\Delta S < 0}}\)
JEE - 2018
CHXI06:THERMODYNAMICS
369265
If the enthalpy change for the transition of liquid water to steam is \(300 \mathrm{~kJ} \mathrm{~mol}^{-1}\) at \(27^{\circ} \mathrm{C}\), the entropy change for the process would be:
369261
The enthalpy of vaporisation of chloroform is \(\mathrm{29.4 \mathrm{~kJ} / \mathrm{mol}}\) at its normal boiling point of \(\mathrm{61.7^{\circ} \mathrm{C}}\). What is the entropy of condensation of chloroform at this temperature ?
1 \({\rm{ - 57}}{\rm{.3}}\,{\rm{R}}\)
2 \({\rm{ - 10}}{\rm{.6}}\,{\rm{R}}\)
3 \({\rm{ - 1}}{\rm{.18}}\,{\rm{R}}\)
4 \({\rm{10}}{\rm{.6}}\,{\rm{R}}\)
Explanation:
\(\mathrm{\Delta_{v a p} H^{\circ}=29.4 \mathrm{~kJ} / \mathrm{mol}}\) and \(\rm{T_{\text {boiling }}=334.7 \mathrm{~K}}\) \(\mathrm{\Rightarrow \Delta_{v a p} S^{\circ}=\dfrac{\Delta_{v a p} H^{\circ}}{T_{\text {boiling }}}=87.84 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}=10.6 \mathrm{R}}\) \(\rm{\left(R=8.314 J^{-1} \mathrm{~mol}^{-1}\right)}\) \(\mathrm{\Rightarrow \Delta_{\text {condensation }} S^{\circ}=-10.6 R}\)
CHXI06:THERMODYNAMICS
369262
One mole of which of the following has the lowest entropy?
1 Mercury
2 Diamond
3 Liquid nitrogen
4 Hydrogen gas
Explanation:
Solids possess least entropy than liquids and gases.
CHXI06:THERMODYNAMICS
369263
In which of the following process, a maximum increase in entropy is observed?
1 Melting of ice
2 Sublimation of naphthalene
3 Condensation of water
4 Dissolution of salt in water
Explanation:
As sublimation involves the conversion of solid into gaseous state directly. Therefore there is a large/maximum increase in entropy as the highly ordered solid goes to highly (most) disordered gaseous state. Therefore \(\mathrm{\Delta \mathrm{S}=}\) maximum for the sublimation of naphthalene.
KCET - 2007
CHXI06:THERMODYNAMICS
369264
For which of the following processes, \(\mathrm{\Delta S}\) is negative?
\(\mathrm{N_{2}(g, 1 \mathrm{~atm}) \rightarrow N_{2}(\mathrm{~g}, 5 \mathrm{~atm})}\) \(\rm{\Delta S=\left(n C_{p} \ln \dfrac{T_{2}}{T_{1}}\right)+n R \ln \dfrac{V_{2}}{V_{1}}}\) For isothermal process, \(\mathrm{T_{1}=T_{2}}\) and \(\mathrm{\dfrac{V_{2}}{V_{1}}=\dfrac{P_{1}}{P_{2}}}\) \({\rm{(\Delta )S = 0 + nRln}}\frac{{{{\rm{P}}_{\rm{1}}}}}{{{{\rm{P}}_{\rm{2}}}}}{\rm{ = nRln}}\frac{{\rm{1}}}{{\rm{5}}}{\rm{ = - nRln}}\,{\rm{5}}\) \(\therefore {\rm{\Delta S < 0}}\)
JEE - 2018
CHXI06:THERMODYNAMICS
369265
If the enthalpy change for the transition of liquid water to steam is \(300 \mathrm{~kJ} \mathrm{~mol}^{-1}\) at \(27^{\circ} \mathrm{C}\), the entropy change for the process would be:
369261
The enthalpy of vaporisation of chloroform is \(\mathrm{29.4 \mathrm{~kJ} / \mathrm{mol}}\) at its normal boiling point of \(\mathrm{61.7^{\circ} \mathrm{C}}\). What is the entropy of condensation of chloroform at this temperature ?
1 \({\rm{ - 57}}{\rm{.3}}\,{\rm{R}}\)
2 \({\rm{ - 10}}{\rm{.6}}\,{\rm{R}}\)
3 \({\rm{ - 1}}{\rm{.18}}\,{\rm{R}}\)
4 \({\rm{10}}{\rm{.6}}\,{\rm{R}}\)
Explanation:
\(\mathrm{\Delta_{v a p} H^{\circ}=29.4 \mathrm{~kJ} / \mathrm{mol}}\) and \(\rm{T_{\text {boiling }}=334.7 \mathrm{~K}}\) \(\mathrm{\Rightarrow \Delta_{v a p} S^{\circ}=\dfrac{\Delta_{v a p} H^{\circ}}{T_{\text {boiling }}}=87.84 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}=10.6 \mathrm{R}}\) \(\rm{\left(R=8.314 J^{-1} \mathrm{~mol}^{-1}\right)}\) \(\mathrm{\Rightarrow \Delta_{\text {condensation }} S^{\circ}=-10.6 R}\)
CHXI06:THERMODYNAMICS
369262
One mole of which of the following has the lowest entropy?
