272727
In which of the following changes entropy decreases ?
1 Crystallization of sucrose from solution
2 Dissolving sucrose in water
3 Melting of ice
4 Vapourisation of camphor
Explanation:
During the crystallization of sucrose solution liquid state is changing into solid state hence entropy decreases. $\text { Gas }>\text { Liquid }>\text { Solid }$ Entropy
UPTU/UPSEE-2018
Thermodynamics
272667
One mole of an ideal gas at $300 \mathrm{~K}$ is expanded isothermally from an initial volume of 1 litre to 10 litres. The value of $\Delta E$ for this process is $\left(\mathbf{R}=\mathbf{2}\right.$ cal $\left.\mathbf{~ m o l}^{-1} \mathrm{~K}^{-1}\right)$
1 $163.7 \mathrm{cal}$
2 zero
3 $138.1 \mathrm{cal}$
4 9 litre atm
Explanation:
$\Delta \mathrm{E}=\mathrm{nC}_{\mathrm{w}} \Delta \mathrm{T}$ For isothermal process, $\Delta \mathrm{T}=0, \Delta \mathrm{E}=0$ Hence, change in internal energy $(\Delta \mathrm{E})$ is zero during isothermal expansion of a gas.
AIIMS-2010
Thermodynamics
272671
For the reaction of one mole of zinc dust with one mole of sulphuric acid in a bomb calorimeter, $\Delta \mathrm{U}$ and $\mathrm{W}$ corresponds to
1 $\Delta \mathrm{U}<0, \mathrm{~W}=0$
2 $\Delta \mathrm{U}<0, \mathrm{~W}<0$
3 $\Delta \mathrm{U}>0, \mathrm{~W}=0$
4 $\Delta \mathrm{U}>0, \mathrm{~W}>0$
Explanation:
$\mathrm{Zn}+\mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{ZnSO}_4+\mathrm{H}_2$ $\Delta \mathrm{n}_{\text {gas }}=0$ $\therefore \quad \mathrm{W}=\Delta \mathrm{n}_{\mathrm{gas}} \mathrm{RT}=0$ $\Delta \mathrm{U}=\mathrm{W}+\mathrm{Q}$ In bomb calorimeter, there is no expansion in volume, so work done will be zero. This reaction is exothermic. So, some heat will evolved which will result in lowering of internal energy (Internal energy are the function of temperature). Hence, $\Delta \mathrm{U}<0$ and $\mathrm{W}=0$
AIIMS-2005
Thermodynamics
272674
During isothermal expansion of one mole of an ideal gas from $10 \mathrm{~atm}$ to $1 \mathrm{~atm}$ at $273 \mathrm{~K}$, the work done is (gas constant=2):
1 $-895.8 \mathrm{cal}$
2 $-1172.6 \mathrm{cal}$
3 $-1257.43 \mathrm{cal}$
4 $-1499.6 \mathrm{cal}$
Explanation:
Work done in expansion of gas during Isothermal process, $\mathrm{W}=-2.303 \mathrm{nRT} \log \frac{\mathrm{P}_1}{\mathrm{P}_2}$ $\mathrm{~W}=-2.303 \times 1 \times 2 \times 273 \log \frac{10}{1}$ $\mathrm{~W}=-1257.43 \mathrm{cal} .$
272727
In which of the following changes entropy decreases ?
1 Crystallization of sucrose from solution
2 Dissolving sucrose in water
3 Melting of ice
4 Vapourisation of camphor
Explanation:
During the crystallization of sucrose solution liquid state is changing into solid state hence entropy decreases. $\text { Gas }>\text { Liquid }>\text { Solid }$ Entropy
UPTU/UPSEE-2018
Thermodynamics
272667
One mole of an ideal gas at $300 \mathrm{~K}$ is expanded isothermally from an initial volume of 1 litre to 10 litres. The value of $\Delta E$ for this process is $\left(\mathbf{R}=\mathbf{2}\right.$ cal $\left.\mathbf{~ m o l}^{-1} \mathrm{~K}^{-1}\right)$
1 $163.7 \mathrm{cal}$
2 zero
3 $138.1 \mathrm{cal}$
4 9 litre atm
Explanation:
$\Delta \mathrm{E}=\mathrm{nC}_{\mathrm{w}} \Delta \mathrm{T}$ For isothermal process, $\Delta \mathrm{T}=0, \Delta \mathrm{E}=0$ Hence, change in internal energy $(\Delta \mathrm{E})$ is zero during isothermal expansion of a gas.
AIIMS-2010
Thermodynamics
272671
For the reaction of one mole of zinc dust with one mole of sulphuric acid in a bomb calorimeter, $\Delta \mathrm{U}$ and $\mathrm{W}$ corresponds to
1 $\Delta \mathrm{U}<0, \mathrm{~W}=0$
2 $\Delta \mathrm{U}<0, \mathrm{~W}<0$
3 $\Delta \mathrm{U}>0, \mathrm{~W}=0$
4 $\Delta \mathrm{U}>0, \mathrm{~W}>0$
Explanation:
$\mathrm{Zn}+\mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{ZnSO}_4+\mathrm{H}_2$ $\Delta \mathrm{n}_{\text {gas }}=0$ $\therefore \quad \mathrm{W}=\Delta \mathrm{n}_{\mathrm{gas}} \mathrm{RT}=0$ $\Delta \mathrm{U}=\mathrm{W}+\mathrm{Q}$ In bomb calorimeter, there is no expansion in volume, so work done will be zero. This reaction is exothermic. So, some heat will evolved which will result in lowering of internal energy (Internal energy are the function of temperature). Hence, $\Delta \mathrm{U}<0$ and $\mathrm{W}=0$
AIIMS-2005
Thermodynamics
272674
During isothermal expansion of one mole of an ideal gas from $10 \mathrm{~atm}$ to $1 \mathrm{~atm}$ at $273 \mathrm{~K}$, the work done is (gas constant=2):
1 $-895.8 \mathrm{cal}$
2 $-1172.6 \mathrm{cal}$
3 $-1257.43 \mathrm{cal}$
4 $-1499.6 \mathrm{cal}$
Explanation:
Work done in expansion of gas during Isothermal process, $\mathrm{W}=-2.303 \mathrm{nRT} \log \frac{\mathrm{P}_1}{\mathrm{P}_2}$ $\mathrm{~W}=-2.303 \times 1 \times 2 \times 273 \log \frac{10}{1}$ $\mathrm{~W}=-1257.43 \mathrm{cal} .$
272727
In which of the following changes entropy decreases ?
1 Crystallization of sucrose from solution
2 Dissolving sucrose in water
3 Melting of ice
4 Vapourisation of camphor
Explanation:
During the crystallization of sucrose solution liquid state is changing into solid state hence entropy decreases. $\text { Gas }>\text { Liquid }>\text { Solid }$ Entropy
UPTU/UPSEE-2018
Thermodynamics
272667
One mole of an ideal gas at $300 \mathrm{~K}$ is expanded isothermally from an initial volume of 1 litre to 10 litres. The value of $\Delta E$ for this process is $\left(\mathbf{R}=\mathbf{2}\right.$ cal $\left.\mathbf{~ m o l}^{-1} \mathrm{~K}^{-1}\right)$
1 $163.7 \mathrm{cal}$
2 zero
3 $138.1 \mathrm{cal}$
4 9 litre atm
Explanation:
$\Delta \mathrm{E}=\mathrm{nC}_{\mathrm{w}} \Delta \mathrm{T}$ For isothermal process, $\Delta \mathrm{T}=0, \Delta \mathrm{E}=0$ Hence, change in internal energy $(\Delta \mathrm{E})$ is zero during isothermal expansion of a gas.
AIIMS-2010
Thermodynamics
272671
For the reaction of one mole of zinc dust with one mole of sulphuric acid in a bomb calorimeter, $\Delta \mathrm{U}$ and $\mathrm{W}$ corresponds to
1 $\Delta \mathrm{U}<0, \mathrm{~W}=0$
2 $\Delta \mathrm{U}<0, \mathrm{~W}<0$
3 $\Delta \mathrm{U}>0, \mathrm{~W}=0$
4 $\Delta \mathrm{U}>0, \mathrm{~W}>0$
Explanation:
$\mathrm{Zn}+\mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{ZnSO}_4+\mathrm{H}_2$ $\Delta \mathrm{n}_{\text {gas }}=0$ $\therefore \quad \mathrm{W}=\Delta \mathrm{n}_{\mathrm{gas}} \mathrm{RT}=0$ $\Delta \mathrm{U}=\mathrm{W}+\mathrm{Q}$ In bomb calorimeter, there is no expansion in volume, so work done will be zero. This reaction is exothermic. So, some heat will evolved which will result in lowering of internal energy (Internal energy are the function of temperature). Hence, $\Delta \mathrm{U}<0$ and $\mathrm{W}=0$
AIIMS-2005
Thermodynamics
272674
During isothermal expansion of one mole of an ideal gas from $10 \mathrm{~atm}$ to $1 \mathrm{~atm}$ at $273 \mathrm{~K}$, the work done is (gas constant=2):
1 $-895.8 \mathrm{cal}$
2 $-1172.6 \mathrm{cal}$
3 $-1257.43 \mathrm{cal}$
4 $-1499.6 \mathrm{cal}$
Explanation:
Work done in expansion of gas during Isothermal process, $\mathrm{W}=-2.303 \mathrm{nRT} \log \frac{\mathrm{P}_1}{\mathrm{P}_2}$ $\mathrm{~W}=-2.303 \times 1 \times 2 \times 273 \log \frac{10}{1}$ $\mathrm{~W}=-1257.43 \mathrm{cal} .$
272727
In which of the following changes entropy decreases ?
1 Crystallization of sucrose from solution
2 Dissolving sucrose in water
3 Melting of ice
4 Vapourisation of camphor
Explanation:
During the crystallization of sucrose solution liquid state is changing into solid state hence entropy decreases. $\text { Gas }>\text { Liquid }>\text { Solid }$ Entropy
UPTU/UPSEE-2018
Thermodynamics
272667
One mole of an ideal gas at $300 \mathrm{~K}$ is expanded isothermally from an initial volume of 1 litre to 10 litres. The value of $\Delta E$ for this process is $\left(\mathbf{R}=\mathbf{2}\right.$ cal $\left.\mathbf{~ m o l}^{-1} \mathrm{~K}^{-1}\right)$
1 $163.7 \mathrm{cal}$
2 zero
3 $138.1 \mathrm{cal}$
4 9 litre atm
Explanation:
$\Delta \mathrm{E}=\mathrm{nC}_{\mathrm{w}} \Delta \mathrm{T}$ For isothermal process, $\Delta \mathrm{T}=0, \Delta \mathrm{E}=0$ Hence, change in internal energy $(\Delta \mathrm{E})$ is zero during isothermal expansion of a gas.
AIIMS-2010
Thermodynamics
272671
For the reaction of one mole of zinc dust with one mole of sulphuric acid in a bomb calorimeter, $\Delta \mathrm{U}$ and $\mathrm{W}$ corresponds to
1 $\Delta \mathrm{U}<0, \mathrm{~W}=0$
2 $\Delta \mathrm{U}<0, \mathrm{~W}<0$
3 $\Delta \mathrm{U}>0, \mathrm{~W}=0$
4 $\Delta \mathrm{U}>0, \mathrm{~W}>0$
Explanation:
$\mathrm{Zn}+\mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{ZnSO}_4+\mathrm{H}_2$ $\Delta \mathrm{n}_{\text {gas }}=0$ $\therefore \quad \mathrm{W}=\Delta \mathrm{n}_{\mathrm{gas}} \mathrm{RT}=0$ $\Delta \mathrm{U}=\mathrm{W}+\mathrm{Q}$ In bomb calorimeter, there is no expansion in volume, so work done will be zero. This reaction is exothermic. So, some heat will evolved which will result in lowering of internal energy (Internal energy are the function of temperature). Hence, $\Delta \mathrm{U}<0$ and $\mathrm{W}=0$
AIIMS-2005
Thermodynamics
272674
During isothermal expansion of one mole of an ideal gas from $10 \mathrm{~atm}$ to $1 \mathrm{~atm}$ at $273 \mathrm{~K}$, the work done is (gas constant=2):
1 $-895.8 \mathrm{cal}$
2 $-1172.6 \mathrm{cal}$
3 $-1257.43 \mathrm{cal}$
4 $-1499.6 \mathrm{cal}$
Explanation:
Work done in expansion of gas during Isothermal process, $\mathrm{W}=-2.303 \mathrm{nRT} \log \frac{\mathrm{P}_1}{\mathrm{P}_2}$ $\mathrm{~W}=-2.303 \times 1 \times 2 \times 273 \log \frac{10}{1}$ $\mathrm{~W}=-1257.43 \mathrm{cal} .$
