NEET Test Series from KOTA - 10 Papers In MS WORD
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Kinetic Theory of Gases
138913
A system of gas at NTP for which $\gamma=1.5$ is suddenly compressed to $\frac{1}{9}$ of its volume. Calculate the final temperature of the gas.
1 $819^{\circ} \mathrm{C}$
2 $546^{\circ} \mathrm{C}$
3 $1095^{\circ} \mathrm{C}$
4 $273{ }^{\circ} \mathrm{C}$
Explanation:
B Given, Initial volume of gas $=\mathrm{V}$ Final volume of gas $=\frac{\mathrm{V}}{9}$ Initial temperature at NTP $=273 \mathrm{~K}$ Specific heat ratio $(\gamma)=1.5$ We know that, Ideal gas equation, $\mathrm{PV}=\mathrm{nRT}$ Adiabatic process $\mathrm{PV}^{\gamma}=\text { Constant }$ From equation (i) and (ii), we get - $\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=\left(\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}\right)^{\gamma-1}$ $\therefore \quad \mathrm{T}_{2}=\mathrm{T}_{1}\left(\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}\right)^{\gamma-1}$ $\mathrm{T}_{2}=273\left(\frac{\mathrm{V}}{\mathrm{V} / 9}\right)^{1.5-1}$ $\mathrm{T}_{2}=273(9)^{0.5}$ $\mathrm{T}_{2}=819 \mathrm{~K}$ $\therefore \quad \mathrm{T}_{2}=(819-273)^{\circ} \mathrm{C}$ $\mathrm{T}_{2}=546^{\circ} \mathrm{C}$
Shift-II
Kinetic Theory of Gases
138914
At constant pressure, the temperature and volume of a gas are increased by $2^{\circ} \mathrm{C}$ and $0.5 \%$ respectively. Find the initial temperature.
1 $400 \mathrm{~K}$
2 $200 \mathrm{~K}$
3 $100 \mathrm{~K}$
4 $150 \mathrm{~K}$
Explanation:
A Given, Initial temperature of gas $=T_{1}$ Final temperature of gas $=T_{2}$ $\mathrm{T}_{2}-\mathrm{T}_{1}=2$ Initial volume of gas $=\mathrm{V}_{1}$ Final volume of gas $=V_{2}$ $\%$ Increase in volume $\mathrm{V}_{2}=\left(1+\frac{0.5}{100}\right) \mathrm{V}$ We know that, Ideal gas equation $\mathrm{PV}=\mathrm{nRT}$ For constant pressure $\mathrm{V} \propto \mathrm{T}$ $\frac{\mathrm{V}_1}{\mathrm{~T}_1}=\frac{\mathrm{V}_2}{\mathrm{~T}_2}$ $\frac{\mathrm{T}_2}{\mathrm{~T}_1}=\frac{\mathrm{V}_2}{\mathrm{~V}_1}$ $\frac{\mathrm{T}_1+2}{\mathrm{~T}_1}=\frac{\left(1+\frac{0.5}{100}\right) \mathrm{V}}{\mathrm{V}}$ $1+\frac{2}{\mathrm{~T}_1}=1+\frac{0.5}{100}$ $\frac{2}{\mathrm{~T}_1}=\frac{0.5}{100}$ $\mathrm{~T}_1=400 \mathrm{~K}$ $\therefore \quad \mathrm{T}_1=400 \mathrm{~K}$
Shift-II
Kinetic Theory of Gases
138916
Pressure versus temperature graph of equal number of moles of an ideal gas of different volumes are plotted as shown in the figure. Choose the correct alternative
C We know that, Ideal gas equation, $\mathrm{PV}=\mathrm{nRT}$ Now, In Pressure - temperature graph Constant volume line is straight line $\mathrm{V}_{1}=\mathrm{V}_{2}$ and $\mathrm{V}_{3}=\mathrm{V}_{4}$ And slope of line 1-2 is smaller than the 3-4 line $\therefore \mathrm{V}_{2}>\mathrm{V}_{3}$
Shift-I]
Kinetic Theory of Gases
138921
The average force applied on the walls of a closed container depends as ' $T$, where ' $T$ ' is the temperature of an ideal gas. The value of ' $x$ ' is
1 1
2 zero
3 2
4 3
Explanation:
A As we know that $\mathrm{F} \propto \mathrm{P}$ According to ideal gas equation, we get- $\mathrm{PV}=\mathrm{nRT}$ $\therefore \quad \mathrm{P}=\frac{\mathrm{nRT}}{\mathrm{V}}$ When $\mathrm{V}$ is constant $\mathrm{P} \propto \mathrm{T}$ From equation (i) and (ii) we can write an equation as$\mathrm{F} \propto \mathrm{T} \Rightarrow \mathrm{F}^{1} \propto \mathrm{T}^{\mathrm{x}}$ Where, $\mathrm{x}=1$
138913
A system of gas at NTP for which $\gamma=1.5$ is suddenly compressed to $\frac{1}{9}$ of its volume. Calculate the final temperature of the gas.
