272343
For the reaction $\mathrm{H}_2 \mathrm{O}_{(l)} \longrightarrow \mathrm{H}_2 \mathrm{O}_{(\mathrm{g})}$ at $\mathrm{T}=100^{\circ}$ and $\mathbf{P}=1$ atm. Choose the correct option:
At $100^{\circ} \mathrm{C}$ and 1 atm. $\mathrm{H}_2 \mathrm{O}(l) \longrightarrow \mathrm{H}_2 \mathrm{O}(\mathrm{g})$
For equilibrium.
$\Delta \mathrm{S}_{\text {total }}=0$
And $\Delta \mathrm{S}_{5 y \mathrm{stem}}+\Delta \mathrm{S}_{\text {surromding }}=0$
$\Delta \mathrm{S}_{\text {sygtem }}>0$ and $\Delta \mathrm{S}_{\text {surrounding }} \lt0$.
AP EAPCET 20.08.2021 Shift-I
Thermodynamics
272314
The volume of $2.8 \mathrm{~g}$ of $\mathrm{CO}$ at $27^{\circ} \mathrm{C}$ and 0.821 atm, pressure is $\left(\mathrm{R}-0.08210\right.$ lit.atm. $\left.\mathrm{K}^{-1} \mathrm{~mol}^{-1}\right)$
1 3 litres
2 30 litres
3 0.3 litres
4 1.5 litres
Explanation:
Given that,
$\mathrm{T}=27^{\circ} \mathrm{C}=27+273=300 \mathrm{~K}$
$\mathrm{R}=0.0821 \mathrm{~atm} \mathrm{~K} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$
$\mathrm{P}=0.821 \mathrm{~atm}$
Molecular mass of $\mathrm{CO}\left(\mathrm{M}_{\mathrm{co}}\right)=12+16=28 \mathrm{~g} / \mathrm{mol}$
No. of mole $(\mathrm{n})=\frac{\mathrm{W}}{\mathrm{M}}=\frac{2.8}{28}=0.1$ moles
According to gas equation-
$\mathrm{PV} =\mathrm{nRT}$
$\therefore \quad \mathrm{V} =\frac{\mathrm{nRT}}{\mathrm{P}}=\frac{0.1 \times 0.0821 \times 300}{0.821}=3 \mathrm{~L}$
Karnataka CET-17.06.2022, Shift-II
Thermodynamics
272315
For spontaneity of a cell, which is correct?
The Gibbs free energy (G) related to the enthalpy and entropy of a system is given as-$\mathrm{G}=\mathrm{H}-\mathrm{TS}$
The free energy change $(\Delta \mathrm{G})$ is a measure of the spontaneity of a process and of the useful energy available from it.
$\Delta \mathrm{G}_{\mathrm{Sy}_{\mathrm{y}}}=\Delta \mathrm{H}_{\mathrm{S}_7}-\mathrm{T} \Delta \mathrm{S}_{\mathrm{S}_7}, \Delta \mathrm{G}_{\mathrm{Syy}_5}=-\mathrm{T} \Delta \mathrm{S}_{\mathrm{Unir} .}$
Therefore, $\Delta \mathrm{G}<0 \rightarrow$ spontaneous process
$\Delta \mathrm{G}>0 \rightarrow$ Non-spontaneous process
$\Delta \mathrm{G}=0 \rightarrow$ Process at a equllibrium
Karnataka CET-17.06.2022, Shift-II
Thermodynamics
272316
At $25^{\circ} \mathrm{C} 1$ mole of butane upon heating forms $\mathrm{CO}_2$ and $\mathrm{H}_2 \mathrm{O}$. The work done is
1 $75.6 \mathrm{~L} \mathrm{~atm}$
2 $85.6 \mathrm{~L}$ atm
3 $50.3 \mathrm{~L} \mathrm{~atm}$
4 None of these
Explanation:
Given that, $\mathrm{C}_4 \mathrm{H}_{10}(\mathrm{~g})+\frac{13}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow 4 \mathrm{CO}_2(\mathrm{~g})+5 \mathrm{H}_2 \mathrm{O}(l)$ $\because \quad \Delta \mathrm{n}_{\mathrm{g}}=4-\left(1+\frac{13}{2}\right)$ $=4-7.5$ $=-3.5$ Work done, $\mathrm{W}=-\Delta \mathrm{n}_{\mathrm{g}} \mathrm{RT}$ Putting the value of $\Delta \mathrm{n}_{\mathrm{g}}$ in equation (i) $\mathrm{W}=-(-3.5 \times 0.0821 \times 298)$ $\mathrm{W}=85.6 \mathrm{~L} \mathrm{~atm} .$
AIIMS 25 May 2019 (Morning)
Thermodynamics
272322
In the complete combustion of butanol $\mathrm{C}_4 \mathrm{H}_9 \mathrm{OH}(l)$, if $\Delta \mathrm{H}$ is enthalpy of combustion at constant pressure and $\Delta E$ is the heat of combustion at constant volume, then
1 $\Delta \mathrm{H}<\Delta \mathrm{E}$
2 $\Delta \mathrm{H}=\Delta \mathrm{E}$
3 $\Delta \mathrm{H}>\Delta \mathrm{E}$
4 $\Delta \mathrm{H}, \Delta \mathrm{E}$ relation cannot be predicted
Explanation:
The complete combustion of Butanol $\mathrm{C}_4 \mathrm{H}_9 \mathrm{OH}$ is- $\mathrm{C}_4 \mathrm{H}_9 \mathrm{OH}(\mathrm{l})+6 \mathrm{O}_2(\mathrm{~g}) \longrightarrow 4 \mathrm{CO}_2(\mathrm{~g})+5 \mathrm{H}_2 \mathrm{O}(l)+\text { Heat }$ In the reaction, $\Delta \mathrm{n}_{\mathrm{g}}\left(\Delta \mathrm{n}_{\mathrm{g}}=\right.$ number of moles of gaseous product - no. of moles of gaseous reactant) is negative. $\Delta \mathrm{n}_{\mathrm{g}}=4-6=-2$ $\Delta \mathrm{H}=\Delta \mathrm{E}+\Delta \mathrm{n}_{\mathrm{g}} \mathrm{RT}$ $\Delta \mathrm{H}=\Delta \mathrm{E}-2 \mathrm{RT}$ $\Delta \mathrm{H}+2 \mathrm{RT}=\Delta \mathrm{E}$ $\therefore \Delta \mathrm{E}>\Delta \mathrm{H}$
272343
For the reaction $\mathrm{H}_2 \mathrm{O}_{(l)} \longrightarrow \mathrm{H}_2 \mathrm{O}_{(\mathrm{g})}$ at $\mathrm{T}=100^{\circ}$ and $\mathbf{P}=1$ atm. Choose the correct option:
At $100^{\circ} \mathrm{C}$ and 1 atm. $\mathrm{H}_2 \mathrm{O}(l) \longrightarrow \mathrm{H}_2 \mathrm{O}(\mathrm{g})$
For equilibrium.
$\Delta \mathrm{S}_{\text {total }}=0$
And $\Delta \mathrm{S}_{5 y \mathrm{stem}}+\Delta \mathrm{S}_{\text {surromding }}=0$
$\Delta \mathrm{S}_{\text {sygtem }}>0$ and $\Delta \mathrm{S}_{\text {surrounding }} \lt0$.
AP EAPCET 20.08.2021 Shift-I
Thermodynamics
272314
The volume of $2.8 \mathrm{~g}$ of $\mathrm{CO}$ at $27^{\circ} \mathrm{C}$ and 0.821 atm, pressure is $\left(\mathrm{R}-0.08210\right.$ lit.atm. $\left.\mathrm{K}^{-1} \mathrm{~mol}^{-1}\right)$
1 3 litres
2 30 litres
3 0.3 litres
4 1.5 litres
Explanation:
Given that,
$\mathrm{T}=27^{\circ} \mathrm{C}=27+273=300 \mathrm{~K}$
$\mathrm{R}=0.0821 \mathrm{~atm} \mathrm{~K} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$
$\mathrm{P}=0.821 \mathrm{~atm}$
Molecular mass of $\mathrm{CO}\left(\mathrm{M}_{\mathrm{co}}\right)=12+16=28 \mathrm{~g} / \mathrm{mol}$
No. of mole $(\mathrm{n})=\frac{\mathrm{W}}{\mathrm{M}}=\frac{2.8}{28}=0.1$ moles
According to gas equation-
$\mathrm{PV} =\mathrm{nRT}$
$\therefore \quad \mathrm{V} =\frac{\mathrm{nRT}}{\mathrm{P}}=\frac{0.1 \times 0.0821 \times 300}{0.821}=3 \mathrm{~L}$
Karnataka CET-17.06.2022, Shift-II
Thermodynamics
272315
For spontaneity of a cell, which is correct?
The Gibbs free energy (G) related to the enthalpy and entropy of a system is given as-$\mathrm{G}=\mathrm{H}-\mathrm{TS}$
The free energy change $(\Delta \mathrm{G})$ is a measure of the spontaneity of a process and of the useful energy available from it.
$\Delta \mathrm{G}_{\mathrm{Sy}_{\mathrm{y}}}=\Delta \mathrm{H}_{\mathrm{S}_7}-\mathrm{T} \Delta \mathrm{S}_{\mathrm{S}_7}, \Delta \mathrm{G}_{\mathrm{Syy}_5}=-\mathrm{T} \Delta \mathrm{S}_{\mathrm{Unir} .}$
Therefore, $\Delta \mathrm{G}<0 \rightarrow$ spontaneous process
$\Delta \mathrm{G}>0 \rightarrow$ Non-spontaneous process
$\Delta \mathrm{G}=0 \rightarrow$ Process at a equllibrium
Karnataka CET-17.06.2022, Shift-II
Thermodynamics
272316
At $25^{\circ} \mathrm{C} 1$ mole of butane upon heating forms $\mathrm{CO}_2$ and $\mathrm{H}_2 \mathrm{O}$. The work done is
1 $75.6 \mathrm{~L} \mathrm{~atm}$
2 $85.6 \mathrm{~L}$ atm
3 $50.3 \mathrm{~L} \mathrm{~atm}$
4 None of these
Explanation:
Given that, $\mathrm{C}_4 \mathrm{H}_{10}(\mathrm{~g})+\frac{13}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow 4 \mathrm{CO}_2(\mathrm{~g})+5 \mathrm{H}_2 \mathrm{O}(l)$ $\because \quad \Delta \mathrm{n}_{\mathrm{g}}=4-\left(1+\frac{13}{2}\right)$ $=4-7.5$ $=-3.5$ Work done, $\mathrm{W}=-\Delta \mathrm{n}_{\mathrm{g}} \mathrm{RT}$ Putting the value of $\Delta \mathrm{n}_{\mathrm{g}}$ in equation (i) $\mathrm{W}=-(-3.5 \times 0.0821 \times 298)$ $\mathrm{W}=85.6 \mathrm{~L} \mathrm{~atm} .$
AIIMS 25 May 2019 (Morning)
Thermodynamics
272322
In the complete combustion of butanol $\mathrm{C}_4 \mathrm{H}_9 \mathrm{OH}(l)$, if $\Delta \mathrm{H}$ is enthalpy of combustion at constant pressure and $\Delta E$ is the heat of combustion at constant volume, then
1 $\Delta \mathrm{H}<\Delta \mathrm{E}$
2 $\Delta \mathrm{H}=\Delta \mathrm{E}$
3 $\Delta \mathrm{H}>\Delta \mathrm{E}$
4 $\Delta \mathrm{H}, \Delta \mathrm{E}$ relation cannot be predicted
Explanation:
The complete combustion of Butanol $\mathrm{C}_4 \mathrm{H}_9 \mathrm{OH}$ is- $\mathrm{C}_4 \mathrm{H}_9 \mathrm{OH}(\mathrm{l})+6 \mathrm{O}_2(\mathrm{~g}) \longrightarrow 4 \mathrm{CO}_2(\mathrm{~g})+5 \mathrm{H}_2 \mathrm{O}(l)+\text { Heat }$ In the reaction, $\Delta \mathrm{n}_{\mathrm{g}}\left(\Delta \mathrm{n}_{\mathrm{g}}=\right.$ number of moles of gaseous product - no. of moles of gaseous reactant) is negative. $\Delta \mathrm{n}_{\mathrm{g}}=4-6=-2$ $\Delta \mathrm{H}=\Delta \mathrm{E}+\Delta \mathrm{n}_{\mathrm{g}} \mathrm{RT}$ $\Delta \mathrm{H}=\Delta \mathrm{E}-2 \mathrm{RT}$ $\Delta \mathrm{H}+2 \mathrm{RT}=\Delta \mathrm{E}$ $\therefore \Delta \mathrm{E}>\Delta \mathrm{H}$
272343
For the reaction $\mathrm{H}_2 \mathrm{O}_{(l)} \longrightarrow \mathrm{H}_2 \mathrm{O}_{(\mathrm{g})}$ at $\mathrm{T}=100^{\circ}$ and $\mathbf{P}=1$ atm. Choose the correct option:
At $100^{\circ} \mathrm{C}$ and 1 atm. $\mathrm{H}_2 \mathrm{O}(l) \longrightarrow \mathrm{H}_2 \mathrm{O}(\mathrm{g})$
For equilibrium.
$\Delta \mathrm{S}_{\text {total }}=0$
And $\Delta \mathrm{S}_{5 y \mathrm{stem}}+\Delta \mathrm{S}_{\text {surromding }}=0$
$\Delta \mathrm{S}_{\text {sygtem }}>0$ and $\Delta \mathrm{S}_{\text {surrounding }} \lt0$.
AP EAPCET 20.08.2021 Shift-I
Thermodynamics
272314
The volume of $2.8 \mathrm{~g}$ of $\mathrm{CO}$ at $27^{\circ} \mathrm{C}$ and 0.821 atm, pressure is $\left(\mathrm{R}-0.08210\right.$ lit.atm. $\left.\mathrm{K}^{-1} \mathrm{~mol}^{-1}\right)$
1 3 litres
2 30 litres
3 0.3 litres
4 1.5 litres
Explanation:
Given that,
$\mathrm{T}=27^{\circ} \mathrm{C}=27+273=300 \mathrm{~K}$
$\mathrm{R}=0.0821 \mathrm{~atm} \mathrm{~K} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$
$\mathrm{P}=0.821 \mathrm{~atm}$
Molecular mass of $\mathrm{CO}\left(\mathrm{M}_{\mathrm{co}}\right)=12+16=28 \mathrm{~g} / \mathrm{mol}$
No. of mole $(\mathrm{n})=\frac{\mathrm{W}}{\mathrm{M}}=\frac{2.8}{28}=0.1$ moles
According to gas equation-
$\mathrm{PV} =\mathrm{nRT}$
$\therefore \quad \mathrm{V} =\frac{\mathrm{nRT}}{\mathrm{P}}=\frac{0.1 \times 0.0821 \times 300}{0.821}=3 \mathrm{~L}$
Karnataka CET-17.06.2022, Shift-II
Thermodynamics
272315
For spontaneity of a cell, which is correct?
The Gibbs free energy (G) related to the enthalpy and entropy of a system is given as-$\mathrm{G}=\mathrm{H}-\mathrm{TS}$
The free energy change $(\Delta \mathrm{G})$ is a measure of the spontaneity of a process and of the useful energy available from it.
$\Delta \mathrm{G}_{\mathrm{Sy}_{\mathrm{y}}}=\Delta \mathrm{H}_{\mathrm{S}_7}-\mathrm{T} \Delta \mathrm{S}_{\mathrm{S}_7}, \Delta \mathrm{G}_{\mathrm{Syy}_5}=-\mathrm{T} \Delta \mathrm{S}_{\mathrm{Unir} .}$
Therefore, $\Delta \mathrm{G}<0 \rightarrow$ spontaneous process
$\Delta \mathrm{G}>0 \rightarrow$ Non-spontaneous process
$\Delta \mathrm{G}=0 \rightarrow$ Process at a equllibrium
Karnataka CET-17.06.2022, Shift-II
Thermodynamics
272316
At $25^{\circ} \mathrm{C} 1$ mole of butane upon heating forms $\mathrm{CO}_2$ and $\mathrm{H}_2 \mathrm{O}$. The work done is
1 $75.6 \mathrm{~L} \mathrm{~atm}$
2 $85.6 \mathrm{~L}$ atm
3 $50.3 \mathrm{~L} \mathrm{~atm}$
4 None of these
Explanation:
Given that, $\mathrm{C}_4 \mathrm{H}_{10}(\mathrm{~g})+\frac{13}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow 4 \mathrm{CO}_2(\mathrm{~g})+5 \mathrm{H}_2 \mathrm{O}(l)$ $\because \quad \Delta \mathrm{n}_{\mathrm{g}}=4-\left(1+\frac{13}{2}\right)$ $=4-7.5$ $=-3.5$ Work done, $\mathrm{W}=-\Delta \mathrm{n}_{\mathrm{g}} \mathrm{RT}$ Putting the value of $\Delta \mathrm{n}_{\mathrm{g}}$ in equation (i) $\mathrm{W}=-(-3.5 \times 0.0821 \times 298)$ $\mathrm{W}=85.6 \mathrm{~L} \mathrm{~atm} .$
AIIMS 25 May 2019 (Morning)
Thermodynamics
272322
In the complete combustion of butanol $\mathrm{C}_4 \mathrm{H}_9 \mathrm{OH}(l)$, if $\Delta \mathrm{H}$ is enthalpy of combustion at constant pressure and $\Delta E$ is the heat of combustion at constant volume, then
1 $\Delta \mathrm{H}<\Delta \mathrm{E}$
2 $\Delta \mathrm{H}=\Delta \mathrm{E}$
3 $\Delta \mathrm{H}>\Delta \mathrm{E}$
4 $\Delta \mathrm{H}, \Delta \mathrm{E}$ relation cannot be predicted
Explanation:
The complete combustion of Butanol $\mathrm{C}_4 \mathrm{H}_9 \mathrm{OH}$ is- $\mathrm{C}_4 \mathrm{H}_9 \mathrm{OH}(\mathrm{l})+6 \mathrm{O}_2(\mathrm{~g}) \longrightarrow 4 \mathrm{CO}_2(\mathrm{~g})+5 \mathrm{H}_2 \mathrm{O}(l)+\text { Heat }$ In the reaction, $\Delta \mathrm{n}_{\mathrm{g}}\left(\Delta \mathrm{n}_{\mathrm{g}}=\right.$ number of moles of gaseous product - no. of moles of gaseous reactant) is negative. $\Delta \mathrm{n}_{\mathrm{g}}=4-6=-2$ $\Delta \mathrm{H}=\Delta \mathrm{E}+\Delta \mathrm{n}_{\mathrm{g}} \mathrm{RT}$ $\Delta \mathrm{H}=\Delta \mathrm{E}-2 \mathrm{RT}$ $\Delta \mathrm{H}+2 \mathrm{RT}=\Delta \mathrm{E}$ $\therefore \Delta \mathrm{E}>\Delta \mathrm{H}$
272343
For the reaction $\mathrm{H}_2 \mathrm{O}_{(l)} \longrightarrow \mathrm{H}_2 \mathrm{O}_{(\mathrm{g})}$ at $\mathrm{T}=100^{\circ}$ and $\mathbf{P}=1$ atm. Choose the correct option:
At $100^{\circ} \mathrm{C}$ and 1 atm. $\mathrm{H}_2 \mathrm{O}(l) \longrightarrow \mathrm{H}_2 \mathrm{O}(\mathrm{g})$
For equilibrium.
$\Delta \mathrm{S}_{\text {total }}=0$
And $\Delta \mathrm{S}_{5 y \mathrm{stem}}+\Delta \mathrm{S}_{\text {surromding }}=0$
$\Delta \mathrm{S}_{\text {sygtem }}>0$ and $\Delta \mathrm{S}_{\text {surrounding }} \lt0$.
AP EAPCET 20.08.2021 Shift-I
Thermodynamics
272314
The volume of $2.8 \mathrm{~g}$ of $\mathrm{CO}$ at $27^{\circ} \mathrm{C}$ and 0.821 atm, pressure is $\left(\mathrm{R}-0.08210\right.$ lit.atm. $\left.\mathrm{K}^{-1} \mathrm{~mol}^{-1}\right)$
1 3 litres
2 30 litres
3 0.3 litres
4 1.5 litres
Explanation:
Given that,
$\mathrm{T}=27^{\circ} \mathrm{C}=27+273=300 \mathrm{~K}$
$\mathrm{R}=0.0821 \mathrm{~atm} \mathrm{~K} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$
$\mathrm{P}=0.821 \mathrm{~atm}$
Molecular mass of $\mathrm{CO}\left(\mathrm{M}_{\mathrm{co}}\right)=12+16=28 \mathrm{~g} / \mathrm{mol}$
No. of mole $(\mathrm{n})=\frac{\mathrm{W}}{\mathrm{M}}=\frac{2.8}{28}=0.1$ moles
According to gas equation-
$\mathrm{PV} =\mathrm{nRT}$
$\therefore \quad \mathrm{V} =\frac{\mathrm{nRT}}{\mathrm{P}}=\frac{0.1 \times 0.0821 \times 300}{0.821}=3 \mathrm{~L}$
Karnataka CET-17.06.2022, Shift-II
Thermodynamics
272315
For spontaneity of a cell, which is correct?
The Gibbs free energy (G) related to the enthalpy and entropy of a system is given as-$\mathrm{G}=\mathrm{H}-\mathrm{TS}$
The free energy change $(\Delta \mathrm{G})$ is a measure of the spontaneity of a process and of the useful energy available from it.
$\Delta \mathrm{G}_{\mathrm{Sy}_{\mathrm{y}}}=\Delta \mathrm{H}_{\mathrm{S}_7}-\mathrm{T} \Delta \mathrm{S}_{\mathrm{S}_7}, \Delta \mathrm{G}_{\mathrm{Syy}_5}=-\mathrm{T} \Delta \mathrm{S}_{\mathrm{Unir} .}$
Therefore, $\Delta \mathrm{G}<0 \rightarrow$ spontaneous process
$\Delta \mathrm{G}>0 \rightarrow$ Non-spontaneous process
$\Delta \mathrm{G}=0 \rightarrow$ Process at a equllibrium
Karnataka CET-17.06.2022, Shift-II
Thermodynamics
272316
At $25^{\circ} \mathrm{C} 1$ mole of butane upon heating forms $\mathrm{CO}_2$ and $\mathrm{H}_2 \mathrm{O}$. The work done is
1 $75.6 \mathrm{~L} \mathrm{~atm}$
2 $85.6 \mathrm{~L}$ atm
3 $50.3 \mathrm{~L} \mathrm{~atm}$
4 None of these
Explanation:
Given that, $\mathrm{C}_4 \mathrm{H}_{10}(\mathrm{~g})+\frac{13}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow 4 \mathrm{CO}_2(\mathrm{~g})+5 \mathrm{H}_2 \mathrm{O}(l)$ $\because \quad \Delta \mathrm{n}_{\mathrm{g}}=4-\left(1+\frac{13}{2}\right)$ $=4-7.5$ $=-3.5$ Work done, $\mathrm{W}=-\Delta \mathrm{n}_{\mathrm{g}} \mathrm{RT}$ Putting the value of $\Delta \mathrm{n}_{\mathrm{g}}$ in equation (i) $\mathrm{W}=-(-3.5 \times 0.0821 \times 298)$ $\mathrm{W}=85.6 \mathrm{~L} \mathrm{~atm} .$
AIIMS 25 May 2019 (Morning)
Thermodynamics
272322
In the complete combustion of butanol $\mathrm{C}_4 \mathrm{H}_9 \mathrm{OH}(l)$, if $\Delta \mathrm{H}$ is enthalpy of combustion at constant pressure and $\Delta E$ is the heat of combustion at constant volume, then
1 $\Delta \mathrm{H}<\Delta \mathrm{E}$
2 $\Delta \mathrm{H}=\Delta \mathrm{E}$
3 $\Delta \mathrm{H}>\Delta \mathrm{E}$
4 $\Delta \mathrm{H}, \Delta \mathrm{E}$ relation cannot be predicted
Explanation:
The complete combustion of Butanol $\mathrm{C}_4 \mathrm{H}_9 \mathrm{OH}$ is- $\mathrm{C}_4 \mathrm{H}_9 \mathrm{OH}(\mathrm{l})+6 \mathrm{O}_2(\mathrm{~g}) \longrightarrow 4 \mathrm{CO}_2(\mathrm{~g})+5 \mathrm{H}_2 \mathrm{O}(l)+\text { Heat }$ In the reaction, $\Delta \mathrm{n}_{\mathrm{g}}\left(\Delta \mathrm{n}_{\mathrm{g}}=\right.$ number of moles of gaseous product - no. of moles of gaseous reactant) is negative. $\Delta \mathrm{n}_{\mathrm{g}}=4-6=-2$ $\Delta \mathrm{H}=\Delta \mathrm{E}+\Delta \mathrm{n}_{\mathrm{g}} \mathrm{RT}$ $\Delta \mathrm{H}=\Delta \mathrm{E}-2 \mathrm{RT}$ $\Delta \mathrm{H}+2 \mathrm{RT}=\Delta \mathrm{E}$ $\therefore \Delta \mathrm{E}>\Delta \mathrm{H}$
272343
For the reaction $\mathrm{H}_2 \mathrm{O}_{(l)} \longrightarrow \mathrm{H}_2 \mathrm{O}_{(\mathrm{g})}$ at $\mathrm{T}=100^{\circ}$ and $\mathbf{P}=1$ atm. Choose the correct option:
At $100^{\circ} \mathrm{C}$ and 1 atm. $\mathrm{H}_2 \mathrm{O}(l) \longrightarrow \mathrm{H}_2 \mathrm{O}(\mathrm{g})$
For equilibrium.
$\Delta \mathrm{S}_{\text {total }}=0$
And $\Delta \mathrm{S}_{5 y \mathrm{stem}}+\Delta \mathrm{S}_{\text {surromding }}=0$
$\Delta \mathrm{S}_{\text {sygtem }}>0$ and $\Delta \mathrm{S}_{\text {surrounding }} \lt0$.
AP EAPCET 20.08.2021 Shift-I
Thermodynamics
272314
The volume of $2.8 \mathrm{~g}$ of $\mathrm{CO}$ at $27^{\circ} \mathrm{C}$ and 0.821 atm, pressure is $\left(\mathrm{R}-0.08210\right.$ lit.atm. $\left.\mathrm{K}^{-1} \mathrm{~mol}^{-1}\right)$
1 3 litres
2 30 litres
3 0.3 litres
4 1.5 litres
Explanation:
Given that,
$\mathrm{T}=27^{\circ} \mathrm{C}=27+273=300 \mathrm{~K}$
$\mathrm{R}=0.0821 \mathrm{~atm} \mathrm{~K} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$
$\mathrm{P}=0.821 \mathrm{~atm}$
Molecular mass of $\mathrm{CO}\left(\mathrm{M}_{\mathrm{co}}\right)=12+16=28 \mathrm{~g} / \mathrm{mol}$
No. of mole $(\mathrm{n})=\frac{\mathrm{W}}{\mathrm{M}}=\frac{2.8}{28}=0.1$ moles
According to gas equation-
$\mathrm{PV} =\mathrm{nRT}$
$\therefore \quad \mathrm{V} =\frac{\mathrm{nRT}}{\mathrm{P}}=\frac{0.1 \times 0.0821 \times 300}{0.821}=3 \mathrm{~L}$
Karnataka CET-17.06.2022, Shift-II
Thermodynamics
272315
For spontaneity of a cell, which is correct?
