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PHXI12:THERMODYNAMICS

371525 A container of volume $4 V_{0}$ made of a perfectly non-conducting material is divided into two equal parts by a fixed rigid wall whose lower half is non-conducting and upper half is purely conducting. The right side of the wall is divided into equal parts (initially) by means of a massless non-conducting piston free to move as shown. Section $A$ contains $2 m o l$ of a gas while the section $B$ and $C$ contain $1 m o l$ each of the same gas $(γ = 1.5)$ at pressure $P_{0}$ . The heater in left part is switched on till the final pressure in section $C$ becomes $125 / 27 P_{0}$ . Calculate: Final temperature in part $A$ :
supporting img

1

\frac{P_{0} V_{0}}{R}

2

\frac{12 P_{0} V_{0}}{13 R}

3

\frac{205 P_{0} V_{0}}{27 R}

4

\frac{105 P_{0} V_{0}}{13 R}

Explanation:

The final volume of the compartment $C$ is
$\begin{aligned} P_{0} V_{0}^{γ} = \frac{125 P_{0}}{27} V_{C}^{γ} \Rightarrow V_{C}^{γ} = \frac{27}{125} V_{0}^{γ} \\ V_{C} = {(\frac{27}{125})}^{\frac{1}{γ}} V_{0} = {(\frac{27}{125})}^{\frac{2}{3}} V_{0} \Rightarrow V_{C} = \frac{9 V_{0}}{25} \end{aligned}$
The final volume of compartment $B$ is
$V_{B} = 2 V_{0} - \frac{9 V_{0}}{25} = \frac{41 V_{0}}{25}$
The final temerature of compartment $B$ is
$\begin{aligned} T_{B} = \frac{P_{B} V_{B}}{n R} \\ P_{B} = P_{C} = \frac{125 P_{0}}{27} (Mechanical equilibrium) \\ V_{B} = \frac{41 V_{0}}{25}, n = 1 \\ \Rightarrow T_{B} = \frac{\frac{125 P_{0}}{27} \times \frac{41 V_{0}}{25}}{R} = \frac{205 P_{0} V_{0}}{27 R} \\ T_{A} = T_{B} = \frac{205 P_{0} V_{0}}{27 R} \end{aligned}$
Since the wall between $A$ & $B$ is conducting in nature.

PHXI12:THERMODYNAMICS

371526 $7$ moles of a monoatomic ideal gas is enclosed in an adiabatic, vertical cylinder fitted with a smooth, light adiabatic piston. The piston is connected to a vertical spring of spring constant $420 N / m .$ The area of cross-section of the cylinder is $35 c m^{2} .$ Initially, the spring is at its natural length and the temperature of the gas is $30^{\circ} C .$ The atmospheric pressure is $10^{5} P a .$ The gas is heated slowly for some time by means of an electric heater so as to move the piston up through $20 c m .$ Find the work done by the gas

1

63.2 J

2

78.4 J

3

47.5 J

4

96.2 J

Explanation:

The work done by the gas as the piston moves through $l = 20 c m$ is
$W = \int_{0}^{1} F d x$
Here the net force acting on the piston is,
$F = P_{a} A + k x$
$∴ W = P_{a} A l + \frac{1}{2} k l^{2}$
$= (10^{5}) \times (35 \times 10^{- 4}) \times (20 \times 10^{- 2}) +$
$\frac{1}{2} (420) \times (400 \times 10^{- 4})$
$= 70 + 8.4$
$∴ W = 78.4 J$

PHXI12:THERMODYNAMICS

371527 A diatomic gas $(γ = 1.4)$ does $200 J$ of work when it is expanded isobarically. The heat given to the gas in the process is

1

800 J

2

600 J

3

700 J

4

850 J

Explanation:

Work done by gas $= 200 J$
Gas is expanding isobarically, so pressure is constant.
$P V = n R T$
$∴ P V_{1} = n R T_{1}$ and $P V_{2} = n R T_{2}$
$\Rightarrow W = P (V_{2} - V_{1}) \Rightarrow n R (T_{2} - T_{1}) = n R Δ T$
$∴ n R Δ T = 200 (1)$
Heat given to gas, $Q = n C_{P} Δ T$
Where, $C_{P} = (\frac{f}{2} + 1) R, f =$ degree of freedom
Using, $\begin{aligned} γ & = 1 + \frac{2}{f} or 1.4 = 1 + \frac{2}{f} \\ \Rightarrow f = 5 \\ ∴ & C_{P} = (\frac{5}{2} + 1) R = \frac{7}{2} R \end{aligned}$
As $Δ Q = n C_{P} Δ T$
$∴ Δ Q = n \cdot \frac{7}{2} R Δ T = \frac{7}{2} (200)$
( $∴$ using eqn. (1))
$= 700 J$
So, correct option is (3)

JEE - 2024

PHXI12:THERMODYNAMICS

371528 When $1 g$ of water is frozen at $0^{\circ} C,$ under normal atmospheric pressure, its volume changes from $1 c c$ to $1.083 c c .$ The change in its internal energy is $- 800.022 \times 10^{x} c a l$ . Find the value of $x$ .

1 5

2

- 3

3

- 1

4

- 7

Explanation:

Given that, $= 1 g$
Heat released by $1 g$ of ice on freezing,
$Δ Q = m L = 1 \times 80 = 80 c a l$
As, this heat is given out by water, it is taken negative.
$∴ Δ Q = - 80 c a l$
Now, $Δ V = V_{2} - V$
$= 1.083 - 1.000$
$= 0.083 c c$
$= 0.083 \times 10^{- 6} m^{3}$
The external work done by water,
$Δ W = P (Δ V)$
$= 1.013 \times 10^{5} \times 0.083 \times 10^{- 6}$
$= 0.0084 J$
$= \frac{0.0084}{4.2} c a l$
$= 0.002 c a l$
From first law of thermodynamics,
$Δ U = Δ Q - Δ W$
$= - 80 - 0.002$
$= - 80.002 c a l$
$= - 800.02 \times 10^{- 1} c a l$
$∴ x = - 1$

PHXI12:THERMODYNAMICS

371529 In which process the $P V$ indicator diagram is straight line parallel to volume axis

1 Isobaric

2 Isothermal

3 Adiabatic

4 Irreversible

PHXI12:THERMODYNAMICS

371525 A container of volume $4 V_{0}$ made of a perfectly non-conducting material is divided into two equal parts by a fixed rigid wall whose lower half is non-conducting and upper half is purely conducting. The right side of the wall is divided into equal parts (initially) by means of a massless non-conducting piston free to move as shown. Section $A$ contains $2 m o l$ of a gas while the section $B$ and $C$ contain $1 m o l$ each of the same gas $(γ = 1.5)$ at pressure $P_{0}$ . The heater in left part is switched on till the final pressure in section $C$ becomes $125 / 27 P_{0}$ . Calculate: Final temperature in part $A$ :
supporting img

1

\frac{P_{0} V_{0}}{R}

2

\frac{12 P_{0} V_{0}}{13 R}

3

\frac{205 P_{0} V_{0}}{27 R}

4

\frac{105 P_{0} V_{0}}{13 R}

Explanation:

The final volume of the compartment $C$ is
$\begin{aligned} P_{0} V_{0}^{γ} = \frac{125 P_{0}}{27} V_{C}^{γ} \Rightarrow V_{C}^{γ} = \frac{27}{125} V_{0}^{γ} \\ V_{C} = {(\frac{27}{125})}^{\frac{1}{γ}} V_{0} = {(\frac{27}{125})}^{\frac{2}{3}} V_{0} \Rightarrow V_{C} = \frac{9 V_{0}}{25} \end{aligned}$
The final volume of compartment $B$ is
$V_{B} = 2 V_{0} - \frac{9 V_{0}}{25} = \frac{41 V_{0}}{25}$
The final temerature of compartment $B$ is
$\begin{aligned} T_{B} = \frac{P_{B} V_{B}}{n R} \\ P_{B} = P_{C} = \frac{125 P_{0}}{27} (Mechanical equilibrium) \\ V_{B} = \frac{41 V_{0}}{25}, n = 1 \\ \Rightarrow T_{B} = \frac{\frac{125 P_{0}}{27} \times \frac{41 V_{0}}{25}}{R} = \frac{205 P_{0} V_{0}}{27 R} \\ T_{A} = T_{B} = \frac{205 P_{0} V_{0}}{27 R} \end{aligned}$
Since the wall between $A$ & $B$ is conducting in nature.

PHXI12:THERMODYNAMICS

371526 $7$ moles of a monoatomic ideal gas is enclosed in an adiabatic, vertical cylinder fitted with a smooth, light adiabatic piston. The piston is connected to a vertical spring of spring constant $420 N / m .$ The area of cross-section of the cylinder is $35 c m^{2} .$ Initially, the spring is at its natural length and the temperature of the gas is $30^{\circ} C .$ The atmospheric pressure is $10^{5} P a .$ The gas is heated slowly for some time by means of an electric heater so as to move the piston up through $20 c m .$ Find the work done by the gas

1

63.2 J

2

78.4 J

3

47.5 J

4

96.2 J

Explanation:

The work done by the gas as the piston moves through $l = 20 c m$ is
$W = \int_{0}^{1} F d x$
Here the net force acting on the piston is,
$F = P_{a} A + k x$
$∴ W = P_{a} A l + \frac{1}{2} k l^{2}$
$= (10^{5}) \times (35 \times 10^{- 4}) \times (20 \times 10^{- 2}) +$
$\frac{1}{2} (420) \times (400 \times 10^{- 4})$
$= 70 + 8.4$
$∴ W = 78.4 J$

PHXI12:THERMODYNAMICS

371527 A diatomic gas $(γ = 1.4)$ does $200 J$ of work when it is expanded isobarically. The heat given to the gas in the process is

1

800 J

2

600 J

3

700 J

4

850 J

Explanation:

Work done by gas $= 200 J$
Gas is expanding isobarically, so pressure is constant.
$P V = n R T$
$∴ P V_{1} = n R T_{1}$ and $P V_{2} = n R T_{2}$
$\Rightarrow W = P (V_{2} - V_{1}) \Rightarrow n R (T_{2} - T_{1}) = n R Δ T$
$∴ n R Δ T = 200 (1)$
Heat given to gas, $Q = n C_{P} Δ T$
Where, $C_{P} = (\frac{f}{2} + 1) R, f =$ degree of freedom
Using, $\begin{aligned} γ & = 1 + \frac{2}{f} or 1.4 = 1 + \frac{2}{f} \\ \Rightarrow f = 5 \\ ∴ & C_{P} = (\frac{5}{2} + 1) R = \frac{7}{2} R \end{aligned}$
As $Δ Q = n C_{P} Δ T$
$∴ Δ Q = n \cdot \frac{7}{2} R Δ T = \frac{7}{2} (200)$
( $∴$ using eqn. (1))
$= 700 J$
So, correct option is (3)

JEE - 2024

PHXI12:THERMODYNAMICS

371528 When $1 g$ of water is frozen at $0^{\circ} C,$ under normal atmospheric pressure, its volume changes from $1 c c$ to $1.083 c c .$ The change in its internal energy is $- 800.022 \times 10^{x} c a l$ . Find the value of $x$ .