1 $819^{\circ} \mathrm{C}$
2 $546^{\circ} \mathrm{C}$
3 $1095^{\circ} \mathrm{C}$
4 $273{ }^{\circ} \mathrm{C}$
Explanation:
B Given, Initial volume of gas $=\mathrm{V}$ Final volume of gas $=\frac{\mathrm{V}}{9}$ Initial temperature at NTP $=273 \mathrm{~K}$ Specific heat ratio $(\gamma)=1.5$ We know that, Ideal gas equation, $\mathrm{PV}=\mathrm{nRT}$ Adiabatic process $\mathrm{PV}^{\gamma}=\text { Constant }$ From equation (i) and (ii), we get - $\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=\left(\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}\right)^{\gamma-1}$ $\therefore \quad \mathrm{T}_{2}=\mathrm{T}_{1}\left(\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}\right)^{\gamma-1}$ $\mathrm{T}_{2}=273\left(\frac{\mathrm{V}}{\mathrm{V} / 9}\right)^{1.5-1}$ $\mathrm{T}_{2}=273(9)^{0.5}$ $\mathrm{T}_{2}=819 \mathrm{~K}$ $\therefore \quad \mathrm{T}_{2}=(819-273)^{\circ} \mathrm{C}$ $\mathrm{T}_{2}=546^{\circ} \mathrm{C}$
Shift-II
Kinetic Theory of Gases
138914
At constant pressure, the temperature and volume of a gas are increased by $2^{\circ} \mathrm{C}$ and $0.5 \%$ respectively. Find the initial temperature.
1 $400 \mathrm{~K}$
2 $200 \mathrm{~K}$
3 $100 \mathrm{~K}$
4 $150 \mathrm{~K}$
Explanation:
A Given, Initial temperature of gas $=T_{1}$ Final temperature of gas $=T_{2}$ $\mathrm{T}_{2}-\mathrm{T}_{1}=2$ Initial volume of gas $=\mathrm{V}_{1}$ Final volume of gas $=V_{2}$ $\%$ Increase in volume $\mathrm{V}_{2}=\left(1+\frac{0.5}{100}\right) \mathrm{V}$ We know that, Ideal gas equation $\mathrm{PV}=\mathrm{nRT}$ For constant pressure $\mathrm{V} \propto \mathrm{T}$ $\frac{\mathrm{V}_1}{\mathrm{~T}_1}=\frac{\mathrm{V}_2}{\mathrm{~T}_2}$ $\frac{\mathrm{T}_2}{\mathrm{~T}_1}=\frac{\mathrm{V}_2}{\mathrm{~V}_1}$ $\frac{\mathrm{T}_1+2}{\mathrm{~T}_1}=\frac{\left(1+\frac{0.5}{100}\right) \mathrm{V}}{\mathrm{V}}$ $1+\frac{2}{\mathrm{~T}_1}=1+\frac{0.5}{100}$ $\frac{2}{\mathrm{~T}_1}=\frac{0.5}{100}$ $\mathrm{~T}_1=400 \mathrm{~K}$ $\therefore \quad \mathrm{T}_1=400 \mathrm{~K}$
Shift-II
Kinetic Theory of Gases
138916
Pressure versus temperature graph of equal number of moles of an ideal gas of different volumes are plotted as shown in the figure. Choose the correct alternative
C We know that, Ideal gas equation, $\mathrm{PV}=\mathrm{nRT}$ Now, In Pressure - temperature graph Constant volume line is straight line $\mathrm{V}_{1}=\mathrm{V}_{2}$ and $\mathrm{V}_{3}=\mathrm{V}_{4}$ And slope of line 1-2 is smaller than the 3-4 line $\therefore \mathrm{V}_{2}>\mathrm{V}_{3}$
Shift-I]
Kinetic Theory of Gases
138921
The average force applied on the walls of a closed container depends as ' $T$, where ' $T$ ' is the temperature of an ideal gas. The value of ' $x$ ' is
1 1
2 zero
3 2
4 3
Explanation:
A As we know that $\mathrm{F} \propto \mathrm{P}$ According to ideal gas equation, we get- $\mathrm{PV}=\mathrm{nRT}$ $\therefore \quad \mathrm{P}=\frac{\mathrm{nRT}}{\mathrm{V}}$ When $\mathrm{V}$ is constant $\mathrm{P} \propto \mathrm{T}$ From equation (i) and (ii) we can write an equation as$\mathrm{F} \propto \mathrm{T} \Rightarrow \mathrm{F}^{1} \propto \mathrm{T}^{\mathrm{x}}$ Where, $\mathrm{x}=1$
138913
A system of gas at NTP for which $\gamma=1.5$ is suddenly compressed to $\frac{1}{9}$ of its volume. Calculate the final temperature of the gas.
