272439
The temperature of a gas in a closed container is $27^{\circ} \mathrm{C}$. If the temperature is raised to $327^{\circ} \mathrm{C}$, the pressure exerted is
1 reduced to half
2 doubled
3 reduced to one-third
4 Cannot be calculated
Explanation:
Closed container mean volume is constant, $\frac{P_1}{T_1}=\frac{P_2}{T_2}$ Given, $T_1=27+273=300 \mathrm{~K}$ $\mathrm{T}_2=327+273=600 \mathrm{~K}$, $\frac{P_1}{300}=\frac{P_2}{600}$ $\therefore \mathrm{P}_2=\frac{600}{300} \times \mathrm{P}_1=2 \mathrm{P}_1$ So, Pressure become double.
Assam CEE-2014
Thermodynamics
272443
For a reaction, $K=100$ the value of $\Delta G$ will be
272465
The decomposition of dinitrogen tetroxide at $300 \mathrm{~K}$ and 1.0 atm pressure is $20 \%$. At equilibrium, the partial pressure (in atm) of nitrogen peroxide is
1 0.2
2 0.67
3 0.33
4 0.8
Explanation:
Molar ratio $=1: 2$ Also, only $20 \%$ of $\mathrm{N}_2 \mathrm{O}_4$ gets decomposed means. Thus, Total number of moles at equilibrium $=0.4+0.8$ $=1.2$ moles $\because$ Total pressure $\left(\mathrm{P}_{\mathrm{t}}\right)=1 \mathrm{~atm}$ (given) According to Daltons law of partial pressure, $\mathrm{P}_{\mathrm{N}_2 \mathrm{O}_4}=\mathrm{X}_{\left(\mathrm{N}_2 \mathrm{O}_4\right)} \times \mathrm{P}_{\mathrm{t}}$ $\mathrm{P}_{\mathrm{NO}_2}=\mathrm{X}_{\left(\mathrm{NO}_2\right)} \times \mathrm{P}_{\mathrm{t}}$ $\mathrm{P}_{\mathrm{NO}_2}=\frac{0.4}{1.2} \times 1=0.33$
272439
The temperature of a gas in a closed container is $27^{\circ} \mathrm{C}$. If the temperature is raised to $327^{\circ} \mathrm{C}$, the pressure exerted is
1 reduced to half
2 doubled
3 reduced to one-third
4 Cannot be calculated
Explanation:
Closed container mean volume is constant, $\frac{P_1}{T_1}=\frac{P_2}{T_2}$ Given, $T_1=27+273=300 \mathrm{~K}$ $\mathrm{T}_2=327+273=600 \mathrm{~K}$, $\frac{P_1}{300}=\frac{P_2}{600}$ $\therefore \mathrm{P}_2=\frac{600}{300} \times \mathrm{P}_1=2 \mathrm{P}_1$ So, Pressure become double.
Assam CEE-2014
Thermodynamics
272443
For a reaction, $K=100$ the value of $\Delta G$ will be
272465
The decomposition of dinitrogen tetroxide at $300 \mathrm{~K}$ and 1.0 atm pressure is $20 \%$. At equilibrium, the partial pressure (in atm) of nitrogen peroxide is
1 0.2
2 0.67
3 0.33
4 0.8
Explanation:
Molar ratio $=1: 2$ Also, only $20 \%$ of $\mathrm{N}_2 \mathrm{O}_4$ gets decomposed means. Thus, Total number of moles at equilibrium $=0.4+0.8$ $=1.2$ moles $\because$ Total pressure $\left(\mathrm{P}_{\mathrm{t}}\right)=1 \mathrm{~atm}$ (given) According to Daltons law of partial pressure, $\mathrm{P}_{\mathrm{N}_2 \mathrm{O}_4}=\mathrm{X}_{\left(\mathrm{N}_2 \mathrm{O}_4\right)} \times \mathrm{P}_{\mathrm{t}}$ $\mathrm{P}_{\mathrm{NO}_2}=\mathrm{X}_{\left(\mathrm{NO}_2\right)} \times \mathrm{P}_{\mathrm{t}}$ $\mathrm{P}_{\mathrm{NO}_2}=\frac{0.4}{1.2} \times 1=0.33$
272439
The temperature of a gas in a closed container is $27^{\circ} \mathrm{C}$. If the temperature is raised to $327^{\circ} \mathrm{C}$, the pressure exerted is
1 reduced to half
2 doubled
3 reduced to one-third
4 Cannot be calculated
Explanation:
Closed container mean volume is constant, $\frac{P_1}{T_1}=\frac{P_2}{T_2}$ Given, $T_1=27+273=300 \mathrm{~K}$ $\mathrm{T}_2=327+273=600 \mathrm{~K}$, $\frac{P_1}{300}=\frac{P_2}{600}$ $\therefore \mathrm{P}_2=\frac{600}{300} \times \mathrm{P}_1=2 \mathrm{P}_1$ So, Pressure become double.
Assam CEE-2014
Thermodynamics
272443
For a reaction, $K=100$ the value of $\Delta G$ will be
272465
The decomposition of dinitrogen tetroxide at $300 \mathrm{~K}$ and 1.0 atm pressure is $20 \%$. At equilibrium, the partial pressure (in atm) of nitrogen peroxide is
1 0.2
2 0.67
3 0.33
4 0.8
Explanation:
Molar ratio $=1: 2$ Also, only $20 \%$ of $\mathrm{N}_2 \mathrm{O}_4$ gets decomposed means. Thus, Total number of moles at equilibrium $=0.4+0.8$ $=1.2$ moles $\because$ Total pressure $\left(\mathrm{P}_{\mathrm{t}}\right)=1 \mathrm{~atm}$ (given) According to Daltons law of partial pressure, $\mathrm{P}_{\mathrm{N}_2 \mathrm{O}_4}=\mathrm{X}_{\left(\mathrm{N}_2 \mathrm{O}_4\right)} \times \mathrm{P}_{\mathrm{t}}$ $\mathrm{P}_{\mathrm{NO}_2}=\mathrm{X}_{\left(\mathrm{NO}_2\right)} \times \mathrm{P}_{\mathrm{t}}$ $\mathrm{P}_{\mathrm{NO}_2}=\frac{0.4}{1.2} \times 1=0.33$
NEET Test Series from KOTA - 10 Papers In MS WORD
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Thermodynamics
272439
The temperature of a gas in a closed container is $27^{\circ} \mathrm{C}$. If the temperature is raised to $327^{\circ} \mathrm{C}$, the pressure exerted is
1 reduced to half
2 doubled
3 reduced to one-third
4 Cannot be calculated
Explanation:
Closed container mean volume is constant, $\frac{P_1}{T_1}=\frac{P_2}{T_2}$ Given, $T_1=27+273=300 \mathrm{~K}$ $\mathrm{T}_2=327+273=600 \mathrm{~K}$, $\frac{P_1}{300}=\frac{P_2}{600}$ $\therefore \mathrm{P}_2=\frac{600}{300} \times \mathrm{P}_1=2 \mathrm{P}_1$ So, Pressure become double.
Assam CEE-2014
Thermodynamics
272443
For a reaction, $K=100$ the value of $\Delta G$ will be
272465
The decomposition of dinitrogen tetroxide at $300 \mathrm{~K}$ and 1.0 atm pressure is $20 \%$. At equilibrium, the partial pressure (in atm) of nitrogen peroxide is
1 0.2
2 0.67
3 0.33
4 0.8
Explanation:
Molar ratio $=1: 2$ Also, only $20 \%$ of $\mathrm{N}_2 \mathrm{O}_4$ gets decomposed means. Thus, Total number of moles at equilibrium $=0.4+0.8$ $=1.2$ moles $\because$ Total pressure $\left(\mathrm{P}_{\mathrm{t}}\right)=1 \mathrm{~atm}$ (given) According to Daltons law of partial pressure, $\mathrm{P}_{\mathrm{N}_2 \mathrm{O}_4}=\mathrm{X}_{\left(\mathrm{N}_2 \mathrm{O}_4\right)} \times \mathrm{P}_{\mathrm{t}}$ $\mathrm{P}_{\mathrm{NO}_2}=\mathrm{X}_{\left(\mathrm{NO}_2\right)} \times \mathrm{P}_{\mathrm{t}}$ $\mathrm{P}_{\mathrm{NO}_2}=\frac{0.4}{1.2} \times 1=0.33$