1 5

2

- 3

3

- 1

4

- 7

Explanation:

Given that, $= 1 g$
Heat released by $1 g$ of ice on freezing,
$Δ Q = m L = 1 \times 80 = 80 c a l$
As, this heat is given out by water, it is taken negative.
$∴ Δ Q = - 80 c a l$
Now, $Δ V = V_{2} - V$
$= 1.083 - 1.000$
$= 0.083 c c$
$= 0.083 \times 10^{- 6} m^{3}$
The external work done by water,
$Δ W = P (Δ V)$
$= 1.013 \times 10^{5} \times 0.083 \times 10^{- 6}$
$= 0.0084 J$
$= \frac{0.0084}{4.2} c a l$
$= 0.002 c a l$
From first law of thermodynamics,
$Δ U = Δ Q - Δ W$
$= - 80 - 0.002$
$= - 80.002 c a l$
$= - 800.02 \times 10^{- 1} c a l$
$∴ x = - 1$

PHXI12:THERMODYNAMICS

371529 In which process the $P V$ indicator diagram is straight line parallel to volume axis

1 Isobaric

2 Isothermal

3 Adiabatic

4 Irreversible

PHXI12:THERMODYNAMICS

371525 A container of volume $4 V_{0}$ made of a perfectly non-conducting material is divided into two equal parts by a fixed rigid wall whose lower half is non-conducting and upper half is purely conducting. The right side of the wall is divided into equal parts (initially) by means of a massless non-conducting piston free to move as shown. Section $A$ contains $2 m o l$ of a gas while the section $B$ and $C$ contain $1 m o l$ each of the same gas $(γ = 1.5)$ at pressure $P_{0}$ . The heater in left part is switched on till the final pressure in section $C$ becomes $125 / 27 P_{0}$ . Calculate: Final temperature in part $A$ :
supporting img

1

\frac{P_{0} V_{0}}{R}

2

\frac{12 P_{0} V_{0}}{13 R}

3

\frac{205 P_{0} V_{0}}{27 R}

4

\frac{105 P_{0} V_{0}}{13 R}

Explanation:

The final volume of the compartment $C$ is
$\begin{aligned} P_{0} V_{0}^{γ} = \frac{125 P_{0}}{27} V_{C}^{γ} \Rightarrow V_{C}^{γ} = \frac{27}{125} V_{0}^{γ} \\ V_{C} = {(\frac{27}{125})}^{\frac{1}{γ}} V_{0} = {(\frac{27}{125})}^{\frac{2}{3}} V_{0} \Rightarrow V_{C} = \frac{9 V_{0}}{25} \end{aligned}$
The final volume of compartment $B$ is
$V_{B} = 2 V_{0} - \frac{9 V_{0}}{25} = \frac{41 V_{0}}{25}$
The final temerature of compartment $B$ is
$\begin{aligned} T_{B} = \frac{P_{B} V_{B}}{n R} \\ P_{B} = P_{C} = \frac{125 P_{0}}{27} (Mechanical equilibrium) \\ V_{B} = \frac{41 V_{0}}{25}, n = 1 \\ \Rightarrow T_{B} = \frac{\frac{125 P_{0}}{27} \times \frac{41 V_{0}}{25}}{R} = \frac{205 P_{0} V_{0}}{27 R} \\ T_{A} = T_{B} = \frac{205 P_{0} V_{0}}{27 R} \end{aligned}$
Since the wall between $A$ & $B$ is conducting in nature.

PHXI12:THERMODYNAMICS

371526 $7$ moles of a monoatomic ideal gas is enclosed in an adiabatic, vertical cylinder fitted with a smooth, light adiabatic piston. The piston is connected to a vertical spring of spring constant $420 N / m .$ The area of cross-section of the cylinder is $35 c m^{2} .$ Initially, the spring is at its natural length and the temperature of the gas is $30^{\circ} C .$ The atmospheric pressure is $10^{5} P a .$ The gas is heated slowly for some time by means of an electric heater so as to move the piston up through $20 c m .$ Find the work done by the gas

1

63.2 J

2

78.4 J

3

47.5 J

4

96.2 J

Explanation:

The work done by the gas as the piston moves through $l = 20 c m$ is
$W = \int_{0}^{1} F d x$
Here the net force acting on the piston is,
$F = P_{a} A + k x$
$∴ W = P_{a} A l + \frac{1}{2} k l^{2}$
$= (10^{5}) \times (35 \times 10^{- 4}) \times (20 \times 10^{- 2}) +$
$\frac{1}{2} (420) \times (400 \times 10^{- 4})$
$= 70 + 8.4$
$∴ W = 78.4 J$

PHXI12:THERMODYNAMICS

371527 A diatomic gas $(γ = 1.4)$ does $200 J$ of work when it is expanded isobarically. The heat given to the gas in the process is

1

800 J

2

600 J

3

700 J

4

850 J

Explanation:

Work done by gas $= 200 J$
Gas is expanding isobarically, so pressure is constant.
$P V = n R T$
$∴ P V_{1} = n R T_{1}$ and $P V_{2} = n R T_{2}$
$\Rightarrow W = P (V_{2} - V_{1}) \Rightarrow n R (T_{2} - T_{1}) = n R Δ T$
$∴ n R Δ T = 200 (1)$
Heat given to gas, $Q = n C_{P} Δ T$
Where, $C_{P} = (\frac{f}{2} + 1) R, f =$ degree of freedom
Using, $\begin{aligned} γ & = 1 + \frac{2}{f} or 1.4 = 1 + \frac{2}{f} \\ \Rightarrow f = 5 \\ ∴ & C_{P} = (\frac{5}{2} + 1) R = \frac{7}{2} R \end{aligned}$
As $Δ Q = n C_{P} Δ T$
$∴ Δ Q = n \cdot \frac{7}{2} R Δ T = \frac{7}{2} (200)$
( $∴$ using eqn. (1))
$= 700 J$
So, correct option is (3)

JEE - 2024

PHXI12:THERMODYNAMICS

371528 When $1 g$ of water is frozen at $0^{\circ} C,$ under normal atmospheric pressure, its volume changes from $1 c c$ to $1.083 c c .$ The change in its internal energy is $- 800.022 \times 10^{x} c a l$ . Find the value of $x$ .

1 5

2

- 3

3

- 1

4

- 7

Explanation:

Given that, $= 1 g$
Heat released by $1 g$ of ice on freezing,
$Δ Q = m L = 1 \times 80 = 80 c a l$
As, this heat is given out by water, it is taken negative.
$∴ Δ Q = - 80 c a l$
Now, $Δ V = V_{2} - V$
$= 1.083 - 1.000$
$= 0.083 c c$
$= 0.083 \times 10^{- 6} m^{3}$
The external work done by water,
$Δ W = P (Δ V)$
$= 1.013 \times 10^{5} \times 0.083 \times 10^{- 6}$
$= 0.0084 J$
$= \frac{0.0084}{4.2} c a l$
$= 0.002 c a l$
From first law of thermodynamics,
$Δ U = Δ Q - Δ W$
$= - 80 - 0.002$
$= - 80.002 c a l$
$= - 800.02 \times 10^{- 1} c a l$
$∴ x = - 1$

PHXI12:THERMODYNAMICS

371529 In which process the $P V$ indicator diagram is straight line parallel to volume axis

1 Isobaric

2 Isothermal

3 Adiabatic

4 Irreversible

PHXI12:THERMODYNAMICS

371525 A container of volume $4 V_{0}$ made of a perfectly non-conducting material is divided into two equal parts by a fixed rigid wall whose lower half is non-conducting and upper half is purely conducting. The right side of the wall is divided into equal parts (initially) by means of a massless non-conducting piston free to move as shown. Section $A$ contains $2 m o l$ of a gas while the section $B$ and $C$ contain $1 m o l$ each of the same gas $(γ = 1.5)$ at pressure $P_{0}$ . The heater in left part is switched on till the final pressure in section $C$ becomes $125 / 27 P_{0}$ . Calculate: Final temperature in part $A$ :
supporting img

