QBManager

CHXI06:THERMODYNAMICS

369283 The molar heat capacity \(\mathrm{\left(C_{p}\right)}\) of \(\mathrm{\mathrm{CD}_{2} \mathrm{O}}\) is 10 cals at \(\mathrm{1000 \mathrm{~K}}\). The change in entropy associated with cooling of \(\mathrm{32 \mathrm{~g}}\) of \(\mathrm{\mathrm{CD}_{2} \mathrm{O}}\) vapour from 1000 \(\mathrm{\mathrm{K}}\) to \(\mathrm{100 \mathrm{~K}}\) at constant pressure will be: \(\mathrm{(\mathrm{D}=}\) deuterium, atomic mass \(\mathrm{=2 \mathrm{u}}\) )

1 \({\text{23}}{\text{.03}}\,{\text{calde}}{{\text{g}}^{{\text{ - 1}}}}\)

2 \({\rm{ - 23}}{\rm{.03}}{\mkern 1mu} {\rm{calde}}{{\rm{g}}^{{\rm{ - 1}}}}\)

3 \({\text{2}}{\text{.303}}\,{\text{calde}}{{\text{g}}^{{\text{ - 1}}}}\)

4 \({\rm{ - 2}}{\rm{.303}}{\mkern 1mu} {\rm{calde}}{{\rm{g}}^{{\rm{ - 1}}}}\)

JEE - 2014

CHXI06:THERMODYNAMICS

369284 For irreversible isothermal expansion of an ideal gas under isothermal condition, the correct option is :

1 \(\mathrm{\Delta U \neq 0, \Delta S_{\text {total }} \neq 0}\)

2 \(\mathrm{\Delta U=0, \Delta S_{\text {total }} \neq 0}\)

3 \(\mathrm{\Delta U \neq 0, \Delta S_{\text {total }}=0}\)

4 \(\mathrm{\Delta U=0, \Delta S_{\text {total }}=0}\)

NEET - 2021

CHXI06:THERMODYNAMICS

369285 When 2 moles of an ideal gas \(\mathrm{\left(C_{p, m}=\left(\dfrac{5}{2}\right) R\right)}\) is heated from \(\mathrm{300 \mathrm{~K}}\) to \(\mathrm{600 \mathrm{~K}}\) at constant pressure, then the change in entropy of gas \(\mathrm{(\Delta S)}\) is

1 \(\mathrm{\dfrac{3}{2} R \ln 2}\)

2 \(\mathrm{-\dfrac{3}{2} R \ln 2}\)

3 \(\mathrm{5 R \ln 2}\)

4 \(\mathrm{\dfrac{5}{2} R \ln 2}\)

CHXI06:THERMODYNAMICS

369286 In the process:
\({{{\rm{H}}_{\rm{2}}}{\rm{O}}({\rm{s}}, - {\rm{10}}^\circ {\rm{ C}},{\rm{1}}\;{\rm{atm}}) \to {{\rm{H}}_{\rm{2}}}{\rm{O}}({\rm{l}},{\rm{10}}^\circ {\rm{ C}},{\rm{1}}\;{\rm{atm}})}\)
\({\mathrm{\mathrm{C}_{P}}}\) for ice \({\mathrm{=9 \, \mathrm{cal} \, \mathrm{deg}^{-1} \mathrm{~mol}^{-1}, \mathrm{C}_{P}}}\) for \({\mathrm{\mathrm{H}_{2} \mathrm{O}=18 \, \mathrm{cal} \, \mathrm{deg}^{-1}}}\) \({\mathrm{\mathrm{mol}^{-1}}}\). Latent heat of fusion of ice \({\mathrm{=1440 \, \mathrm{cal} \, \mathrm{mol}^{-1}}}\) at \({\mathrm{0^{\circ} \mathrm{C}}}\). The entropy change for the above process is \({\rm{6}}.{\rm{258}}{\mkern 1mu} \,{\rm{cal}}\,{\rm{de}}{{\rm{g}}^{ - 1}}\) \({\mathrm{\mathrm{mol}^{-1}}}\).
Give the total number of steps in which the third law of thermodynamics is used.

