QBManager

Kinetic Theory of Gases

139001 An ideal gas initially at pressure 1 bar is being compressed from 30 m3 to 10 m3 volume and its
temperature decreases from 320 K to 280 K then fine final pressure of gas.

1 2.625 bar

2 3.4 bar

3 1.325 bar

4 4.5 bar

AIIMS-25.05.2019(E) Shift-2

Kinetic Theory of Gases

139002 An ideal gas is placed in a tank at $27^{\circ} \mathrm{C}$. The pressure is initially $600 \mathrm{kPa}$. One fourth of the gas is then released from the tank and thermal equilibrium is established. What will be the pressure if the temperature is $327^{\circ} \mathrm{C}$ ?

1 $900 \mathrm{kPa}$

2 $1000 \mathrm{kPa}$

3 $1050 \mathrm{kPa}$

4 $1250 \mathrm{kPa}$

Shift-2]

Kinetic Theory of Gases

139003 A gas expands with temperature according to the relation, $\mathrm{V}=\mathrm{KT}^{2 / 3}$, where $\mathrm{k}$ is a constant. Work done when the temperature changes by $60 \mathrm{~K}$ is $(R=$ universal gas constant. $)$

1 $10 \mathrm{R}$

2 $20 \mathrm{R}$

3 $50 \mathrm{R}$

4 $40 \mathrm{R}$

AP EAMCET (20.04.2019) Shift-1

Kinetic Theory of Gases

139004 A cylinder closed at both ends is separated into two equal parts $(45 \mathrm{~cm}$ each) by a piston impermeable to heat. Both the parts contain the same masses of gas at a temperature of 300 $K$ and a pressure of $1 \mathrm{~atm}$. How much the gas should be heated in one part of the cylinder to shift the piston by $5 \mathrm{~cm}$ and the pressure of the gas after shifting the piston?

1 $\mathrm{T}=365 \mathrm{~K}$ and $\mathrm{P}=1.125 \mathrm{~atm}$

2 $\mathrm{t}=350 \mathrm{~K}$ and $\mathrm{P}=1.125 \mathrm{~atm}$

3 $\mathrm{T}=375 \mathrm{~K}$ and $\mathrm{P}=2.125 \mathrm{~atm}$

4 $\mathrm{T}=350 \mathrm{~K}$ and $\mathrm{P}=2.125 \mathrm{~atm}$

5 $\mathrm{T}=375 \mathrm{~K}$ and $\mathrm{P}=1.125 \mathrm{~atm}$

Explanation:

E Let the cross - sectional area of cylinder is A.
original image
$\mathrm{P}_1=1 \mathrm{~atm} \mathrm{P}_1^{\prime}=1 \mathrm{~atm}$
$\mathrm{~T}_1=300 \mathrm{~K} \mathrm{~T}_1^{\prime}=300 \mathrm{~K}$
$\mathrm{~V}_1=45 \mathrm{~A} \mathrm{~V}^{\prime}=45 \mathrm{~A}$
Where $\mathrm{A}$ is cross section of closed cylinder.
Let in chamber $\mathrm{C}$, providing heat input then the gas will expand and it will push to chamber $\mathrm{C}^{\prime}$, therefore $\mathrm{C}^{\prime}$ will get compressed and $\mathrm{C}$ chamber will shift to $5 \mathrm{~cm}$.
Let the pressure be $\mathrm{P}_{2}$ after heating-
$\mathrm{P}_{2}=? \mathrm{P}_{2}^{\prime}=?$
$\mathrm{~T}_{2}=? \mathrm{~T}_{2}^{\prime}=300 \mathrm{~K}$
$\mathrm{~V}_{2}=(45+5) \mathrm{A} \mathrm{V}_{2}^{\prime}=(45-5) \mathrm{A}$
$\mathrm{V}_{2}=50 \mathrm{~A} \mathrm{~V}_{2}^{\prime}=40 \mathrm{~A}$
Now, $\quad \mathrm{P}_{2} \mathrm{~A}=\mathrm{P}_{2}^{\prime} \mathrm{A}=$ Pressure force
$\mathrm{P}_{2}=\mathrm{P}_{2}^{\prime}$
$\frac{\mathrm{nRT}_{2}}{\mathrm{~V}_{2}}=\frac{\mathrm{nRT}_{2}^{\prime}}{\mathrm{V}_{2}^{\prime}}$
$\mathrm{T}_{2}=\frac{\mathrm{T}_{2}^{\prime} \mathrm{V}_{2}}{\mathrm{~V}_{2}^{\prime}}=\frac{50 \mathrm{~A} \times 300}{40 \mathrm{~A}}$
$\mathrm{T}_{2}=1.25 \times 300=375 \mathrm{~K}$
Though the mass will be the same after heating and before heating the pressure will be -
$\mathrm{PV}=\mathrm{nRT}$
$\mathrm{PV}=\frac{\mathrm{m}}{\mathrm{M}} \mathrm{RT}$
$\mathrm{PV} \times \mathrm{M}=\mathrm{mRT}$
$\mathrm{m}=\frac{\mathrm{PVM}}{\mathrm{RT}}$
$\therefore \quad \frac{\mathrm{P}_{1} \mathrm{~V}_{1} \mathrm{M}}{\mathrm{RT}_{1}}=\frac{\mathrm{P}_{2} \mathrm{~V}_{2} \mathrm{M}}{\mathrm{RT}_{2}}$
$\frac{\mathrm{P}_{1} \mathrm{~V}_{1}}{\mathrm{~T}_{1}}=\frac{\mathrm{P}_{2} \mathrm{~V}_{2}}{\mathrm{~T}_{2}}$
$\mathrm{P}_{1} \mathrm{~V}_{1} \mathrm{~T}_{2}=\mathrm{P}_{2} \mathrm{~V}_{2} \mathrm{~T}_{1}$
$\mathrm{P}_{2}=\frac{\mathrm{P}_{1} \mathrm{~V}_{1} \mathrm{~T}_{2}}{\mathrm{~V}_{2} \mathrm{~T}_{1}}$
$=\frac{1 \times 45 \mathrm{~A} \times 375}{300 \times 50 \mathrm{~A}}$
$=\frac{3375}{3000}$
$\mathrm{P}_{2}=1.125 \mathrm{~atm} .$
Therefore, the temperature will be $375 \mathrm{~K}$ and the pressure will be $1.125 \mathrm{~atm}$.

Kerala CEE -2018

Kinetic Theory of Gases

139005 Different curves in the figure showing the behavior of different gases

1 $1,2,4$

2 $1,3,4$

3 $1,3,5$

4 $1,2,5$

AMU-2018

Kinetic Theory of Gases

139001 An ideal gas initially at pressure 1 bar is being compressed from 30 m3 to 10 m3 volume and its
temperature decreases from 320 K to 280 K then fine final pressure of gas.

1 2.625 bar

2 3.4 bar

3 1.325 bar

4 4.5 bar

AIIMS-25.05.2019(E) Shift-2

Kinetic Theory of Gases

139002 An ideal gas is placed in a tank at $27^{\circ} \mathrm{C}$. The pressure is initially $600 \mathrm{kPa}$. One fourth of the gas is then released from the tank and thermal equilibrium is established. What will be the pressure if the temperature is $327^{\circ} \mathrm{C}$ ?

1 $900 \mathrm{kPa}$

2 $1000 \mathrm{kPa}$

3 $1050 \mathrm{kPa}$

4 $1250 \mathrm{kPa}$

Shift-2]

Kinetic Theory of Gases

139003 A gas expands with temperature according to the relation, $\mathrm{V}=\mathrm{KT}^{2 / 3}$, where $\mathrm{k}$ is a constant. Work done when the temperature changes by $60 \mathrm{~K}$ is $(R=$ universal gas constant. $)$

1 $10 \mathrm{R}$

2 $20 \mathrm{R}$

3 $50 \mathrm{R}$

4 $40 \mathrm{R}$

AP EAMCET (20.04.2019) Shift-1

Kinetic Theory of Gases

139004 A cylinder closed at both ends is separated into two equal parts $(45 \mathrm{~cm}$ each) by a piston impermeable to heat. Both the parts contain the same masses of gas at a temperature of 300 $K$ and a pressure of $1 \mathrm{~atm}$. How much the gas should be heated in one part of the cylinder to shift the piston by $5 \mathrm{~cm}$ and the pressure of the gas after shifting the piston?