1

\frac{P_{0} V_{0}}{R}

2

\frac{12 P_{0} V_{0}}{13 R}

3

\frac{205 P_{0} V_{0}}{27 R}

4

\frac{105 P_{0} V_{0}}{13 R}

Explanation:

The final volume of the compartment $C$ is
$\begin{aligned} P_{0} V_{0}^{γ} = \frac{125 P_{0}}{27} V_{C}^{γ} \Rightarrow V_{C}^{γ} = \frac{27}{125} V_{0}^{γ} \\ V_{C} = {(\frac{27}{125})}^{\frac{1}{γ}} V_{0} = {(\frac{27}{125})}^{\frac{2}{3}} V_{0} \Rightarrow V_{C} = \frac{9 V_{0}}{25} \end{aligned}$
The final volume of compartment $B$ is
$V_{B} = 2 V_{0} - \frac{9 V_{0}}{25} = \frac{41 V_{0}}{25}$
The final temerature of compartment $B$ is
$\begin{aligned} T_{B} = \frac{P_{B} V_{B}}{n R} \\ P_{B} = P_{C} = \frac{125 P_{0}}{27} (Mechanical equilibrium) \\ V_{B} = \frac{41 V_{0}}{25}, n = 1 \\ \Rightarrow T_{B} = \frac{\frac{125 P_{0}}{27} \times \frac{41 V_{0}}{25}}{R} = \frac{205 P_{0} V_{0}}{27 R} \\ T_{A} = T_{B} = \frac{205 P_{0} V_{0}}{27 R} \end{aligned}$
Since the wall between $A$ & $B$ is conducting in nature.

PHXI12:THERMODYNAMICS

371526 $7$ moles of a monoatomic ideal gas is enclosed in an adiabatic, vertical cylinder fitted with a smooth, light adiabatic piston. The piston is connected to a vertical spring of spring constant $420 N / m .$ The area of cross-section of the cylinder is $35 c m^{2} .$ Initially, the spring is at its natural length and the temperature of the gas is $30^{\circ} C .$ The atmospheric pressure is $10^{5} P a .$ The gas is heated slowly for some time by means of an electric heater so as to move the piston up through $20 c m .$ Find the work done by the gas

1

63.2 J

2

78.4 J

3

47.5 J

4

96.2 J

Explanation:

The work done by the gas as the piston moves through $l = 20 c m$ is
$W = \int_{0}^{1} F d x$
Here the net force acting on the piston is,
$F = P_{a} A + k x$
$∴ W = P_{a} A l + \frac{1}{2} k l^{2}$
$= (10^{5}) \times (35 \times 10^{- 4}) \times (20 \times 10^{- 2}) +$
$\frac{1}{2} (420) \times (400 \times 10^{- 4})$
$= 70 + 8.4$
$∴ W = 78.4 J$

PHXI12:THERMODYNAMICS

371527 A diatomic gas $(γ = 1.4)$ does $200 J$ of work when it is expanded isobarically. The heat given to the gas in the process is

1

800 J

2

600 J

3

700 J

4

850 J

Explanation:

Work done by gas $= 200 J$
Gas is expanding isobarically, so pressure is constant.
$P V = n R T$
$∴ P V_{1} = n R T_{1}$ and $P V_{2} = n R T_{2}$
$\Rightarrow W = P (V_{2} - V_{1}) \Rightarrow n R (T_{2} - T_{1}) = n R Δ T$
$∴ n R Δ T = 200 (1)$
Heat given to gas, $Q = n C_{P} Δ T$
Where, $C_{P} = (\frac{f}{2} + 1) R, f =$ degree of freedom
Using, $\begin{aligned} γ & = 1 + \frac{2}{f} or 1.4 = 1 + \frac{2}{f} \\ \Rightarrow f = 5 \\ ∴ & C_{P} = (\frac{5}{2} + 1) R = \frac{7}{2} R \end{aligned}$
As $Δ Q = n C_{P} Δ T$
$∴ Δ Q = n \cdot \frac{7}{2} R Δ T = \frac{7}{2} (200)$
( $∴$ using eqn. (1))
$= 700 J$
So, correct option is (3)

JEE - 2024

PHXI12:THERMODYNAMICS

371528 When $1 g$ of water is frozen at $0^{\circ} C,$ under normal atmospheric pressure, its volume changes from $1 c c$ to $1.083 c c .$ The change in its internal energy is $- 800.022 \times 10^{x} c a l$ . Find the value of $x$ .