1 $819^{\circ} \mathrm{C}$
2 $546^{\circ} \mathrm{C}$
3 $1095^{\circ} \mathrm{C}$
4 $273{ }^{\circ} \mathrm{C}$
Explanation:
B Given, Initial volume of gas $=\mathrm{V}$ Final volume of gas $=\frac{\mathrm{V}}{9}$ Initial temperature at NTP $=273 \mathrm{~K}$ Specific heat ratio $(\gamma)=1.5$ We know that, Ideal gas equation, $\mathrm{PV}=\mathrm{nRT}$ Adiabatic process $\mathrm{PV}^{\gamma}=\text { Constant }$ From equation (i) and (ii), we get - $\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=\left(\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}\right)^{\gamma-1}$ $\therefore \quad \mathrm{T}_{2}=\mathrm{T}_{1}\left(\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}\right)^{\gamma-1}$ $\mathrm{T}_{2}=273\left(\frac{\mathrm{V}}{\mathrm{V} / 9}\right)^{1.5-1}$ $\mathrm{T}_{2}=273(9)^{0.5}$ $\mathrm{T}_{2}=819 \mathrm{~K}$ $\therefore \quad \mathrm{T}_{2}=(819-273)^{\circ} \mathrm{C}$ $\mathrm{T}_{2}=546^{\circ} \mathrm{C}$
Shift-II
Kinetic Theory of Gases
138914
At constant pressure, the temperature and volume of a gas are increased by $2^{\circ} \mathrm{C}$ and $0.5 \%$ respectively. Find the initial temperature.
1 $400 \mathrm{~K}$
2 $200 \mathrm{~K}$
3 $100 \mathrm{~K}$
4 $150 \mathrm{~K}$
Explanation:
A Given, Initial temperature of gas $=T_{1}$ Final temperature of gas $=T_{2}$ $\mathrm{T}_{2}-\mathrm{T}_{1}=2$ Initial volume of gas $=\mathrm{V}_{1}$ Final volume of gas $=V_{2}$ $\%$ Increase in volume $\mathrm{V}_{2}=\left(1+\frac{0.5}{100}\right) \mathrm{V}$ We know that, Ideal gas equation $\mathrm{PV}=\mathrm{nRT}$ For constant pressure $\mathrm{V} \propto \mathrm{T}$ $\frac{\mathrm{V}_1}{\mathrm{~T}_1}=\frac{\mathrm{V}_2}{\mathrm{~T}_2}$ $\frac{\mathrm{T}_2}{\mathrm{~T}_1}=\frac{\mathrm{V}_2}{\mathrm{~V}_1}$ $\frac{\mathrm{T}_1+2}{\mathrm{~T}_1}=\frac{\left(1+\frac{0.5}{100}\right) \mathrm{V}}{\mathrm{V}}$ $1+\frac{2}{\mathrm{~T}_1}=1+\frac{0.5}{100}$ $\frac{2}{\mathrm{~T}_1}=\frac{0.5}{100}$ $\mathrm{~T}_1=400 \mathrm{~K}$ $\therefore \quad \mathrm{T}_1=400 \mathrm{~K}$
Shift-II
Kinetic Theory of Gases
138916
Pressure versus temperature graph of equal number of moles of an ideal gas of different volumes are plotted as shown in the figure. Choose the correct alternative
C We know that, Ideal gas equation, $\mathrm{PV}=\mathrm{nRT}$ Now, In Pressure - temperature graph Constant volume line is straight line $\mathrm{V}_{1}=\mathrm{V}_{2}$ and $\mathrm{V}_{3}=\mathrm{V}_{4}$ And slope of line 1-2 is smaller than the 3-4 line $\therefore \mathrm{V}_{2}>\mathrm{V}_{3}$
Shift-I]
Kinetic Theory of Gases
138921
The average force applied on the walls of a closed container depends as ' $T$, where ' $T$ ' is the temperature of an ideal gas. The value of ' $x$ ' is
1 1
2 zero
3 2
4 3
Explanation:
A As we know that $\mathrm{F} \propto \mathrm{P}$ According to ideal gas equation, we get- $\mathrm{PV}=\mathrm{nRT}$ $\therefore \quad \mathrm{P}=\frac{\mathrm{nRT}}{\mathrm{V}}$ When $\mathrm{V}$ is constant $\mathrm{P} \propto \mathrm{T}$ From equation (i) and (ii) we can write an equation as$\mathrm{F} \propto \mathrm{T} \Rightarrow \mathrm{F}^{1} \propto \mathrm{T}^{\mathrm{x}}$ Where, $\mathrm{x}=1$
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Kinetic Theory of Gases
138913
A system of gas at NTP for which $\gamma=1.5$ is suddenly compressed to $\frac{1}{9}$ of its volume. Calculate the final temperature of the gas.