The Gibbs free energy (G) related to the enthalpy and entropy of a system is given as-$\mathrm{G}=\mathrm{H}-\mathrm{TS}$
The free energy change $(\Delta \mathrm{G})$ is a measure of the spontaneity of a process and of the useful energy available from it.
$\Delta \mathrm{G}_{\mathrm{Sy}_{\mathrm{y}}}=\Delta \mathrm{H}_{\mathrm{S}_7}-\mathrm{T} \Delta \mathrm{S}_{\mathrm{S}_7}, \Delta \mathrm{G}_{\mathrm{Syy}_5}=-\mathrm{T} \Delta \mathrm{S}_{\mathrm{Unir} .}$
Therefore, $\Delta \mathrm{G}<0 \rightarrow$ spontaneous process
$\Delta \mathrm{G}>0 \rightarrow$ Non-spontaneous process
$\Delta \mathrm{G}=0 \rightarrow$ Process at a equllibrium
Karnataka CET-17.06.2022, Shift-II
Thermodynamics
272316
At $25^{\circ} \mathrm{C} 1$ mole of butane upon heating forms $\mathrm{CO}_2$ and $\mathrm{H}_2 \mathrm{O}$. The work done is
1 $75.6 \mathrm{~L} \mathrm{~atm}$
2 $85.6 \mathrm{~L}$ atm
3 $50.3 \mathrm{~L} \mathrm{~atm}$
4 None of these
Explanation:
Given that, $\mathrm{C}_4 \mathrm{H}_{10}(\mathrm{~g})+\frac{13}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow 4 \mathrm{CO}_2(\mathrm{~g})+5 \mathrm{H}_2 \mathrm{O}(l)$ $\because \quad \Delta \mathrm{n}_{\mathrm{g}}=4-\left(1+\frac{13}{2}\right)$ $=4-7.5$ $=-3.5$ Work done, $\mathrm{W}=-\Delta \mathrm{n}_{\mathrm{g}} \mathrm{RT}$ Putting the value of $\Delta \mathrm{n}_{\mathrm{g}}$ in equation (i) $\mathrm{W}=-(-3.5 \times 0.0821 \times 298)$ $\mathrm{W}=85.6 \mathrm{~L} \mathrm{~atm} .$
AIIMS 25 May 2019 (Morning)
Thermodynamics
272322
In the complete combustion of butanol $\mathrm{C}_4 \mathrm{H}_9 \mathrm{OH}(l)$, if $\Delta \mathrm{H}$ is enthalpy of combustion at constant pressure and $\Delta E$ is the heat of combustion at constant volume, then
1 $\Delta \mathrm{H}<\Delta \mathrm{E}$
2 $\Delta \mathrm{H}=\Delta \mathrm{E}$
3 $\Delta \mathrm{H}>\Delta \mathrm{E}$
4 $\Delta \mathrm{H}, \Delta \mathrm{E}$ relation cannot be predicted
Explanation:
The complete combustion of Butanol $\mathrm{C}_4 \mathrm{H}_9 \mathrm{OH}$ is- $\mathrm{C}_4 \mathrm{H}_9 \mathrm{OH}(\mathrm{l})+6 \mathrm{O}_2(\mathrm{~g}) \longrightarrow 4 \mathrm{CO}_2(\mathrm{~g})+5 \mathrm{H}_2 \mathrm{O}(l)+\text { Heat }$ In the reaction, $\Delta \mathrm{n}_{\mathrm{g}}\left(\Delta \mathrm{n}_{\mathrm{g}}=\right.$ number of moles of gaseous product - no. of moles of gaseous reactant) is negative. $\Delta \mathrm{n}_{\mathrm{g}}=4-6=-2$ $\Delta \mathrm{H}=\Delta \mathrm{E}+\Delta \mathrm{n}_{\mathrm{g}} \mathrm{RT}$ $\Delta \mathrm{H}=\Delta \mathrm{E}-2 \mathrm{RT}$ $\Delta \mathrm{H}+2 \mathrm{RT}=\Delta \mathrm{E}$ $\therefore \Delta \mathrm{E}>\Delta \mathrm{H}$
