QBManager

7 RBTS PAPER

164191 If 2 moles of an ideal monatomic gas at temperature $T_{0}$ is mixed with 4 moles of another ideal monatomic gas at temperature $2 T_{0}$ , then the temperature of the mixture is:

1

\frac{5}{3} T_{0}

2

\frac{3}{2} T_{0}

3

\frac{4}{3} T_{0}

4

\frac{5}{4} T_{0}

Explanation:

$T_{mix} = \frac{μ_{1} T_{1} + μ_{2} T_{2}}{μ_{1} + μ_{2}}$
$= \frac{2 \times T_{0} + 4 \times 2 T_{0}}{2 + 4} = \frac{5}{3} T_{0}$

NCERT-XI-II-246

7 RBTS PAPER

164192 If pressure of a gas contained in a closed vessel is increased by $0.4 %$ when heated by $1^{\circ} C$ , the initial temperature must be :

1

250 K

2

250^{\circ} C

3

2500 K

4

25^{\circ} C

.

Explanation:

$\frac{Δ P}{P_{0}} \times 100 = \frac{Δ T}{T_{0}} \times 100$
$0.4 = \frac{1}{T_{0}} \times 100$
$T_{0} = \frac{1000}{4}$
$T_{0} = 250 K .$

NCERT-XI-II-247

7 RBTS PAPER

164193 A balloon contains $1500 m^{3}$ of helium at $27^{\circ} C$ and 4 atmospheric pressure. The volume of helium at $- 3^{\circ} C$ temperature and 2 atmosphere pressure will be :

1

1500 m^{3}

2

1700 m^{3}

3

1900 m^{3}

4

2700 m^{3}

.

Explanation:

Here, $V_{1} = 1500 m^{3}, T_{1} = 27^{\circ} C \Rightarrow 300 K$
$P_{1} = 4 atm, T_{2} = - 3^{\circ} C \Rightarrow 270 K$
$P_{2} = 2 atm .$

According to ideal gas equation
$\begin{matrix} \frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}} \\ ∴ V_{2} = \frac{P_{1} V_{1}}{T_{1}} \times \frac{T_{2}}{P_{2}} = \frac{4 \times 1500 \times 270}{300 \times 2} = 2700 m^{3} \end{matrix}$

NCERT-XI-II-247

7 RBTS PAPER

164194 An air bubble of volume $1.0 {cm}^{3}$ rises from the bottom of a lake $40 m$ deep at a temperature of $12^{\circ} C$ . To what volume does it grow when it reaches the surface, which is of a temperature of $35^{\circ} C$ (atm. pressure $= 10^{5} N / m)$ :

1

10.6 \times 10^{- 6} m^{3}

2

5.4 \times 10^{- 6} m^{3}

3

2.8 \times 10^{- 6} m^{3}

4

15.6 \times 10^{- 6} m^{3}

.

Explanation:

$\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$
$\frac{(h d g + P_{0})}{(12 + 273)} = \frac{P_{0} \times V_{2}}{(35 + 273)}$
$V_{2} = (h d g + P_{0}) (\frac{308}{285}) \times \frac{1}{P_{0}}$
$= \frac{(40 \times 10^{3} \times 10 + 10^{5})}{10^{5}} \times \frac{308}{285}$
$5 \times \frac{308}{285}$
$= 5.4 {cm}^{3} .$

NCERT-XI-II-258

7 RBTS PAPER

164191 If 2 moles of an ideal monatomic gas at temperature $T_{0}$ is mixed with 4 moles of another ideal monatomic gas at temperature $2 T_{0}$ , then the temperature of the mixture is:

1

\frac{5}{3} T_{0}

2

\frac{3}{2} T_{0}

3

\frac{4}{3} T_{0}

4

\frac{5}{4} T_{0}

Explanation:

$T_{mix} = \frac{μ_{1} T_{1} + μ_{2} T_{2}}{μ_{1} + μ_{2}}$
$= \frac{2 \times T_{0} + 4 \times 2 T_{0}}{2 + 4} = \frac{5}{3} T_{0}$

NCERT-XI-II-246

7 RBTS PAPER

164192 If pressure of a gas contained in a closed vessel is increased by $0.4 %$ when heated by $1^{\circ} C$ , the initial temperature must be :

1

250 K

2

250^{\circ} C

3

2500 K

4

25^{\circ} C

.

Explanation:

$\frac{Δ P}{P_{0}} \times 100 = \frac{Δ T}{T_{0}} \times 100$
$0.4 = \frac{1}{T_{0}} \times 100$
$T_{0} = \frac{1000}{4}$
$T_{0} = 250 K .$

NCERT-XI-II-247

7 RBTS PAPER

164193 A balloon contains $1500 m^{3}$ of helium at $27^{\circ} C$ and 4 atmospheric pressure. The volume of helium at $- 3^{\circ} C$ temperature and 2 atmosphere pressure will be :

1

1500 m^{3}

2

1700 m^{3}

3

1900 m^{3}

4

2700 m^{3}

.

Explanation:

Here, $V_{1} = 1500 m^{3}, T_{1} = 27^{\circ} C \Rightarrow 300 K$
$P_{1} = 4 atm, T_{2} = - 3^{\circ} C \Rightarrow 270 K$
$P_{2} = 2 atm .$

According to ideal gas equation
$\begin{matrix} \frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}} \\ ∴ V_{2} = \frac{P_{1} V_{1}}{T_{1}} \times \frac{T_{2}}{P_{2}} = \frac{4 \times 1500 \times 270}{300 \times 2} = 2700 m^{3} \end{matrix}$

NCERT-XI-II-247

7 RBTS PAPER

164194 An air bubble of volume $1.0 {cm}^{3}$ rises from the bottom of a lake $40 m$ deep at a temperature of $12^{\circ} C$ . To what volume does it grow when it reaches the surface, which is of a temperature of $35^{\circ} C$ (atm. pressure $= 10^{5} N / m)$ :

1

10.6 \times 10^{- 6} m^{3}

2

5.4 \times 10^{- 6} m^{3}

3

2.8 \times 10^{- 6} m^{3}

4

15.6 \times 10^{- 6} m^{3}

.

Explanation:

$\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$
$\frac{(h d g + P_{0})}{(12 + 273)} = \frac{P_{0} \times V_{2}}{(35 + 273)}$
$V_{2} = (h d g + P_{0}) (\frac{308}{285}) \times \frac{1}{P_{0}}$
$= \frac{(40 \times 10^{3} \times 10 + 10^{5})}{10^{5}} \times \frac{308}{285}$
$5 \times \frac{308}{285}$
$= 5.4 {cm}^{3} .$

NCERT-XI-II-258

7 RBTS PAPER

164191 If 2 moles of an ideal monatomic gas at temperature $T_{0}$ is mixed with 4 moles of another ideal monatomic gas at temperature $2 T_{0}$ , then the temperature of the mixture is:

1

\frac{5}{3} T_{0}

2

\frac{3}{2} T_{0}

3

\frac{4}{3} T_{0}

4

\frac{5}{4} T_{0}

Explanation:

$T_{mix} = \frac{μ_{1} T_{1} + μ_{2} T_{2}}{μ_{1} + μ_{2}}$
$= \frac{2 \times T_{0} + 4 \times 2 T_{0}}{2 + 4} = \frac{5}{3} T_{0}$

NCERT-XI-II-246

7 RBTS PAPER

164192 If pressure of a gas contained in a closed vessel is increased by $0.4 %$ when heated by $1^{\circ} C$ , the initial temperature must be :

1

250 K

2

250^{\circ} C

3

2500 K

4

25^{\circ} C

.

Explanation:

$\frac{Δ P}{P_{0}} \times 100 = \frac{Δ T}{T_{0}} \times 100$
$0.4 = \frac{1}{T_{0}} \times 100$
$T_{0} = \frac{1000}{4}$
$T_{0} = 250 K .$

NCERT-XI-II-247

7 RBTS PAPER

164193 A balloon contains $1500 m^{3}$ of helium at $27^{\circ} C$ and 4 atmospheric pressure. The volume of helium at $- 3^{\circ} C$ temperature and 2 atmosphere pressure will be :

1

1500 m^{3}

2

1700 m^{3}

3

1900 m^{3}

4

2700 m^{3}

.

Explanation:

Here, $V_{1} = 1500 m^{3}, T_{1} = 27^{\circ} C \Rightarrow 300 K$
$P_{1} = 4 atm, T_{2} = - 3^{\circ} C \Rightarrow 270 K$
$P_{2} = 2 atm .$

According to ideal gas equation
$\begin{matrix} \frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}} \\ ∴ V_{2} = \frac{P_{1} V_{1}}{T_{1}} \times \frac{T_{2}}{P_{2}} = \frac{4 \times 1500 \times 270}{300 \times 2} = 2700 m^{3} \end{matrix}$

NCERT-XI-II-247

7 RBTS PAPER

164194 An air bubble of volume $1.0 {cm}^{3}$ rises from the bottom of a lake $40 m$ deep at a temperature of $12^{\circ} C$ . To what volume does it grow when it reaches the surface, which is of a temperature of $35^{\circ} C$ (atm. pressure $= 10^{5} N / m)$ :

1

10.6 \times 10^{- 6} m^{3}

2

5.4 \times 10^{- 6} m^{3}

3

2.8 \times 10^{- 6} m^{3}

4

15.6 \times 10^{- 6} m^{3}

.

Explanation:

$\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$
$\frac{(h d g + P_{0})}{(12 + 273)} = \frac{P_{0} \times V_{2}}{(35 + 273)}$
$V_{2} = (h d g + P_{0}) (\frac{308}{285}) \times \frac{1}{P_{0}}$
$= \frac{(40 \times 10^{3} \times 10 + 10^{5})}{10^{5}} \times \frac{308}{285}$
$5 \times \frac{308}{285}$
$= 5.4 {cm}^{3} .$

NCERT-XI-II-258

7 RBTS PAPER

164191 If 2 moles of an ideal monatomic gas at temperature $T_{0}$ is mixed with 4 moles of another ideal monatomic gas at temperature $2 T_{0}$ , then the temperature of the mixture is:

1

\frac{5}{3} T_{0}

2

\frac{3}{2} T_{0}

3

\frac{4}{3} T_{0}

4

\frac{5}{4} T_{0}

Explanation:

$T_{mix} = \frac{μ_{1} T_{1} + μ_{2} T_{2}}{μ_{1} + μ_{2}}$
$= \frac{2 \times T_{0} + 4 \times 2 T_{0}}{2 + 4} = \frac{5}{3} T_{0}$

NCERT-XI-II-246

7 RBTS PAPER

164192 If pressure of a gas contained in a closed vessel is increased by $0.4 %$ when heated by $1^{\circ} C$ , the initial temperature must be :

1

250 K

2

250^{\circ} C

3

2500 K

4

25^{\circ} C

.

Explanation:

$\frac{Δ P}{P_{0}} \times 100 = \frac{Δ T}{T_{0}} \times 100$
$0.4 = \frac{1}{T_{0}} \times 100$
$T_{0} = \frac{1000}{4}$
$T_{0} = 250 K .$

NCERT-XI-II-247

7 RBTS PAPER

164193 A balloon contains $1500 m^{3}$ of helium at $27^{\circ} C$ and 4 atmospheric pressure. The volume of helium at $- 3^{\circ} C$ temperature and 2 atmosphere pressure will be :

1

1500 m^{3}

2

1700 m^{3}

3

1900 m^{3}

4

2700 m^{3}

.

Explanation:

Here, $V_{1} = 1500 m^{3}, T_{1} = 27^{\circ} C \Rightarrow 300 K$
$P_{1} = 4 atm, T_{2} = - 3^{\circ} C \Rightarrow 270 K$
$P_{2} = 2 atm .$

According to ideal gas equation
$\begin{matrix} \frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}} \\ ∴ V_{2} = \frac{P_{1} V_{1}}{T_{1}} \times \frac{T_{2}}{P_{2}} = \frac{4 \times 1500 \times 270}{300 \times 2} = 2700 m^{3} \end{matrix}$

NCERT-XI-II-247

7 RBTS PAPER

164194 An air bubble of volume $1.0 {cm}^{3}$ rises from the bottom of a lake $40 m$ deep at a temperature of $12^{\circ} C$ . To what volume does it grow when it reaches the surface, which is of a temperature of $35^{\circ} C$ (atm. pressure $= 10^{5} N / m)$ :

1

10.6 \times 10^{- 6} m^{3}

2

5.4 \times 10^{- 6} m^{3}

3

2.8 \times 10^{- 6} m^{3}

4

15.6 \times 10^{- 6} m^{3}

.

Explanation:

$\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$
$\frac{(h d g + P_{0})}{(12 + 273)} = \frac{P_{0} \times V_{2}}{(35 + 273)}$
$V_{2} = (h d g + P_{0}) (\frac{308}{285}) \times \frac{1}{P_{0}}$
$= \frac{(40 \times 10^{3} \times 10 + 10^{5})}{10^{5}} \times \frac{308}{285}$
$5 \times \frac{308}{285}$
$= 5.4 {cm}^{3} .$

NCERT-XI-II-258