1 1

2 2

3 3

4 5

Explanation:

Step 1. (using the third law of thermodynamics):
(For changing
\({{\rm{H}}_{\rm{2}}}{\rm{O(s,}}\left( { - {\rm{1}}{{\rm{0}}^{\rm{^\circ }}}{\rm{C,1\;atm}}} \right) \to {{\rm{H}}_{\rm{2}}}{\rm{O}}\left( {{\rm{s,}}{{\rm{0}}^{\rm{^\circ }}}{\rm{C,}}\,{\rm{1\;atm}}} \right)\)
\({\rm{\Delta }}{{\rm{S}}_{\rm{1}}} = \int_{{\rm{ - 10}}}^{\rm{0}} {\rm{n}} \frac{{{{\rm{C}}_{\rm{P}}}}}{{\rm{T}}}\;{\rm{dT}} = 1 \times 9 \times 2.3 \times {\rm{log}}\frac{{273}}{{263}}\)
\(\,\,\,\,\,\,\,\,\,\, = 0.336\,\,{\rm{cal}}\,{\rm{de}}{{\rm{g}}^{ - 1}}\;{\rm{mo}}{{\rm{l}}^{ - 1}}\)
Step 2 (using the second law of thermodynamics):
\({\mathrm{\mathrm{H}_{2} \mathrm{O}(\mathrm{s})\left(0^{\circ} \mathrm{C}, 1 \mathrm{~atm}\right) \longrightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{l})\left(0^{\circ} \mathrm{C}, 1 \mathrm{~atm}\right)}}\)
\({\mathrm{\Delta S_{2}=\dfrac{q_{\text {rev }}}{T}=\dfrac{1440}{273}=5.258 \, \mathrm{cal} \, \mathrm{deg}^{-1} \mathrm{~mol}^{-1}}}\)
Step 3 (using the third law of themrodynamics):
\({\mathrm{\mathrm{H}_{2} \mathrm{O}(\mathrm{l})\left(0^{\circ} \mathrm{C}, 1 \mathrm{~atm}\right) \longrightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{l})\left(10^{\circ} \mathrm{C}, 1 \mathrm{~atm}\right)}}\)
\({\rm{\Delta }}{{\rm{S}}_{\rm{3}}} = \int_{\rm{0}}^{{\rm{10}}} {\rm{n}} \frac{{{{\rm{C}}_{\rm{P}}}}}{{\rm{T}}}\;{\rm{dT}} = 1 \times 18 \times 2.3 \times {\rm{log}}{\mkern 1mu} {\mkern 1mu} \frac{{283}}{{273}}\)
\(\,\,\,\,\,\,\,\,\,\,\, = 0.647\,{\rm{cal}}\,{\rm{de}}{{\rm{g}}^{ - 1}}\;{\rm{mo}}{{\rm{l}}^{ - 1}}\)
\({\rm{\Delta S}}\,{\rm{ = }}\,{\rm{\Delta }}{{\rm{S}}_{\rm{1}}}{\rm{ + }}\,{\rm{\Delta }}{{\rm{S}}_{\rm{2}}}\,{\rm{ + }}\,{\rm{\Delta }}{{\rm{S}}_{\rm{3}}}\)
\( = 0.336 + 5.258 + 0.647\)
\( = 6.258\,\,{\rm{cal}}\,\,{\rm{de}}{{\rm{g}}^{ - 1}}\;{\rm{mo}}{{\rm{l}}^{ - 1}}\)
From above calculations, number of steps which is based on third law of thermodynamics are "2".

CHXI06:THERMODYNAMICS

369283 The molar heat capacity \(\mathrm{\left(C_{p}\right)}\) of \(\mathrm{\mathrm{CD}_{2} \mathrm{O}}\) is 10 cals at \(\mathrm{1000 \mathrm{~K}}\). The change in entropy associated with cooling of \(\mathrm{32 \mathrm{~g}}\) of \(\mathrm{\mathrm{CD}_{2} \mathrm{O}}\) vapour from 1000 \(\mathrm{\mathrm{K}}\) to \(\mathrm{100 \mathrm{~K}}\) at constant pressure will be: \(\mathrm{(\mathrm{D}=}\) deuterium, atomic mass \(\mathrm{=2 \mathrm{u}}\) )

1 \({\text{23}}{\text{.03}}\,{\text{calde}}{{\text{g}}^{{\text{ - 1}}}}\)

2 \({\rm{ - 23}}{\rm{.03}}{\mkern 1mu} {\rm{calde}}{{\rm{g}}^{{\rm{ - 1}}}}\)