1 $\mathrm{T}=365 \mathrm{~K}$ and $\mathrm{P}=1.125 \mathrm{~atm}$

2 $\mathrm{t}=350 \mathrm{~K}$ and $\mathrm{P}=1.125 \mathrm{~atm}$

3 $\mathrm{T}=375 \mathrm{~K}$ and $\mathrm{P}=2.125 \mathrm{~atm}$

4 $\mathrm{T}=350 \mathrm{~K}$ and $\mathrm{P}=2.125 \mathrm{~atm}$

5 $\mathrm{T}=375 \mathrm{~K}$ and $\mathrm{P}=1.125 \mathrm{~atm}$

Explanation:

E Let the cross - sectional area of cylinder is A.
original image
$\mathrm{P}_1=1 \mathrm{~atm} \mathrm{P}_1^{\prime}=1 \mathrm{~atm}$
$\mathrm{~T}_1=300 \mathrm{~K} \mathrm{~T}_1^{\prime}=300 \mathrm{~K}$
$\mathrm{~V}_1=45 \mathrm{~A} \mathrm{~V}^{\prime}=45 \mathrm{~A}$
Where $\mathrm{A}$ is cross section of closed cylinder.
Let in chamber $\mathrm{C}$, providing heat input then the gas will expand and it will push to chamber $\mathrm{C}^{\prime}$, therefore $\mathrm{C}^{\prime}$ will get compressed and $\mathrm{C}$ chamber will shift to $5 \mathrm{~cm}$.
Let the pressure be $\mathrm{P}_{2}$ after heating-
$\mathrm{P}_{2}=? \mathrm{P}_{2}^{\prime}=?$
$\mathrm{~T}_{2}=? \mathrm{~T}_{2}^{\prime}=300 \mathrm{~K}$
$\mathrm{~V}_{2}=(45+5) \mathrm{A} \mathrm{V}_{2}^{\prime}=(45-5) \mathrm{A}$
$\mathrm{V}_{2}=50 \mathrm{~A} \mathrm{~V}_{2}^{\prime}=40 \mathrm{~A}$
Now, $\quad \mathrm{P}_{2} \mathrm{~A}=\mathrm{P}_{2}^{\prime} \mathrm{A}=$ Pressure force
$\mathrm{P}_{2}=\mathrm{P}_{2}^{\prime}$
$\frac{\mathrm{nRT}_{2}}{\mathrm{~V}_{2}}=\frac{\mathrm{nRT}_{2}^{\prime}}{\mathrm{V}_{2}^{\prime}}$
$\mathrm{T}_{2}=\frac{\mathrm{T}_{2}^{\prime} \mathrm{V}_{2}}{\mathrm{~V}_{2}^{\prime}}=\frac{50 \mathrm{~A} \times 300}{40 \mathrm{~A}}$
$\mathrm{T}_{2}=1.25 \times 300=375 \mathrm{~K}$
Though the mass will be the same after heating and before heating the pressure will be -
$\mathrm{PV}=\mathrm{nRT}$
$\mathrm{PV}=\frac{\mathrm{m}}{\mathrm{M}} \mathrm{RT}$
$\mathrm{PV} \times \mathrm{M}=\mathrm{mRT}$
$\mathrm{m}=\frac{\mathrm{PVM}}{\mathrm{RT}}$
$\therefore \quad \frac{\mathrm{P}_{1} \mathrm{~V}_{1} \mathrm{M}}{\mathrm{RT}_{1}}=\frac{\mathrm{P}_{2} \mathrm{~V}_{2} \mathrm{M}}{\mathrm{RT}_{2}}$
$\frac{\mathrm{P}_{1} \mathrm{~V}_{1}}{\mathrm{~T}_{1}}=\frac{\mathrm{P}_{2} \mathrm{~V}_{2}}{\mathrm{~T}_{2}}$
$\mathrm{P}_{1} \mathrm{~V}_{1} \mathrm{~T}_{2}=\mathrm{P}_{2} \mathrm{~V}_{2} \mathrm{~T}_{1}$
$\mathrm{P}_{2}=\frac{\mathrm{P}_{1} \mathrm{~V}_{1} \mathrm{~T}_{2}}{\mathrm{~V}_{2} \mathrm{~T}_{1}}$
$=\frac{1 \times 45 \mathrm{~A} \times 375}{300 \times 50 \mathrm{~A}}$
$=\frac{3375}{3000}$
$\mathrm{P}_{2}=1.125 \mathrm{~atm} .$
Therefore, the temperature will be $375 \mathrm{~K}$ and the pressure will be $1.125 \mathrm{~atm}$.

Kerala CEE -2018

Kinetic Theory of Gases

139005 Different curves in the figure showing the behavior of different gases

1 $1,2,4$

2 $1,3,4$

3 $1,3,5$

4 $1,2,5$

AMU-2018

Kinetic Theory of Gases

139001 An ideal gas initially at pressure 1 bar is being compressed from 30 m3 to 10 m3 volume and its
temperature decreases from 320 K to 280 K then fine final pressure of gas.

1 2.625 bar

2 3.4 bar

3 1.325 bar

4 4.5 bar

AIIMS-25.05.2019(E) Shift-2

Kinetic Theory of Gases

139002 An ideal gas is placed in a tank at $27^{\circ} \mathrm{C}$. The pressure is initially $600 \mathrm{kPa}$. One fourth of the gas is then released from the tank and thermal equilibrium is established. What will be the pressure if the temperature is $327^{\circ} \mathrm{C}$ ?

1 $900 \mathrm{kPa}$

2 $1000 \mathrm{kPa}$

3 $1050 \mathrm{kPa}$

4 $1250 \mathrm{kPa}$

Shift-2]

Kinetic Theory of Gases

139003 A gas expands with temperature according to the relation, $\mathrm{V}=\mathrm{KT}^{2 / 3}$, where $\mathrm{k}$ is a constant. Work done when the temperature changes by $60 \mathrm{~K}$ is $(R=$ universal gas constant. $)$

1 $10 \mathrm{R}$

2 $20 \mathrm{R}$

3 $50 \mathrm{R}$

4 $40 \mathrm{R}$

AP EAMCET (20.04.2019) Shift-1

Kinetic Theory of Gases

139004 A cylinder closed at both ends is separated into two equal parts $(45 \mathrm{~cm}$ each) by a piston impermeable to heat. Both the parts contain the same masses of gas at a temperature of 300 $K$ and a pressure of $1 \mathrm{~atm}$. How much the gas should be heated in one part of the cylinder to shift the piston by $5 \mathrm{~cm}$ and the pressure of the gas after shifting the piston?

1 $\mathrm{T}=365 \mathrm{~K}$ and $\mathrm{P}=1.125 \mathrm{~atm}$

2 $\mathrm{t}=350 \mathrm{~K}$ and $\mathrm{P}=1.125 \mathrm{~atm}$

3 $\mathrm{T}=375 \mathrm{~K}$ and $\mathrm{P}=2.125 \mathrm{~atm}$

4 $\mathrm{T}=350 \mathrm{~K}$ and $\mathrm{P}=2.125 \mathrm{~atm}$

5 $\mathrm{T}=375 \mathrm{~K}$ and $\mathrm{P}=1.125 \mathrm{~atm}$

Explanation:

E Let the cross - sectional area of cylinder is A.
original image
$\mathrm{P}_1=1 \mathrm{~atm} \mathrm{P}_1^{\prime}=1 \mathrm{~atm}$
$\mathrm{~T}_1=300 \mathrm{~K} \mathrm{~T}_1^{\prime}=300 \mathrm{~K}$
$\mathrm{~V}_1=45 \mathrm{~A} \mathrm{~V}^{\prime}=45 \mathrm{~A}$
Where $\mathrm{A}$ is cross section of closed cylinder.
Let in chamber $\mathrm{C}$, providing heat input then the gas will expand and it will push to chamber $\mathrm{C}^{\prime}$, therefore $\mathrm{C}^{\prime}$ will get compressed and $\mathrm{C}$ chamber will shift to $5 \mathrm{~cm}$.
Let the pressure be $\mathrm{P}_{2}$ after heating-
$\mathrm{P}_{2}=? \mathrm{P}_{2}^{\prime}=?$
$\mathrm{~T}_{2}=? \mathrm{~T}_{2}^{\prime}=300 \mathrm{~K}$
$\mathrm{~V}_{2}=(45+5) \mathrm{A} \mathrm{V}_{2}^{\prime}=(45-5) \mathrm{A}$
$\mathrm{V}_{2}=50 \mathrm{~A} \mathrm{~V}_{2}^{\prime}=40 \mathrm{~A}$
Now, $\quad \mathrm{P}_{2} \mathrm{~A}=\mathrm{P}_{2}^{\prime} \mathrm{A}=$ Pressure force
$\mathrm{P}_{2}=\mathrm{P}_{2}^{\prime}$
$\frac{\mathrm{nRT}_{2}}{\mathrm{~V}_{2}}=\frac{\mathrm{nRT}_{2}^{\prime}}{\mathrm{V}_{2}^{\prime}}$
$\mathrm{T}_{2}=\frac{\mathrm{T}_{2}^{\prime} \mathrm{V}_{2}}{\mathrm{~V}_{2}^{\prime}}=\frac{50 \mathrm{~A} \times 300}{40 \mathrm{~A}}$
$\mathrm{T}_{2}=1.25 \times 300=375 \mathrm{~K}$
Though the mass will be the same after heating and before heating the pressure will be -
$\mathrm{PV}=\mathrm{nRT}$
$\mathrm{PV}=\frac{\mathrm{m}}{\mathrm{M}} \mathrm{RT}$
$\mathrm{PV} \times \mathrm{M}=\mathrm{mRT}$
$\mathrm{m}=\frac{\mathrm{PVM}}{\mathrm{RT}}$
$\therefore \quad \frac{\mathrm{P}_{1} \mathrm{~V}_{1} \mathrm{M}}{\mathrm{RT}_{1}}=\frac{\mathrm{P}_{2} \mathrm{~V}_{2} \mathrm{M}}{\mathrm{RT}_{2}}$
$\frac{\mathrm{P}_{1} \mathrm{~V}_{1}}{\mathrm{~T}_{1}}=\frac{\mathrm{P}_{2} \mathrm{~V}_{2}}{\mathrm{~T}_{2}}$
$\mathrm{P}_{1} \mathrm{~V}_{1} \mathrm{~T}_{2}=\mathrm{P}_{2} \mathrm{~V}_{2} \mathrm{~T}_{1}$
$\mathrm{P}_{2}=\frac{\mathrm{P}_{1} \mathrm{~V}_{1} \mathrm{~T}_{2}}{\mathrm{~V}_{2} \mathrm{~T}_{1}}$
$=\frac{1 \times 45 \mathrm{~A} \times 375}{300 \times 50 \mathrm{~A}}$
$=\frac{3375}{3000}$
$\mathrm{P}_{2}=1.125 \mathrm{~atm} .$
Therefore, the temperature will be $375 \mathrm{~K}$ and the pressure will be $1.125 \mathrm{~atm}$.

Kerala CEE -2018

Kinetic Theory of Gases

139005 Different curves in the figure showing the behavior of different gases

1 $1,2,4$

2 $1,3,4$

3 $1,3,5$

4 $1,2,5$

AMU-2018

Kinetic Theory of Gases

139001 An ideal gas initially at pressure 1 bar is being compressed from 30 m3 to 10 m3 volume and its
temperature decreases from 320 K to 280 K then fine final pressure of gas.

1 2.625 bar

2 3.4 bar

3 1.325 bar

4 4.5 bar

AIIMS-25.05.2019(E) Shift-2

Kinetic Theory of Gases

139002 An ideal gas is placed in a tank at $27^{\circ} \mathrm{C}$. The pressure is initially $600 \mathrm{kPa}$. One fourth of the gas is then released from the tank and thermal equilibrium is established. What will be the pressure if the temperature is $327^{\circ} \mathrm{C}$ ?

1 $900 \mathrm{kPa}$

2 $1000 \mathrm{kPa}$

3 $1050 \mathrm{kPa}$

4 $1250 \mathrm{kPa}$

Shift-2]

Kinetic Theory of Gases

139003 A gas expands with temperature according to the relation, $\mathrm{V}=\mathrm{KT}^{2 / 3}$, where $\mathrm{k}$ is a constant. Work done when the temperature changes by $60 \mathrm{~K}$ is $(R=$ universal gas constant. $)$

1 $10 \mathrm{R}$

2 $20 \mathrm{R}$

3 $50 \mathrm{R}$

4 $40 \mathrm{R}$

AP EAMCET (20.04.2019) Shift-1

Kinetic Theory of Gases

139004 A cylinder closed at both ends is separated into two equal parts $(45 \mathrm{~cm}$ each) by a piston impermeable to heat. Both the parts contain the same masses of gas at a temperature of 300 $K$ and a pressure of $1 \mathrm{~atm}$. How much the gas should be heated in one part of the cylinder to shift the piston by $5 \mathrm{~cm}$ and the pressure of the gas after shifting the piston?

1 $\mathrm{T}=365 \mathrm{~K}$ and $\mathrm{P}=1.125 \mathrm{~atm}$

2 $\mathrm{t}=350 \mathrm{~K}$ and $\mathrm{P}=1.125 \mathrm{~atm}$

3 $\mathrm{T}=375 \mathrm{~K}$ and $\mathrm{P}=2.125 \mathrm{~atm}$

4 $\mathrm{T}=350 \mathrm{~K}$ and $\mathrm{P}=2.125 \mathrm{~atm}$

5 $\mathrm{T}=375 \mathrm{~K}$ and $\mathrm{P}=1.125 \mathrm{~atm}$

Explanation:

E Let the cross - sectional area of cylinder is A.
original image
$\mathrm{P}_1=1 \mathrm{~atm} \mathrm{P}_1^{\prime}=1 \mathrm{~atm}$
$\mathrm{~T}_1=300 \mathrm{~K} \mathrm{~T}_1^{\prime}=300 \mathrm{~K}$
$\mathrm{~V}_1=45 \mathrm{~A} \mathrm{~V}^{\prime}=45 \mathrm{~A}$
Where $\mathrm{A}$ is cross section of closed cylinder.
Let in chamber $\mathrm{C}$, providing heat input then the gas will expand and it will push to chamber $\mathrm{C}^{\prime}$, therefore $\mathrm{C}^{\prime}$ will get compressed and $\mathrm{C}$ chamber will shift to $5 \mathrm{~cm}$.
Let the pressure be $\mathrm{P}_{2}$ after heating-
$\mathrm{P}_{2}=? \mathrm{P}_{2}^{\prime}=?$
$\mathrm{~T}_{2}=? \mathrm{~T}_{2}^{\prime}=300 \mathrm{~K}$
$\mathrm{~V}_{2}=(45+5) \mathrm{A} \mathrm{V}_{2}^{\prime}=(45-5) \mathrm{A}$
$\mathrm{V}_{2}=50 \mathrm{~A} \mathrm{~V}_{2}^{\prime}=40 \mathrm{~A}$
Now, $\quad \mathrm{P}_{2} \mathrm{~A}=\mathrm{P}_{2}^{\prime} \mathrm{A}=$ Pressure force
$\mathrm{P}_{2}=\mathrm{P}_{2}^{\prime}$
$\frac{\mathrm{nRT}_{2}}{\mathrm{~V}_{2}}=\frac{\mathrm{nRT}_{2}^{\prime}}{\mathrm{V}_{2}^{\prime}}$
$\mathrm{T}_{2}=\frac{\mathrm{T}_{2}^{\prime} \mathrm{V}_{2}}{\mathrm{~V}_{2}^{\prime}}=\frac{50 \mathrm{~A} \times 300}{40 \mathrm{~A}}$
$\mathrm{T}_{2}=1.25 \times 300=375 \mathrm{~K}$
Though the mass will be the same after heating and before heating the pressure will be -
$\mathrm{PV}=\mathrm{nRT}$
$\mathrm{PV}=\frac{\mathrm{m}}{\mathrm{M}} \mathrm{RT}$
$\mathrm{PV} \times \mathrm{M}=\mathrm{mRT}$
$\mathrm{m}=\frac{\mathrm{PVM}}{\mathrm{RT}}$
$\therefore \quad \frac{\mathrm{P}_{1} \mathrm{~V}_{1} \mathrm{M}}{\mathrm{RT}_{1}}=\frac{\mathrm{P}_{2} \mathrm{~V}_{2} \mathrm{M}}{\mathrm{RT}_{2}}$
$\frac{\mathrm{P}_{1} \mathrm{~V}_{1}}{\mathrm{~T}_{1}}=\frac{\mathrm{P}_{2} \mathrm{~V}_{2}}{\mathrm{~T}_{2}}$
$\mathrm{P}_{1} \mathrm{~V}_{1} \mathrm{~T}_{2}=\mathrm{P}_{2} \mathrm{~V}_{2} \mathrm{~T}_{1}$
$\mathrm{P}_{2}=\frac{\mathrm{P}_{1} \mathrm{~V}_{1} \mathrm{~T}_{2}}{\mathrm{~V}_{2} \mathrm{~T}_{1}}$
$=\frac{1 \times 45 \mathrm{~A} \times 375}{300 \times 50 \mathrm{~A}}$
$=\frac{3375}{3000}$
$\mathrm{P}_{2}=1.125 \mathrm{~atm} .$
Therefore, the temperature will be $375 \mathrm{~K}$ and the pressure will be $1.125 \mathrm{~atm}$.