1 Mercury
2 Diamond
3 Liquid nitrogen
4 Hydrogen gas
Explanation:
Solids possess least entropy than liquids and gases.
CHXI06:THERMODYNAMICS
369263
In which of the following process, a maximum increase in entropy is observed?
1 Melting of ice
2 Sublimation of naphthalene
3 Condensation of water
4 Dissolution of salt in water
Explanation:
As sublimation involves the conversion of solid into gaseous state directly. Therefore there is a large/maximum increase in entropy as the highly ordered solid goes to highly (most) disordered gaseous state. Therefore \(\mathrm{\Delta \mathrm{S}=}\) maximum for the sublimation of naphthalene.
KCET - 2007
CHXI06:THERMODYNAMICS
369264
For which of the following processes, \(\mathrm{\Delta S}\) is negative?
\(\mathrm{N_{2}(g, 1 \mathrm{~atm}) \rightarrow N_{2}(\mathrm{~g}, 5 \mathrm{~atm})}\) \(\rm{\Delta S=\left(n C_{p} \ln \dfrac{T_{2}}{T_{1}}\right)+n R \ln \dfrac{V_{2}}{V_{1}}}\) For isothermal process, \(\mathrm{T_{1}=T_{2}}\) and \(\mathrm{\dfrac{V_{2}}{V_{1}}=\dfrac{P_{1}}{P_{2}}}\) \({\rm{(\Delta )S = 0 + nRln}}\frac{{{{\rm{P}}_{\rm{1}}}}}{{{{\rm{P}}_{\rm{2}}}}}{\rm{ = nRln}}\frac{{\rm{1}}}{{\rm{5}}}{\rm{ = - nRln}}\,{\rm{5}}\) \(\therefore {\rm{\Delta S < 0}}\)
JEE - 2018
CHXI06:THERMODYNAMICS
369265
If the enthalpy change for the transition of liquid water to steam is \(300 \mathrm{~kJ} \mathrm{~mol}^{-1}\) at \(27^{\circ} \mathrm{C}\), the entropy change for the process would be:
369261
The enthalpy of vaporisation of chloroform is \(\mathrm{29.4 \mathrm{~kJ} / \mathrm{mol}}\) at its normal boiling point of \(\mathrm{61.7^{\circ} \mathrm{C}}\). What is the entropy of condensation of chloroform at this temperature ?
1 \({\rm{ - 57}}{\rm{.3}}\,{\rm{R}}\)
2 \({\rm{ - 10}}{\rm{.6}}\,{\rm{R}}\)
3 \({\rm{ - 1}}{\rm{.18}}\,{\rm{R}}\)
4 \({\rm{10}}{\rm{.6}}\,{\rm{R}}\)
Explanation:
\(\mathrm{\Delta_{v a p} H^{\circ}=29.4 \mathrm{~kJ} / \mathrm{mol}}\) and \(\rm{T_{\text {boiling }}=334.7 \mathrm{~K}}\) \(\mathrm{\Rightarrow \Delta_{v a p} S^{\circ}=\dfrac{\Delta_{v a p} H^{\circ}}{T_{\text {boiling }}}=87.84 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}=10.6 \mathrm{R}}\) \(\rm{\left(R=8.314 J^{-1} \mathrm{~mol}^{-1}\right)}\) \(\mathrm{\Rightarrow \Delta_{\text {condensation }} S^{\circ}=-10.6 R}\)
CHXI06:THERMODYNAMICS
369262
One mole of which of the following has the lowest entropy?
1 Mercury
2 Diamond
3 Liquid nitrogen
4 Hydrogen gas
Explanation:
Solids possess least entropy than liquids and gases.
CHXI06:THERMODYNAMICS
369263
In which of the following process, a maximum increase in entropy is observed?
1 Melting of ice
2 Sublimation of naphthalene
3 Condensation of water
4 Dissolution of salt in water
Explanation:
As sublimation involves the conversion of solid into gaseous state directly. Therefore there is a large/maximum increase in entropy as the highly ordered solid goes to highly (most) disordered gaseous state. Therefore \(\mathrm{\Delta \mathrm{S}=}\) maximum for the sublimation of naphthalene.
KCET - 2007
CHXI06:THERMODYNAMICS
369264
For which of the following processes, \(\mathrm{\Delta S}\) is negative?
\(\mathrm{N_{2}(g, 1 \mathrm{~atm}) \rightarrow N_{2}(\mathrm{~g}, 5 \mathrm{~atm})}\) \(\rm{\Delta S=\left(n C_{p} \ln \dfrac{T_{2}}{T_{1}}\right)+n R \ln \dfrac{V_{2}}{V_{1}}}\) For isothermal process, \(\mathrm{T_{1}=T_{2}}\) and \(\mathrm{\dfrac{V_{2}}{V_{1}}=\dfrac{P_{1}}{P_{2}}}\) \({\rm{(\Delta )S = 0 + nRln}}\frac{{{{\rm{P}}_{\rm{1}}}}}{{{{\rm{P}}_{\rm{2}}}}}{\rm{ = nRln}}\frac{{\rm{1}}}{{\rm{5}}}{\rm{ = - nRln}}\,{\rm{5}}\) \(\therefore {\rm{\Delta S < 0}}\)
JEE - 2018
CHXI06:THERMODYNAMICS
369265
If the enthalpy change for the transition of liquid water to steam is \(300 \mathrm{~kJ} \mathrm{~mol}^{-1}\) at \(27^{\circ} \mathrm{C}\), the entropy change for the process would be:
369261
The enthalpy of vaporisation of chloroform is \(\mathrm{29.4 \mathrm{~kJ} / \mathrm{mol}}\) at its normal boiling point of \(\mathrm{61.7^{\circ} \mathrm{C}}\). What is the entropy of condensation of chloroform at this temperature ?
1 \({\rm{ - 57}}{\rm{.3}}\,{\rm{R}}\)
2 \({\rm{ - 10}}{\rm{.6}}\,{\rm{R}}\)
3 \({\rm{ - 1}}{\rm{.18}}\,{\rm{R}}\)
4 \({\rm{10}}{\rm{.6}}\,{\rm{R}}\)
Explanation:
\(\mathrm{\Delta_{v a p} H^{\circ}=29.4 \mathrm{~kJ} / \mathrm{mol}}\) and \(\rm{T_{\text {boiling }}=334.7 \mathrm{~K}}\) \(\mathrm{\Rightarrow \Delta_{v a p} S^{\circ}=\dfrac{\Delta_{v a p} H^{\circ}}{T_{\text {boiling }}}=87.84 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}=10.6 \mathrm{R}}\) \(\rm{\left(R=8.314 J^{-1} \mathrm{~mol}^{-1}\right)}\) \(\mathrm{\Rightarrow \Delta_{\text {condensation }} S^{\circ}=-10.6 R}\)
CHXI06:THERMODYNAMICS
369262
One mole of which of the following has the lowest entropy?
1 Mercury
2 Diamond
3 Liquid nitrogen
4 Hydrogen gas
Explanation:
Solids possess least entropy than liquids and gases.
CHXI06:THERMODYNAMICS
369263
In which of the following process, a maximum increase in entropy is observed?
1 Melting of ice
2 Sublimation of naphthalene
3 Condensation of water
4 Dissolution of salt in water
Explanation:
As sublimation involves the conversion of solid into gaseous state directly. Therefore there is a large/maximum increase in entropy as the highly ordered solid goes to highly (most) disordered gaseous state. Therefore \(\mathrm{\Delta \mathrm{S}=}\) maximum for the sublimation of naphthalene.
KCET - 2007
CHXI06:THERMODYNAMICS
369264
For which of the following processes, \(\mathrm{\Delta S}\) is negative?
\(\mathrm{N_{2}(g, 1 \mathrm{~atm}) \rightarrow N_{2}(\mathrm{~g}, 5 \mathrm{~atm})}\) \(\rm{\Delta S=\left(n C_{p} \ln \dfrac{T_{2}}{T_{1}}\right)+n R \ln \dfrac{V_{2}}{V_{1}}}\) For isothermal process, \(\mathrm{T_{1}=T_{2}}\) and \(\mathrm{\dfrac{V_{2}}{V_{1}}=\dfrac{P_{1}}{P_{2}}}\) \({\rm{(\Delta )S = 0 + nRln}}\frac{{{{\rm{P}}_{\rm{1}}}}}{{{{\rm{P}}_{\rm{2}}}}}{\rm{ = nRln}}\frac{{\rm{1}}}{{\rm{5}}}{\rm{ = - nRln}}\,{\rm{5}}\) \(\therefore {\rm{\Delta S < 0}}\)
JEE - 2018
CHXI06:THERMODYNAMICS
369265
If the enthalpy change for the transition of liquid water to steam is \(300 \mathrm{~kJ} \mathrm{~mol}^{-1}\) at \(27^{\circ} \mathrm{C}\), the entropy change for the process would be:
