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PHXI13:KINETIC THEORY

360068 Boyle's law is applicable in

1 Isobaric process

2 Isochoric process

3 Isothermal process

4 Adiabatic process

PHXI13:KINETIC THEORY

360069 Two monoatomic gas $A$ and $B$ occupying the same volume $V$ are at the same temperature $T$ and pressure $P$ . If they are mixed, the resulting mixture has volume $V$ and temperature $T$ . The pressure of the mixture is

1

P

2

\frac{P}{2}

3

4 P

4

2 P

Explanation:

$\frac{P_{1} V_{1}}{V_{1}} = \frac{P_{m i x} V_{m i x}}{T_{m i x}}$
$T_{1} = T_{m i x} = T$ and $V_{1} = V + V, V_{m i x} = V$
$\frac{P (2 V)}{T} = \frac{P_{m i x} V}{T}$
$P_{m i x} = 2 P$

PHXI13:KINETIC THEORY

360070 If $P = 10^{6} K T$ , Then the number of molecules per unit volume of the gas is [The letters have usual meaning.]

1

10^{9}

2

10^{5}

3

10^{6}

4

10^{8}

Explanation:

The ideal gas equation is
$P V = n R T = \frac{N}{N_{A}} R T = N K T$
Where $K$ is the Boltzmann's constant
Given that $P = 10^{6} K T$
$\Rightarrow \frac{N}{V} = 10^{6}$

PHXI13:KINETIC THEORY

360071 Air is pumped into an automobile tube upto a pressure of $200 k P a$ in the morning at temperature $22^{\circ} C$ . When the air temperature rises to $42^{\circ} C$ and the tube expands by $2 %$ . The pressure of the air in the tube at this temperature, will be approximately

1

206 k P a

2

209 k P a

3

200 k P a

4

212 k P a

Explanation:

$\frac{P V}{T} = R$ (constant) $\Rightarrow \frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$
$\Rightarrow \frac{200 \times V}{(273 + 22)} = \frac{P_{2} \times 1.02 V}{(273 + 42)}$
$[V_{2} = V + 0.02 V]$
$\Rightarrow P_{2} = \frac{200 \times 315}{295 \times 1.02} = 209 k P a$

PHXI13:KINETIC THEORY

360072 A bottle is inverted and dipped into a tank. At the bottom of the tank, $3 / 5$ th part of the bottle is filled with water. The depth of the tank is

1

20 m

2

15 m

3

10 m

4

5 m

Explanation:

We assume that the pressure on the surface of water is $1 a t m$ and process is isothermal.
$P_{1} V_{1} = P_{2} V_{2}$
$1 \times V_{1} = P_{2} \frac{2}{5} V$
$P_{2} = \frac{5}{2} = 2.5 a t m$
$Δ P = 1.5 a t m$
$1.5 \times 10^{5} \frac{N}{m^{2}} = ρ g h$
$h = \frac{1.5 \times 10^{5}}{10^{3} \times 10} = 15 m$

PHXI13:KINETIC THEORY

360068 Boyle's law is applicable in

1 Isobaric process

2 Isochoric process

3 Isothermal process

4 Adiabatic process

PHXI13:KINETIC THEORY

360069 Two monoatomic gas $A$ and $B$ occupying the same volume $V$ are at the same temperature $T$ and pressure $P$ . If they are mixed, the resulting mixture has volume $V$ and temperature $T$ . The pressure of the mixture is

1

P

2

\frac{P}{2}

3

4 P

4

2 P

Explanation:

$\frac{P_{1} V_{1}}{V_{1}} = \frac{P_{m i x} V_{m i x}}{T_{m i x}}$
$T_{1} = T_{m i x} = T$ and $V_{1} = V + V, V_{m i x} = V$
$\frac{P (2 V)}{T} = \frac{P_{m i x} V}{T}$
$P_{m i x} = 2 P$

PHXI13:KINETIC THEORY

360070 If $P = 10^{6} K T$ , Then the number of molecules per unit volume of the gas is [The letters have usual meaning.]

1

10^{9}

2

10^{5}

3

10^{6}

4

10^{8}

Explanation:

The ideal gas equation is
$P V = n R T = \frac{N}{N_{A}} R T = N K T$
Where $K$ is the Boltzmann's constant
Given that $P = 10^{6} K T$
$\Rightarrow \frac{N}{V} = 10^{6}$

PHXI13:KINETIC THEORY

360071 Air is pumped into an automobile tube upto a pressure of $200 k P a$ in the morning at temperature $22^{\circ} C$ . When the air temperature rises to $42^{\circ} C$ and the tube expands by $2 %$ . The pressure of the air in the tube at this temperature, will be approximately

1

206 k P a

2

209 k P a

3

200 k P a

4

212 k P a

Explanation:

$\frac{P V}{T} = R$ (constant) $\Rightarrow \frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$
$\Rightarrow \frac{200 \times V}{(273 + 22)} = \frac{P_{2} \times 1.02 V}{(273 + 42)}$
$[V_{2} = V + 0.02 V]$
$\Rightarrow P_{2} = \frac{200 \times 315}{295 \times 1.02} = 209 k P a$

PHXI13:KINETIC THEORY

360072 A bottle is inverted and dipped into a tank. At the bottom of the tank, $3 / 5$ th part of the bottle is filled with water. The depth of the tank is

1

20 m

2

15 m

3

10 m

4

5 m

Explanation:

We assume that the pressure on the surface of water is $1 a t m$ and process is isothermal.
$P_{1} V_{1} = P_{2} V_{2}$
$1 \times V_{1} = P_{2} \frac{2}{5} V$
$P_{2} = \frac{5}{2} = 2.5 a t m$
$Δ P = 1.5 a t m$
$1.5 \times 10^{5} \frac{N}{m^{2}} = ρ g h$
$h = \frac{1.5 \times 10^{5}}{10^{3} \times 10} = 15 m$

PHXI13:KINETIC THEORY

360068 Boyle's law is applicable in

1 Isobaric process

2 Isochoric process

3 Isothermal process

4 Adiabatic process

PHXI13:KINETIC THEORY

360069 Two monoatomic gas $A$ and $B$ occupying the same volume $V$ are at the same temperature $T$ and pressure $P$ . If they are mixed, the resulting mixture has volume $V$ and temperature $T$ . The pressure of the mixture is

1

P

2

\frac{P}{2}

3

4 P

4

2 P

Explanation:

$\frac{P_{1} V_{1}}{V_{1}} = \frac{P_{m i x} V_{m i x}}{T_{m i x}}$
$T_{1} = T_{m i x} = T$ and $V_{1} = V + V, V_{m i x} = V$
$\frac{P (2 V)}{T} = \frac{P_{m i x} V}{T}$
$P_{m i x} = 2 P$

PHXI13:KINETIC THEORY

360070 If $P = 10^{6} K T$ , Then the number of molecules per unit volume of the gas is [The letters have usual meaning.]

1

10^{9}

2

10^{5}

3

10^{6}

4

10^{8}

Explanation:

The ideal gas equation is
$P V = n R T = \frac{N}{N_{A}} R T = N K T$
Where $K$ is the Boltzmann's constant
Given that $P = 10^{6} K T$
$\Rightarrow \frac{N}{V} = 10^{6}$

PHXI13:KINETIC THEORY

360071 Air is pumped into an automobile tube upto a pressure of $200 k P a$ in the morning at temperature $22^{\circ} C$ . When the air temperature rises to $42^{\circ} C$ and the tube expands by $2 %$ . The pressure of the air in the tube at this temperature, will be approximately

1

206 k P a

2

209 k P a

3

200 k P a

4

212 k P a

Explanation:

$\frac{P V}{T} = R$ (constant) $\Rightarrow \frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$
$\Rightarrow \frac{200 \times V}{(273 + 22)} = \frac{P_{2} \times 1.02 V}{(273 + 42)}$
$[V_{2} = V + 0.02 V]$
$\Rightarrow P_{2} = \frac{200 \times 315}{295 \times 1.02} = 209 k P a$

PHXI13:KINETIC THEORY

360072 A bottle is inverted and dipped into a tank. At the bottom of the tank, $3 / 5$ th part of the bottle is filled with water. The depth of the tank is

1

20 m

2

15 m

3

10 m

4

5 m

Explanation:

We assume that the pressure on the surface of water is $1 a t m$ and process is isothermal.
$P_{1} V_{1} = P_{2} V_{2}$
$1 \times V_{1} = P_{2} \frac{2}{5} V$
$P_{2} = \frac{5}{2} = 2.5 a t m$
$Δ P = 1.5 a t m$
$1.5 \times 10^{5} \frac{N}{m^{2}} = ρ g h$
$h = \frac{1.5 \times 10^{5}}{10^{3} \times 10} = 15 m$

PHXI13:KINETIC THEORY

360068 Boyle's law is applicable in

1 Isobaric process

2 Isochoric process

3 Isothermal process

4 Adiabatic process

PHXI13:KINETIC THEORY

360069 Two monoatomic gas $A$ and $B$ occupying the same volume $V$ are at the same temperature $T$ and pressure $P$ . If they are mixed, the resulting mixture has volume $V$ and temperature $T$ . The pressure of the mixture is

1

P

2

\frac{P}{2}

3

4 P

4

2 P

Explanation:

$\frac{P_{1} V_{1}}{V_{1}} = \frac{P_{m i x} V_{m i x}}{T_{m i x}}$
$T_{1} = T_{m i x} = T$ and $V_{1} = V + V, V_{m i x} = V$
$\frac{P (2 V)}{T} = \frac{P_{m i x} V}{T}$
$P_{m i x} = 2 P$

PHXI13:KINETIC THEORY

360070 If $P = 10^{6} K T$ , Then the number of molecules per unit volume of the gas is [The letters have usual meaning.]

1

10^{9}

2

10^{5}

3

10^{6}

4

10^{8}

Explanation:

The ideal gas equation is
$P V = n R T = \frac{N}{N_{A}} R T = N K T$
Where $K$ is the Boltzmann's constant
Given that $P = 10^{6} K T$
$\Rightarrow \frac{N}{V} = 10^{6}$

PHXI13:KINETIC THEORY

360071 Air is pumped into an automobile tube upto a pressure of $200 k P a$ in the morning at temperature $22^{\circ} C$ . When the air temperature rises to $42^{\circ} C$ and the tube expands by $2 %$ . The pressure of the air in the tube at this temperature, will be approximately

1

206 k P a

2

209 k P a

3

200 k P a

4

212 k P a

Explanation:

$\frac{P V}{T} = R$ (constant) $\Rightarrow \frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$
$\Rightarrow \frac{200 \times V}{(273 + 22)} = \frac{P_{2} \times 1.02 V}{(273 + 42)}$
$[V_{2} = V + 0.02 V]$
$\Rightarrow P_{2} = \frac{200 \times 315}{295 \times 1.02} = 209 k P a$

PHXI13:KINETIC THEORY

360072 A bottle is inverted and dipped into a tank. At the bottom of the tank, $3 / 5$ th part of the bottle is filled with water. The depth of the tank is

1

20 m

2

15 m

3

10 m

4

5 m

Explanation:

We assume that the pressure on the surface of water is $1 a t m$ and process is isothermal.
$P_{1} V_{1} = P_{2} V_{2}$
$1 \times V_{1} = P_{2} \frac{2}{5} V$
$P_{2} = \frac{5}{2} = 2.5 a t m$
$Δ P = 1.5 a t m$
$1.5 \times 10^{5} \frac{N}{m^{2}} = ρ g h$
$h = \frac{1.5 \times 10^{5}}{10^{3} \times 10} = 15 m$

PHXI13:KINETIC THEORY

360068 Boyle's law is applicable in

1 Isobaric process

2 Isochoric process

3 Isothermal process

4 Adiabatic process

PHXI13:KINETIC THEORY

360069 Two monoatomic gas $A$ and $B$ occupying the same volume $V$ are at the same temperature $T$ and pressure $P$ . If they are mixed, the resulting mixture has volume $V$ and temperature $T$ . The pressure of the mixture is

1

P

2

\frac{P}{2}

3

4 P

4

2 P

Explanation:

$\frac{P_{1} V_{1}}{V_{1}} = \frac{P_{m i x} V_{m i x}}{T_{m i x}}$
$T_{1} = T_{m i x} = T$ and $V_{1} = V + V, V_{m i x} = V$
$\frac{P (2 V)}{T} = \frac{P_{m i x} V}{T}$
$P_{m i x} = 2 P$

PHXI13:KINETIC THEORY

360070 If $P = 10^{6} K T$ , Then the number of molecules per unit volume of the gas is [The letters have usual meaning.]

1

10^{9}

2

10^{5}

3

10^{6}

4

10^{8}

Explanation:

The ideal gas equation is
$P V = n R T = \frac{N}{N_{A}} R T = N K T$
Where $K$ is the Boltzmann's constant
Given that $P = 10^{6} K T$
$\Rightarrow \frac{N}{V} = 10^{6}$

PHXI13:KINETIC THEORY

360071 Air is pumped into an automobile tube upto a pressure of $200 k P a$ in the morning at temperature $22^{\circ} C$ . When the air temperature rises to $42^{\circ} C$ and the tube expands by $2 %$ . The pressure of the air in the tube at this temperature, will be approximately

1

206 k P a

2

209 k P a

3

200 k P a

4

212 k P a

Explanation:

$\frac{P V}{T} = R$ (constant) $\Rightarrow \frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$
$\Rightarrow \frac{200 \times V}{(273 + 22)} = \frac{P_{2} \times 1.02 V}{(273 + 42)}$
$[V_{2} = V + 0.02 V]$
$\Rightarrow P_{2} = \frac{200 \times 315}{295 \times 1.02} = 209 k P a$

PHXI13:KINETIC THEORY

360072 A bottle is inverted and dipped into a tank. At the bottom of the tank, $3 / 5$ th part of the bottle is filled with water. The depth of the tank is

1

20 m

2

15 m

3

10 m

4

5 m

Explanation:

We assume that the pressure on the surface of water is $1 a t m$ and process is isothermal.
$P_{1} V_{1} = P_{2} V_{2}$
$1 \times V_{1} = P_{2} \frac{2}{5} V$
$P_{2} = \frac{5}{2} = 2.5 a t m$
$Δ P = 1.5 a t m$
$1.5 \times 10^{5} \frac{N}{m^{2}} = ρ g h$
$h = \frac{1.5 \times 10^{5}}{10^{3} \times 10} = 15 m$