Kerala CEE -2018

Kinetic Theory of Gases

139005 Different curves in the figure showing the behavior of different gases

1 $1,2,4$

2 $1,3,4$

3 $1,3,5$

4 $1,2,5$

AMU-2018

Kinetic Theory of Gases

139001 An ideal gas initially at pressure 1 bar is being compressed from 30 m3 to 10 m3 volume and its
temperature decreases from 320 K to 280 K then fine final pressure of gas.

1 2.625 bar

2 3.4 bar

3 1.325 bar

4 4.5 bar

AIIMS-25.05.2019(E) Shift-2

Kinetic Theory of Gases

139002 An ideal gas is placed in a tank at $27^{\circ} \mathrm{C}$. The pressure is initially $600 \mathrm{kPa}$. One fourth of the gas is then released from the tank and thermal equilibrium is established. What will be the pressure if the temperature is $327^{\circ} \mathrm{C}$ ?

1 $900 \mathrm{kPa}$

2 $1000 \mathrm{kPa}$

3 $1050 \mathrm{kPa}$

4 $1250 \mathrm{kPa}$

Shift-2]

Kinetic Theory of Gases

139003 A gas expands with temperature according to the relation, $\mathrm{V}=\mathrm{KT}^{2 / 3}$, where $\mathrm{k}$ is a constant. Work done when the temperature changes by $60 \mathrm{~K}$ is $(R=$ universal gas constant. $)$

1 $10 \mathrm{R}$

2 $20 \mathrm{R}$

3 $50 \mathrm{R}$

4 $40 \mathrm{R}$

AP EAMCET (20.04.2019) Shift-1

Kinetic Theory of Gases

139004 A cylinder closed at both ends is separated into two equal parts $(45 \mathrm{~cm}$ each) by a piston impermeable to heat. Both the parts contain the same masses of gas at a temperature of 300 $K$ and a pressure of $1 \mathrm{~atm}$. How much the gas should be heated in one part of the cylinder to shift the piston by $5 \mathrm{~cm}$ and the pressure of the gas after shifting the piston?

1 $\mathrm{T}=365 \mathrm{~K}$ and $\mathrm{P}=1.125 \mathrm{~atm}$

2 $\mathrm{t}=350 \mathrm{~K}$ and $\mathrm{P}=1.125 \mathrm{~atm}$

3 $\mathrm{T}=375 \mathrm{~K}$ and $\mathrm{P}=2.125 \mathrm{~atm}$

4 $\mathrm{T}=350 \mathrm{~K}$ and $\mathrm{P}=2.125 \mathrm{~atm}$

5 $\mathrm{T}=375 \mathrm{~K}$ and $\mathrm{P}=1.125 \mathrm{~atm}$

Explanation:

E Let the cross - sectional area of cylinder is A.
original image
$\mathrm{P}_1=1 \mathrm{~atm} \mathrm{P}_1^{\prime}=1 \mathrm{~atm}$
$\mathrm{~T}_1=300 \mathrm{~K} \mathrm{~T}_1^{\prime}=300 \mathrm{~K}$
$\mathrm{~V}_1=45 \mathrm{~A} \mathrm{~V}^{\prime}=45 \mathrm{~A}$
Where $\mathrm{A}$ is cross section of closed cylinder.
Let in chamber $\mathrm{C}$, providing heat input then the gas will expand and it will push to chamber $\mathrm{C}^{\prime}$, therefore $\mathrm{C}^{\prime}$ will get compressed and $\mathrm{C}$ chamber will shift to $5 \mathrm{~cm}$.
Let the pressure be $\mathrm{P}_{2}$ after heating-
$\mathrm{P}_{2}=? \mathrm{P}_{2}^{\prime}=?$
$\mathrm{~T}_{2}=? \mathrm{~T}_{2}^{\prime}=300 \mathrm{~K}$
$\mathrm{~V}_{2}=(45+5) \mathrm{A} \mathrm{V}_{2}^{\prime}=(45-5) \mathrm{A}$
$\mathrm{V}_{2}=50 \mathrm{~A} \mathrm{~V}_{2}^{\prime}=40 \mathrm{~A}$
Now, $\quad \mathrm{P}_{2} \mathrm{~A}=\mathrm{P}_{2}^{\prime} \mathrm{A}=$ Pressure force
$\mathrm{P}_{2}=\mathrm{P}_{2}^{\prime}$
$\frac{\mathrm{nRT}_{2}}{\mathrm{~V}_{2}}=\frac{\mathrm{nRT}_{2}^{\prime}}{\mathrm{V}_{2}^{\prime}}$
$\mathrm{T}_{2}=\frac{\mathrm{T}_{2}^{\prime} \mathrm{V}_{2}}{\mathrm{~V}_{2}^{\prime}}=\frac{50 \mathrm{~A} \times 300}{40 \mathrm{~A}}$
$\mathrm{T}_{2}=1.25 \times 300=375 \mathrm{~K}$
Though the mass will be the same after heating and before heating the pressure will be -
$\mathrm{PV}=\mathrm{nRT}$
$\mathrm{PV}=\frac{\mathrm{m}}{\mathrm{M}} \mathrm{RT}$
$\mathrm{PV} \times \mathrm{M}=\mathrm{mRT}$
$\mathrm{m}=\frac{\mathrm{PVM}}{\mathrm{RT}}$
$\therefore \quad \frac{\mathrm{P}_{1} \mathrm{~V}_{1} \mathrm{M}}{\mathrm{RT}_{1}}=\frac{\mathrm{P}_{2} \mathrm{~V}_{2} \mathrm{M}}{\mathrm{RT}_{2}}$
$\frac{\mathrm{P}_{1} \mathrm{~V}_{1}}{\mathrm{~T}_{1}}=\frac{\mathrm{P}_{2} \mathrm{~V}_{2}}{\mathrm{~T}_{2}}$
$\mathrm{P}_{1} \mathrm{~V}_{1} \mathrm{~T}_{2}=\mathrm{P}_{2} \mathrm{~V}_{2} \mathrm{~T}_{1}$
$\mathrm{P}_{2}=\frac{\mathrm{P}_{1} \mathrm{~V}_{1} \mathrm{~T}_{2}}{\mathrm{~V}_{2} \mathrm{~T}_{1}}$
$=\frac{1 \times 45 \mathrm{~A} \times 375}{300 \times 50 \mathrm{~A}}$
$=\frac{3375}{3000}$
$\mathrm{P}_{2}=1.125 \mathrm{~atm} .$
Therefore, the temperature will be $375 \mathrm{~K}$ and the pressure will be $1.125 \mathrm{~atm}$.

Kerala CEE -2018

Kinetic Theory of Gases

139005 Different curves in the figure showing the behavior of different gases

1 $1,2,4$

2 $1,3,4$

3 $1,3,5$

4 $1,2,5$

AMU-2018

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