3 \({\text{2}}{\text{.303}}\,{\text{calde}}{{\text{g}}^{{\text{ - 1}}}}\)

4 \({\rm{ - 2}}{\rm{.303}}{\mkern 1mu} {\rm{calde}}{{\rm{g}}^{{\rm{ - 1}}}}\)

JEE - 2014

CHXI06:THERMODYNAMICS

369284 For irreversible isothermal expansion of an ideal gas under isothermal condition, the correct option is :

1 \(\mathrm{\Delta U \neq 0, \Delta S_{\text {total }} \neq 0}\)

2 \(\mathrm{\Delta U=0, \Delta S_{\text {total }} \neq 0}\)

3 \(\mathrm{\Delta U \neq 0, \Delta S_{\text {total }}=0}\)

4 \(\mathrm{\Delta U=0, \Delta S_{\text {total }}=0}\)

NEET - 2021

CHXI06:THERMODYNAMICS

369285 When 2 moles of an ideal gas \(\mathrm{\left(C_{p, m}=\left(\dfrac{5}{2}\right) R\right)}\) is heated from \(\mathrm{300 \mathrm{~K}}\) to \(\mathrm{600 \mathrm{~K}}\) at constant pressure, then the change in entropy of gas \(\mathrm{(\Delta S)}\) is

1 \(\mathrm{\dfrac{3}{2} R \ln 2}\)

2 \(\mathrm{-\dfrac{3}{2} R \ln 2}\)

3 \(\mathrm{5 R \ln 2}\)

4 \(\mathrm{\dfrac{5}{2} R \ln 2}\)

CHXI06:THERMODYNAMICS

369286 In the process:
\({{{\rm{H}}_{\rm{2}}}{\rm{O}}({\rm{s}}, - {\rm{10}}^\circ {\rm{ C}},{\rm{1}}\;{\rm{atm}}) \to {{\rm{H}}_{\rm{2}}}{\rm{O}}({\rm{l}},{\rm{10}}^\circ {\rm{ C}},{\rm{1}}\;{\rm{atm}})}\)
\({\mathrm{\mathrm{C}_{P}}}\) for ice \({\mathrm{=9 \, \mathrm{cal} \, \mathrm{deg}^{-1} \mathrm{~mol}^{-1}, \mathrm{C}_{P}}}\) for \({\mathrm{\mathrm{H}_{2} \mathrm{O}=18 \, \mathrm{cal} \, \mathrm{deg}^{-1}}}\) \({\mathrm{\mathrm{mol}^{-1}}}\). Latent heat of fusion of ice \({\mathrm{=1440 \, \mathrm{cal} \, \mathrm{mol}^{-1}}}\) at \({\mathrm{0^{\circ} \mathrm{C}}}\). The entropy change for the above process is \({\rm{6}}.{\rm{258}}{\mkern 1mu} \,{\rm{cal}}\,{\rm{de}}{{\rm{g}}^{ - 1}}\) \({\mathrm{\mathrm{mol}^{-1}}}\).
Give the total number of steps in which the third law of thermodynamics is used.

1 1

2 2

3 3

4 5

Explanation:

Step 1. (using the third law of thermodynamics):
(For changing
\({{\rm{H}}_{\rm{2}}}{\rm{O(s,}}\left( { - {\rm{1}}{{\rm{0}}^{\rm{^\circ }}}{\rm{C,1\;atm}}} \right) \to {{\rm{H}}_{\rm{2}}}{\rm{O}}\left( {{\rm{s,}}{{\rm{0}}^{\rm{^\circ }}}{\rm{C,}}\,{\rm{1\;atm}}} \right)\)
\({\rm{\Delta }}{{\rm{S}}_{\rm{1}}} = \int_{{\rm{ - 10}}}^{\rm{0}} {\rm{n}} \frac{{{{\rm{C}}_{\rm{P}}}}}{{\rm{T}}}\;{\rm{dT}} = 1 \times 9 \times 2.3 \times {\rm{log}}\frac{{273}}{{263}}\)
\(\,\,\,\,\,\,\,\,\,\, = 0.336\,\,{\rm{cal}}\,{\rm{de}}{{\rm{g}}^{ - 1}}\;{\rm{mo}}{{\rm{l}}^{ - 1}}\)
Step 2 (using the second law of thermodynamics):
\({\mathrm{\mathrm{H}_{2} \mathrm{O}(\mathrm{s})\left(0^{\circ} \mathrm{C}, 1 \mathrm{~atm}\right) \longrightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{l})\left(0^{\circ} \mathrm{C}, 1 \mathrm{~atm}\right)}}\)
\({\mathrm{\Delta S_{2}=\dfrac{q_{\text {rev }}}{T}=\dfrac{1440}{273}=5.258 \, \mathrm{cal} \, \mathrm{deg}^{-1} \mathrm{~mol}^{-1}}}\)
Step 3 (using the third law of themrodynamics):
\({\mathrm{\mathrm{H}_{2} \mathrm{O}(\mathrm{l})\left(0^{\circ} \mathrm{C}, 1 \mathrm{~atm}\right) \longrightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{l})\left(10^{\circ} \mathrm{C}, 1 \mathrm{~atm}\right)}}\)
\({\rm{\Delta }}{{\rm{S}}_{\rm{3}}} = \int_{\rm{0}}^{{\rm{10}}} {\rm{n}} \frac{{{{\rm{C}}_{\rm{P}}}}}{{\rm{T}}}\;{\rm{dT}} = 1 \times 18 \times 2.3 \times {\rm{log}}{\mkern 1mu} {\mkern 1mu} \frac{{283}}{{273}}\)
\(\,\,\,\,\,\,\,\,\,\,\, = 0.647\,{\rm{cal}}\,{\rm{de}}{{\rm{g}}^{ - 1}}\;{\rm{mo}}{{\rm{l}}^{ - 1}}\)
\({\rm{\Delta S}}\,{\rm{ = }}\,{\rm{\Delta }}{{\rm{S}}_{\rm{1}}}{\rm{ + }}\,{\rm{\Delta }}{{\rm{S}}_{\rm{2}}}\,{\rm{ + }}\,{\rm{\Delta }}{{\rm{S}}_{\rm{3}}}\)
\( = 0.336 + 5.258 + 0.647\)
\( = 6.258\,\,{\rm{cal}}\,\,{\rm{de}}{{\rm{g}}^{ - 1}}\;{\rm{mo}}{{\rm{l}}^{ - 1}}\)
From above calculations, number of steps which is based on third law of thermodynamics are "2".

CHXI06:THERMODYNAMICS

369283 The molar heat capacity \(\mathrm{\left(C_{p}\right)}\) of \(\mathrm{\mathrm{CD}_{2} \mathrm{O}}\) is 10 cals at \(\mathrm{1000 \mathrm{~K}}\). The change in entropy associated with cooling of \(\mathrm{32 \mathrm{~g}}\) of \(\mathrm{\mathrm{CD}_{2} \mathrm{O}}\) vapour from 1000 \(\mathrm{\mathrm{K}}\) to \(\mathrm{100 \mathrm{~K}}\) at constant pressure will be: \(\mathrm{(\mathrm{D}=}\) deuterium, atomic mass \(\mathrm{=2 \mathrm{u}}\) )

1 \({\text{23}}{\text{.03}}\,{\text{calde}}{{\text{g}}^{{\text{ - 1}}}}\)

2 \({\rm{ - 23}}{\rm{.03}}{\mkern 1mu} {\rm{calde}}{{\rm{g}}^{{\rm{ - 1}}}}\)

3 \({\text{2}}{\text{.303}}\,{\text{calde}}{{\text{g}}^{{\text{ - 1}}}}\)

4 \({\rm{ - 2}}{\rm{.303}}{\mkern 1mu} {\rm{calde}}{{\rm{g}}^{{\rm{ - 1}}}}\)

JEE - 2014

CHXI06:THERMODYNAMICS

369284 For irreversible isothermal expansion of an ideal gas under isothermal condition, the correct option is :

1 \(\mathrm{\Delta U \neq 0, \Delta S_{\text {total }} \neq 0}\)

2 \(\mathrm{\Delta U=0, \Delta S_{\text {total }} \neq 0}\)

3 \(\mathrm{\Delta U \neq 0, \Delta S_{\text {total }}=0}\)

4 \(\mathrm{\Delta U=0, \Delta S_{\text {total }}=0}\)

NEET - 2021

CHXI06:THERMODYNAMICS

369285 When 2 moles of an ideal gas \(\mathrm{\left(C_{p, m}=\left(\dfrac{5}{2}\right) R\right)}\) is heated from \(\mathrm{300 \mathrm{~K}}\) to \(\mathrm{600 \mathrm{~K}}\) at constant pressure, then the change in entropy of gas \(\mathrm{(\Delta S)}\) is