1 $819^{\circ} \mathrm{C}$
2 $546^{\circ} \mathrm{C}$
3 $1095^{\circ} \mathrm{C}$
4 $273{ }^{\circ} \mathrm{C}$
Explanation:
B Given, Initial volume of gas $=\mathrm{V}$ Final volume of gas $=\frac{\mathrm{V}}{9}$ Initial temperature at NTP $=273 \mathrm{~K}$ Specific heat ratio $(\gamma)=1.5$ We know that, Ideal gas equation, $\mathrm{PV}=\mathrm{nRT}$ Adiabatic process $\mathrm{PV}^{\gamma}=\text { Constant }$ From equation (i) and (ii), we get - $\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=\left(\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}\right)^{\gamma-1}$ $\therefore \quad \mathrm{T}_{2}=\mathrm{T}_{1}\left(\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}\right)^{\gamma-1}$ $\mathrm{T}_{2}=273\left(\frac{\mathrm{V}}{\mathrm{V} / 9}\right)^{1.5-1}$ $\mathrm{T}_{2}=273(9)^{0.5}$ $\mathrm{T}_{2}=819 \mathrm{~K}$ $\therefore \quad \mathrm{T}_{2}=(819-273)^{\circ} \mathrm{C}$ $\mathrm{T}_{2}=546^{\circ} \mathrm{C}$
Shift-II
Kinetic Theory of Gases
138914
At constant pressure, the temperature and volume of a gas are increased by $2^{\circ} \mathrm{C}$ and $0.5 \%$ respectively. Find the initial temperature.
1 $400 \mathrm{~K}$
2 $200 \mathrm{~K}$
3 $100 \mathrm{~K}$
4 $150 \mathrm{~K}$
Explanation:
A Given, Initial temperature of gas $=T_{1}$ Final temperature of gas $=T_{2}$ $\mathrm{T}_{2}-\mathrm{T}_{1}=2$ Initial volume of gas $=\mathrm{V}_{1}$ Final volume of gas $=V_{2}$ $\%$ Increase in volume $\mathrm{V}_{2}=\left(1+\frac{0.5}{100}\right) \mathrm{V}$ We know that, Ideal gas equation $\mathrm{PV}=\mathrm{nRT}$ For constant pressure $\mathrm{V} \propto \mathrm{T}$ $\frac{\mathrm{V}_1}{\mathrm{~T}_1}=\frac{\mathrm{V}_2}{\mathrm{~T}_2}$ $\frac{\mathrm{T}_2}{\mathrm{~T}_1}=\frac{\mathrm{V}_2}{\mathrm{~V}_1}$ $\frac{\mathrm{T}_1+2}{\mathrm{~T}_1}=\frac{\left(1+\frac{0.5}{100}\right) \mathrm{V}}{\mathrm{V}}$ $1+\frac{2}{\mathrm{~T}_1}=1+\frac{0.5}{100}$ $\frac{2}{\mathrm{~T}_1}=\frac{0.5}{100}$ $\mathrm{~T}_1=400 \mathrm{~K}$ $\therefore \quad \mathrm{T}_1=400 \mathrm{~K}$
Shift-II
Kinetic Theory of Gases
138916
Pressure versus temperature graph of equal number of moles of an ideal gas of different volumes are plotted as shown in the figure. Choose the correct alternative
C We know that, Ideal gas equation, $\mathrm{PV}=\mathrm{nRT}$ Now, In Pressure - temperature graph Constant volume line is straight line $\mathrm{V}_{1}=\mathrm{V}_{2}$ and $\mathrm{V}_{3}=\mathrm{V}_{4}$ And slope of line 1-2 is smaller than the 3-4 line $\therefore \mathrm{V}_{2}>\mathrm{V}_{3}$
Shift-I]
Kinetic Theory of Gases
138921
The average force applied on the walls of a closed container depends as ' $T$, where ' $T$ ' is the temperature of an ideal gas. The value of ' $x$ ' is
1 1
2 zero
3 2
4 3
Explanation:
A As we know that $\mathrm{F} \propto \mathrm{P}$ According to ideal gas equation, we get- $\mathrm{PV}=\mathrm{nRT}$ $\therefore \quad \mathrm{P}=\frac{\mathrm{nRT}}{\mathrm{V}}$ When $\mathrm{V}$ is constant $\mathrm{P} \propto \mathrm{T}$ From equation (i) and (ii) we can write an equation as$\mathrm{F} \propto \mathrm{T} \Rightarrow \mathrm{F}^{1} \propto \mathrm{T}^{\mathrm{x}}$ Where, $\mathrm{x}=1$