1 5

2

- 3

3

- 1

4

- 7

Explanation:

Given that, $= 1 g$
Heat released by $1 g$ of ice on freezing,
$Δ Q = m L = 1 \times 80 = 80 c a l$
As, this heat is given out by water, it is taken negative.
$∴ Δ Q = - 80 c a l$
Now, $Δ V = V_{2} - V$
$= 1.083 - 1.000$
$= 0.083 c c$
$= 0.083 \times 10^{- 6} m^{3}$
The external work done by water,
$Δ W = P (Δ V)$
$= 1.013 \times 10^{5} \times 0.083 \times 10^{- 6}$
$= 0.0084 J$
$= \frac{0.0084}{4.2} c a l$
$= 0.002 c a l$
From first law of thermodynamics,
$Δ U = Δ Q - Δ W$
$= - 80 - 0.002$
$= - 80.002 c a l$
$= - 800.02 \times 10^{- 1} c a l$
$∴ x = - 1$

PHXI12:THERMODYNAMICS

371529 In which process the $P V$ indicator diagram is straight line parallel to volume axis

1 Isobaric

2 Isothermal

3 Adiabatic

4 Irreversible

PHXI12:THERMODYNAMICS

371525 A container of volume $4 V_{0}$ made of a perfectly non-conducting material is divided into two equal parts by a fixed rigid wall whose lower half is non-conducting and upper half is purely conducting. The right side of the wall is divided into equal parts (initially) by means of a massless non-conducting piston free to move as shown. Section $A$ contains $2 m o l$ of a gas while the section $B$ and $C$ contain $1 m o l$ each of the same gas $(γ = 1.5)$ at pressure $P_{0}$ . The heater in left part is switched on till the final pressure in section $C$ becomes $125 / 27 P_{0}$ . Calculate: Final temperature in part $A$ :
supporting img

1

\frac{P_{0} V_{0}}{R}

2

\frac{12 P_{0} V_{0}}{13 R}

3

\frac{205 P_{0} V_{0}}{27 R}

4

\frac{105 P_{0} V_{0}}{13 R}

Explanation:

The final volume of the compartment $C$ is
$\begin{aligned} P_{0} V_{0}^{γ} = \frac{125 P_{0}}{27} V_{C}^{γ} \Rightarrow V_{C}^{γ} = \frac{27}{125} V_{0}^{γ} \\ V_{C} = {(\frac{27}{125})}^{\frac{1}{γ}} V_{0} = {(\frac{27}{125})}^{\frac{2}{3}} V_{0} \Rightarrow V_{C} = \frac{9 V_{0}}{25} \end{aligned}$
The final volume of compartment $B$ is
$V_{B} = 2 V_{0} - \frac{9 V_{0}}{25} = \frac{41 V_{0}}{25}$
The final temerature of compartment $B$ is
$\begin{aligned} T_{B} = \frac{P_{B} V_{B}}{n R} \\ P_{B} = P_{C} = \frac{125 P_{0}}{27} (Mechanical equilibrium) \\ V_{B} = \frac{41 V_{0}}{25}, n = 1 \\ \Rightarrow T_{B} = \frac{\frac{125 P_{0}}{27} \times \frac{41 V_{0}}{25}}{R} = \frac{205 P_{0} V_{0}}{27 R} \\ T_{A} = T_{B} = \frac{205 P_{0} V_{0}}{27 R} \end{aligned}$
Since the wall between $A$ & $B$ is conducting in nature.

PHXI12:THERMODYNAMICS

371526 $7$ moles of a monoatomic ideal gas is enclosed in an adiabatic, vertical cylinder fitted with a smooth, light adiabatic piston. The piston is connected to a vertical spring of spring constant $420 N / m .$ The area of cross-section of the cylinder is $35 c m^{2} .$ Initially, the spring is at its natural length and the temperature of the gas is $30^{\circ} C .$ The atmospheric pressure is $10^{5} P a .$ The gas is heated slowly for some time by means of an electric heater so as to move the piston up through $20 c m .$ Find the work done by the gas

1

63.2 J

2

78.4 J

3

47.5 J

4

96.2 J

Explanation:

The work done by the gas as the piston moves through $l = 20 c m$ is
$W = \int_{0}^{1} F d x$
Here the net force acting on the piston is,
$F = P_{a} A + k x$
$∴ W = P_{a} A l + \frac{1}{2} k l^{2}$
$= (10^{5}) \times (35 \times 10^{- 4}) \times (20 \times 10^{- 2}) +$
$\frac{1}{2} (420) \times (400 \times 10^{- 4})$
$= 70 + 8.4$
$∴ W = 78.4 J$

PHXI12:THERMODYNAMICS

371527 A diatomic gas $(γ = 1.4)$ does $200 J$ of work when it is expanded isobarically. The heat given to the gas in the process is

1

800 J

2

600 J

3

700 J

4

850 J

Explanation:

Work done by gas $= 200 J$
Gas is expanding isobarically, so pressure is constant.
$P V = n R T$
$∴ P V_{1} = n R T_{1}$ and $P V_{2} = n R T_{2}$
$\Rightarrow W = P (V_{2} - V_{1}) \Rightarrow n R (T_{2} - T_{1}) = n R Δ T$
$∴ n R Δ T = 200 (1)$
Heat given to gas, $Q = n C_{P} Δ T$
Where, $C_{P} = (\frac{f}{2} + 1) R, f =$ degree of freedom
Using, $\begin{aligned} γ & = 1 + \frac{2}{f} or 1.4 = 1 + \frac{2}{f} \\ \Rightarrow f = 5 \\ ∴ & C_{P} = (\frac{5}{2} + 1) R = \frac{7}{2} R \end{aligned}$
As $Δ Q = n C_{P} Δ T$
$∴ Δ Q = n \cdot \frac{7}{2} R Δ T = \frac{7}{2} (200)$
( $∴$ using eqn. (1))
$= 700 J$
So, correct option is (3)

JEE - 2024

PHXI12:THERMODYNAMICS

371528 When $1 g$ of water is frozen at $0^{\circ} C,$ under normal atmospheric pressure, its volume changes from $1 c c$ to $1.083 c c .$ The change in its internal energy is $- 800.022 \times 10^{x} c a l$ . Find the value of $x$ .

1 5

2

- 3

3

- 1

4

- 7

Explanation:

Given that, $= 1 g$
Heat released by $1 g$ of ice on freezing,
$Δ Q = m L = 1 \times 80 = 80 c a l$
As, this heat is given out by water, it is taken negative.
$∴ Δ Q = - 80 c a l$
Now, $Δ V = V_{2} - V$
$= 1.083 - 1.000$
$= 0.083 c c$
$= 0.083 \times 10^{- 6} m^{3}$
The external work done by water,
$Δ W = P (Δ V)$
$= 1.013 \times 10^{5} \times 0.083 \times 10^{- 6}$
$= 0.0084 J$
$= \frac{0.0084}{4.2} c a l$
$= 0.002 c a l$
From first law of thermodynamics,
$Δ U = Δ Q - Δ W$
$= - 80 - 0.002$
$= - 80.002 c a l$
$= - 800.02 \times 10^{- 1} c a l$
$∴ x = - 1$

PHXI12:THERMODYNAMICS

371529 In which process the $P V$ indicator diagram is straight line parallel to volume axis

1 Isobaric

2 Isothermal

3 Adiabatic

4 Irreversible