1 \(\mathrm{\dfrac{3}{2} R \ln 2}\)

2 \(\mathrm{-\dfrac{3}{2} R \ln 2}\)

3 \(\mathrm{5 R \ln 2}\)

4 \(\mathrm{\dfrac{5}{2} R \ln 2}\)

CHXI06:THERMODYNAMICS

369286 In the process:
\({{{\rm{H}}_{\rm{2}}}{\rm{O}}({\rm{s}}, - {\rm{10}}^\circ {\rm{ C}},{\rm{1}}\;{\rm{atm}}) \to {{\rm{H}}_{\rm{2}}}{\rm{O}}({\rm{l}},{\rm{10}}^\circ {\rm{ C}},{\rm{1}}\;{\rm{atm}})}\)
\({\mathrm{\mathrm{C}_{P}}}\) for ice \({\mathrm{=9 \, \mathrm{cal} \, \mathrm{deg}^{-1} \mathrm{~mol}^{-1}, \mathrm{C}_{P}}}\) for \({\mathrm{\mathrm{H}_{2} \mathrm{O}=18 \, \mathrm{cal} \, \mathrm{deg}^{-1}}}\) \({\mathrm{\mathrm{mol}^{-1}}}\). Latent heat of fusion of ice \({\mathrm{=1440 \, \mathrm{cal} \, \mathrm{mol}^{-1}}}\) at \({\mathrm{0^{\circ} \mathrm{C}}}\). The entropy change for the above process is \({\rm{6}}.{\rm{258}}{\mkern 1mu} \,{\rm{cal}}\,{\rm{de}}{{\rm{g}}^{ - 1}}\) \({\mathrm{\mathrm{mol}^{-1}}}\).
Give the total number of steps in which the third law of thermodynamics is used.

1 1

2 2

3 3

4 5

Explanation:

Step 1. (using the third law of thermodynamics):
(For changing
\({{\rm{H}}_{\rm{2}}}{\rm{O(s,}}\left( { - {\rm{1}}{{\rm{0}}^{\rm{^\circ }}}{\rm{C,1\;atm}}} \right) \to {{\rm{H}}_{\rm{2}}}{\rm{O}}\left( {{\rm{s,}}{{\rm{0}}^{\rm{^\circ }}}{\rm{C,}}\,{\rm{1\;atm}}} \right)\)
\({\rm{\Delta }}{{\rm{S}}_{\rm{1}}} = \int_{{\rm{ - 10}}}^{\rm{0}} {\rm{n}} \frac{{{{\rm{C}}_{\rm{P}}}}}{{\rm{T}}}\;{\rm{dT}} = 1 \times 9 \times 2.3 \times {\rm{log}}\frac{{273}}{{263}}\)
\(\,\,\,\,\,\,\,\,\,\, = 0.336\,\,{\rm{cal}}\,{\rm{de}}{{\rm{g}}^{ - 1}}\;{\rm{mo}}{{\rm{l}}^{ - 1}}\)
Step 2 (using the second law of thermodynamics):
\({\mathrm{\mathrm{H}_{2} \mathrm{O}(\mathrm{s})\left(0^{\circ} \mathrm{C}, 1 \mathrm{~atm}\right) \longrightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{l})\left(0^{\circ} \mathrm{C}, 1 \mathrm{~atm}\right)}}\)
\({\mathrm{\Delta S_{2}=\dfrac{q_{\text {rev }}}{T}=\dfrac{1440}{273}=5.258 \, \mathrm{cal} \, \mathrm{deg}^{-1} \mathrm{~mol}^{-1}}}\)
Step 3 (using the third law of themrodynamics):
\({\mathrm{\mathrm{H}_{2} \mathrm{O}(\mathrm{l})\left(0^{\circ} \mathrm{C}, 1 \mathrm{~atm}\right) \longrightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{l})\left(10^{\circ} \mathrm{C}, 1 \mathrm{~atm}\right)}}\)
\({\rm{\Delta }}{{\rm{S}}_{\rm{3}}} = \int_{\rm{0}}^{{\rm{10}}} {\rm{n}} \frac{{{{\rm{C}}_{\rm{P}}}}}{{\rm{T}}}\;{\rm{dT}} = 1 \times 18 \times 2.3 \times {\rm{log}}{\mkern 1mu} {\mkern 1mu} \frac{{283}}{{273}}\)
\(\,\,\,\,\,\,\,\,\,\,\, = 0.647\,{\rm{cal}}\,{\rm{de}}{{\rm{g}}^{ - 1}}\;{\rm{mo}}{{\rm{l}}^{ - 1}}\)
\({\rm{\Delta S}}\,{\rm{ = }}\,{\rm{\Delta }}{{\rm{S}}_{\rm{1}}}{\rm{ + }}\,{\rm{\Delta }}{{\rm{S}}_{\rm{2}}}\,{\rm{ + }}\,{\rm{\Delta }}{{\rm{S}}_{\rm{3}}}\)
\( = 0.336 + 5.258 + 0.647\)
\( = 6.258\,\,{\rm{cal}}\,\,{\rm{de}}{{\rm{g}}^{ - 1}}\;{\rm{mo}}{{\rm{l}}^{ - 1}}\)
From above calculations, number of steps which is based on third law of thermodynamics are "2".

CHXI06:THERMODYNAMICS

369283 The molar heat capacity \(\mathrm{\left(C_{p}\right)}\) of \(\mathrm{\mathrm{CD}_{2} \mathrm{O}}\) is 10 cals at \(\mathrm{1000 \mathrm{~K}}\). The change in entropy associated with cooling of \(\mathrm{32 \mathrm{~g}}\) of \(\mathrm{\mathrm{CD}_{2} \mathrm{O}}\) vapour from 1000 \(\mathrm{\mathrm{K}}\) to \(\mathrm{100 \mathrm{~K}}\) at constant pressure will be: \(\mathrm{(\mathrm{D}=}\) deuterium, atomic mass \(\mathrm{=2 \mathrm{u}}\) )

1 \({\text{23}}{\text{.03}}\,{\text{calde}}{{\text{g}}^{{\text{ - 1}}}}\)

2 \({\rm{ - 23}}{\rm{.03}}{\mkern 1mu} {\rm{calde}}{{\rm{g}}^{{\rm{ - 1}}}}\)

3 \({\text{2}}{\text{.303}}\,{\text{calde}}{{\text{g}}^{{\text{ - 1}}}}\)

4 \({\rm{ - 2}}{\rm{.303}}{\mkern 1mu} {\rm{calde}}{{\rm{g}}^{{\rm{ - 1}}}}\)

JEE - 2014

CHXI06:THERMODYNAMICS

369284 For irreversible isothermal expansion of an ideal gas under isothermal condition, the correct option is :

1 \(\mathrm{\Delta U \neq 0, \Delta S_{\text {total }} \neq 0}\)

2 \(\mathrm{\Delta U=0, \Delta S_{\text {total }} \neq 0}\)

3 \(\mathrm{\Delta U \neq 0, \Delta S_{\text {total }}=0}\)

4 \(\mathrm{\Delta U=0, \Delta S_{\text {total }}=0}\)

NEET - 2021

CHXI06:THERMODYNAMICS

369285 When 2 moles of an ideal gas \(\mathrm{\left(C_{p, m}=\left(\dfrac{5}{2}\right) R\right)}\) is heated from \(\mathrm{300 \mathrm{~K}}\) to \(\mathrm{600 \mathrm{~K}}\) at constant pressure, then the change in entropy of gas \(\mathrm{(\Delta S)}\) is

1 \(\mathrm{\dfrac{3}{2} R \ln 2}\)

2 \(\mathrm{-\dfrac{3}{2} R \ln 2}\)

3 \(\mathrm{5 R \ln 2}\)

4 \(\mathrm{\dfrac{5}{2} R \ln 2}\)

CHXI06:THERMODYNAMICS

369286 In the process:
\({{{\rm{H}}_{\rm{2}}}{\rm{O}}({\rm{s}}, - {\rm{10}}^\circ {\rm{ C}},{\rm{1}}\;{\rm{atm}}) \to {{\rm{H}}_{\rm{2}}}{\rm{O}}({\rm{l}},{\rm{10}}^\circ {\rm{ C}},{\rm{1}}\;{\rm{atm}})}\)
\({\mathrm{\mathrm{C}_{P}}}\) for ice \({\mathrm{=9 \, \mathrm{cal} \, \mathrm{deg}^{-1} \mathrm{~mol}^{-1}, \mathrm{C}_{P}}}\) for \({\mathrm{\mathrm{H}_{2} \mathrm{O}=18 \, \mathrm{cal} \, \mathrm{deg}^{-1}}}\) \({\mathrm{\mathrm{mol}^{-1}}}\). Latent heat of fusion of ice \({\mathrm{=1440 \, \mathrm{cal} \, \mathrm{mol}^{-1}}}\) at \({\mathrm{0^{\circ} \mathrm{C}}}\). The entropy change for the above process is \({\rm{6}}.{\rm{258}}{\mkern 1mu} \,{\rm{cal}}\,{\rm{de}}{{\rm{g}}^{ - 1}}\) \({\mathrm{\mathrm{mol}^{-1}}}\).
Give the total number of steps in which the third law of thermodynamics is used.

1 1

2 2

3 3

4 5

Explanation:

Step 1. (using the third law of thermodynamics):
(For changing
\({{\rm{H}}_{\rm{2}}}{\rm{O(s,}}\left( { - {\rm{1}}{{\rm{0}}^{\rm{^\circ }}}{\rm{C,1\;atm}}} \right) \to {{\rm{H}}_{\rm{2}}}{\rm{O}}\left( {{\rm{s,}}{{\rm{0}}^{\rm{^\circ }}}{\rm{C,}}\,{\rm{1\;atm}}} \right)\)
\({\rm{\Delta }}{{\rm{S}}_{\rm{1}}} = \int_{{\rm{ - 10}}}^{\rm{0}} {\rm{n}} \frac{{{{\rm{C}}_{\rm{P}}}}}{{\rm{T}}}\;{\rm{dT}} = 1 \times 9 \times 2.3 \times {\rm{log}}\frac{{273}}{{263}}\)
\(\,\,\,\,\,\,\,\,\,\, = 0.336\,\,{\rm{cal}}\,{\rm{de}}{{\rm{g}}^{ - 1}}\;{\rm{mo}}{{\rm{l}}^{ - 1}}\)
Step 2 (using the second law of thermodynamics):
\({\mathrm{\mathrm{H}_{2} \mathrm{O}(\mathrm{s})\left(0^{\circ} \mathrm{C}, 1 \mathrm{~atm}\right) \longrightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{l})\left(0^{\circ} \mathrm{C}, 1 \mathrm{~atm}\right)}}\)
\({\mathrm{\Delta S_{2}=\dfrac{q_{\text {rev }}}{T}=\dfrac{1440}{273}=5.258 \, \mathrm{cal} \, \mathrm{deg}^{-1} \mathrm{~mol}^{-1}}}\)
Step 3 (using the third law of themrodynamics):
\({\mathrm{\mathrm{H}_{2} \mathrm{O}(\mathrm{l})\left(0^{\circ} \mathrm{C}, 1 \mathrm{~atm}\right) \longrightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{l})\left(10^{\circ} \mathrm{C}, 1 \mathrm{~atm}\right)}}\)
\({\rm{\Delta }}{{\rm{S}}_{\rm{3}}} = \int_{\rm{0}}^{{\rm{10}}} {\rm{n}} \frac{{{{\rm{C}}_{\rm{P}}}}}{{\rm{T}}}\;{\rm{dT}} = 1 \times 18 \times 2.3 \times {\rm{log}}{\mkern 1mu} {\mkern 1mu} \frac{{283}}{{273}}\)
\(\,\,\,\,\,\,\,\,\,\,\, = 0.647\,{\rm{cal}}\,{\rm{de}}{{\rm{g}}^{ - 1}}\;{\rm{mo}}{{\rm{l}}^{ - 1}}\)
\({\rm{\Delta S}}\,{\rm{ = }}\,{\rm{\Delta }}{{\rm{S}}_{\rm{1}}}{\rm{ + }}\,{\rm{\Delta }}{{\rm{S}}_{\rm{2}}}\,{\rm{ + }}\,{\rm{\Delta }}{{\rm{S}}_{\rm{3}}}\)
\( = 0.336 + 5.258 + 0.647\)
\( = 6.258\,\,{\rm{cal}}\,\,{\rm{de}}{{\rm{g}}^{ - 1}}\;{\rm{mo}}{{\rm{l}}^{ - 1}}\)
From above calculations, number of steps which is based on third law of thermodynamics are "2".

Explanation:

Explanation:

Explanation:

Explanation:

Question Bank Series

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation: