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PHXI12:THERMODYNAMICS

371258 A gas can be taken from $A$ to $B$ via two different processes $A C B$ and $A D B$ . When path $A C B$ is used $60 J$ of heat flows into the system and $30 J$ of work is done by the system. If path $A D B$ is used work done by the system is $10 J$ . The heat flow into the system in path $A D B$ is
supporting img

1

100 J

2

80 J

3

40 J

4

20 J

Explanation:

$Δ Q = Δ U + W$
$Δ U = Δ Q - W$
$(Δ U)_{A C B} = (Δ U)_{A D B}$
$\Rightarrow 60 - 30 = Δ Q - 10$
$\Rightarrow Δ Q = 40 J$

JEE - 2019

PHXI12:THERMODYNAMICS

371259 A cyclic Process of an ideal gas is shown in the figure. Let $W_{a \to b}, W_{b \to c}, W_{c \to d}$ and $W_{d \to a}$
represents the work done by the system during the processes $a \to b, b \to c, c \to d$ and $d \to a$ respectively. Consider the following relations
supporting img

I. $W_{a \to b} + W_{b \to c} + W_{c \to d} + W_{d \to a} > 0$
II. $W_{a \to b} + W_{b \to c} + W_{c \to a} + W_{d \to a} < 0$
III. $W_{a \to b} + W_{c \to d} > 0$
IV. $W_{b \to c} + W_{d \to a} = 0$
Which of the above relations is/are true?

1 I and III

2 III and IV

3 I, III and IV

4 II, III and IV

Explanation:

In Carnot cycle, in order to obtain continuous supply of work, net work done
As, $W_{a \to b}$ is isothermal
$W_{b \to c}$ is adiabatic
$W_{c \to d}$ is isothermal (negative)
$W_{d \to a}$ is adiabatic (negative)
Net work done during the complete cycle,
$W = W_{a \to b} + W_{b \to c} + (- W_{c \to d}) + (- W_{d \to a})$
$W = W_{a \to b} - W_{c \to d}$
$[as W_{b \to c} = W_{d \to a}]$
i.e. $W_{a \to b} + W_{c \to d} > 0$
$W_{a \to b} + W_{b \to c} + W_{c \to d} + W_{d \to a} > 0$
$W_{b \to c} = W_{d \to a}$
$∴ W_{b \to c} + W_{d \to a} = 0$

PHXI12:THERMODYNAMICS

371260 A thermodynamic system is taken through the cycle $P Q R S P$ process. The net work done by the system is
supporting img

1

20 J

2

- 40 J

3

400 J

4

- 374 J

Explanation:

Work done $=$ area of the cycle.
$W = - 200 c c \times 200 K P a ((-)$ sign indicates the path is in anticlockwise direction). $= - 200 \times 10^{- 6} \times 200 \times 10^{3} = - 40 J$

PHXI12:THERMODYNAMICS

371261 A thermodynamic system is taken through the cycle $A B C D$ as shown in the figure. Heat rejected by the gas during the cycle is
supporting img

1

P V

2

2 P V

3

4 P V

4

1 / 2 P V

Explanation:

From the first law of thermodynamics $Q = Δ U + Δ W$
For cyclic process $Δ U = 0$
Hence $Q = W$
Work done $W =$ area under $P - V$ diagram
$\Rightarrow Q = - 2 P V .$

PHXI12:THERMODYNAMICS

371262 In the following indicator diagram, the net amount of work done will be :
supporting img

1 positive

2 Negative

3 zero

4 Infinity

PHXI12:THERMODYNAMICS

371258 A gas can be taken from $A$ to $B$ via two different processes $A C B$ and $A D B$ . When path $A C B$ is used $60 J$ of heat flows into the system and $30 J$ of work is done by the system. If path $A D B$ is used work done by the system is $10 J$ . The heat flow into the system in path $A D B$ is
supporting img

1

100 J

2

80 J

3

40 J

4

20 J

Explanation:

$Δ Q = Δ U + W$
$Δ U = Δ Q - W$
$(Δ U)_{A C B} = (Δ U)_{A D B}$
$\Rightarrow 60 - 30 = Δ Q - 10$
$\Rightarrow Δ Q = 40 J$

JEE - 2019

PHXI12:THERMODYNAMICS

371259 A cyclic Process of an ideal gas is shown in the figure. Let $W_{a \to b}, W_{b \to c}, W_{c \to d}$ and $W_{d \to a}$
represents the work done by the system during the processes $a \to b, b \to c, c \to d$ and $d \to a$ respectively. Consider the following relations
supporting img

I. $W_{a \to b} + W_{b \to c} + W_{c \to d} + W_{d \to a} > 0$
II. $W_{a \to b} + W_{b \to c} + W_{c \to a} + W_{d \to a} < 0$
III. $W_{a \to b} + W_{c \to d} > 0$
IV. $W_{b \to c} + W_{d \to a} = 0$
Which of the above relations is/are true?

1 I and III

2 III and IV

3 I, III and IV

4 II, III and IV

Explanation:

In Carnot cycle, in order to obtain continuous supply of work, net work done
As, $W_{a \to b}$ is isothermal
$W_{b \to c}$ is adiabatic
$W_{c \to d}$ is isothermal (negative)
$W_{d \to a}$ is adiabatic (negative)
Net work done during the complete cycle,
$W = W_{a \to b} + W_{b \to c} + (- W_{c \to d}) + (- W_{d \to a})$
$W = W_{a \to b} - W_{c \to d}$
$[as W_{b \to c} = W_{d \to a}]$
i.e. $W_{a \to b} + W_{c \to d} > 0$
$W_{a \to b} + W_{b \to c} + W_{c \to d} + W_{d \to a} > 0$
$W_{b \to c} = W_{d \to a}$
$∴ W_{b \to c} + W_{d \to a} = 0$

PHXI12:THERMODYNAMICS

371260 A thermodynamic system is taken through the cycle $P Q R S P$ process. The net work done by the system is
supporting img

1

20 J

2

- 40 J

3

400 J

4

- 374 J

Explanation:

Work done $=$ area of the cycle.
$W = - 200 c c \times 200 K P a ((-)$ sign indicates the path is in anticlockwise direction). $= - 200 \times 10^{- 6} \times 200 \times 10^{3} = - 40 J$

PHXI12:THERMODYNAMICS

371261 A thermodynamic system is taken through the cycle $A B C D$ as shown in the figure. Heat rejected by the gas during the cycle is
supporting img

1

P V

2

2 P V

3

4 P V

4

1 / 2 P V

Explanation:

From the first law of thermodynamics $Q = Δ U + Δ W$
For cyclic process $Δ U = 0$
Hence $Q = W$
Work done $W =$ area under $P - V$ diagram
$\Rightarrow Q = - 2 P V .$

PHXI12:THERMODYNAMICS

371262 In the following indicator diagram, the net amount of work done will be :
supporting img

1 positive

2 Negative

3 zero

4 Infinity

PHXI12:THERMODYNAMICS

371258 A gas can be taken from $A$ to $B$ via two different processes $A C B$ and $A D B$ . When path $A C B$ is used $60 J$ of heat flows into the system and $30 J$ of work is done by the system. If path $A D B$ is used work done by the system is $10 J$ . The heat flow into the system in path $A D B$ is
supporting img

1

100 J

2

80 J

3

40 J

4

20 J

Explanation:

$Δ Q = Δ U + W$
$Δ U = Δ Q - W$
$(Δ U)_{A C B} = (Δ U)_{A D B}$
$\Rightarrow 60 - 30 = Δ Q - 10$
$\Rightarrow Δ Q = 40 J$

JEE - 2019

PHXI12:THERMODYNAMICS

371259 A cyclic Process of an ideal gas is shown in the figure. Let $W_{a \to b}, W_{b \to c}, W_{c \to d}$ and $W_{d \to a}$
represents the work done by the system during the processes $a \to b, b \to c, c \to d$ and $d \to a$ respectively. Consider the following relations
supporting img

I. $W_{a \to b} + W_{b \to c} + W_{c \to d} + W_{d \to a} > 0$
II. $W_{a \to b} + W_{b \to c} + W_{c \to a} + W_{d \to a} < 0$
III. $W_{a \to b} + W_{c \to d} > 0$
IV. $W_{b \to c} + W_{d \to a} = 0$
Which of the above relations is/are true?

1 I and III

2 III and IV

3 I, III and IV

4 II, III and IV

Explanation:

In Carnot cycle, in order to obtain continuous supply of work, net work done
As, $W_{a \to b}$ is isothermal
$W_{b \to c}$ is adiabatic
$W_{c \to d}$ is isothermal (negative)
$W_{d \to a}$ is adiabatic (negative)
Net work done during the complete cycle,
$W = W_{a \to b} + W_{b \to c} + (- W_{c \to d}) + (- W_{d \to a})$
$W = W_{a \to b} - W_{c \to d}$
$[as W_{b \to c} = W_{d \to a}]$
i.e. $W_{a \to b} + W_{c \to d} > 0$
$W_{a \to b} + W_{b \to c} + W_{c \to d} + W_{d \to a} > 0$
$W_{b \to c} = W_{d \to a}$
$∴ W_{b \to c} + W_{d \to a} = 0$

PHXI12:THERMODYNAMICS

371260 A thermodynamic system is taken through the cycle $P Q R S P$ process. The net work done by the system is
supporting img

1

20 J

2

- 40 J

3

400 J

4

- 374 J

Explanation:

Work done $=$ area of the cycle.
$W = - 200 c c \times 200 K P a ((-)$ sign indicates the path is in anticlockwise direction). $= - 200 \times 10^{- 6} \times 200 \times 10^{3} = - 40 J$

PHXI12:THERMODYNAMICS

371261 A thermodynamic system is taken through the cycle $A B C D$ as shown in the figure. Heat rejected by the gas during the cycle is
supporting img

1

P V

2

2 P V

3

4 P V

4

1 / 2 P V

Explanation:

From the first law of thermodynamics $Q = Δ U + Δ W$
For cyclic process $Δ U = 0$
Hence $Q = W$
Work done $W =$ area under $P - V$ diagram
$\Rightarrow Q = - 2 P V .$

PHXI12:THERMODYNAMICS

371262 In the following indicator diagram, the net amount of work done will be :
supporting img

1 positive

2 Negative

3 zero

4 Infinity

PHXI12:THERMODYNAMICS

371258 A gas can be taken from $A$ to $B$ via two different processes $A C B$ and $A D B$ . When path $A C B$ is used $60 J$ of heat flows into the system and $30 J$ of work is done by the system. If path $A D B$ is used work done by the system is $10 J$ . The heat flow into the system in path $A D B$ is
supporting img

1

100 J

2

80 J

3

40 J

4

20 J

Explanation:

$Δ Q = Δ U + W$
$Δ U = Δ Q - W$
$(Δ U)_{A C B} = (Δ U)_{A D B}$
$\Rightarrow 60 - 30 = Δ Q - 10$
$\Rightarrow Δ Q = 40 J$

JEE - 2019

PHXI12:THERMODYNAMICS

371259 A cyclic Process of an ideal gas is shown in the figure. Let $W_{a \to b}, W_{b \to c}, W_{c \to d}$ and $W_{d \to a}$
represents the work done by the system during the processes $a \to b, b \to c, c \to d$ and $d \to a$ respectively. Consider the following relations
supporting img

I. $W_{a \to b} + W_{b \to c} + W_{c \to d} + W_{d \to a} > 0$
II. $W_{a \to b} + W_{b \to c} + W_{c \to a} + W_{d \to a} < 0$
III. $W_{a \to b} + W_{c \to d} > 0$
IV. $W_{b \to c} + W_{d \to a} = 0$
Which of the above relations is/are true?

1 I and III

2 III and IV

3 I, III and IV

4 II, III and IV

Explanation:

In Carnot cycle, in order to obtain continuous supply of work, net work done
As, $W_{a \to b}$ is isothermal
$W_{b \to c}$ is adiabatic
$W_{c \to d}$ is isothermal (negative)
$W_{d \to a}$ is adiabatic (negative)
Net work done during the complete cycle,
$W = W_{a \to b} + W_{b \to c} + (- W_{c \to d}) + (- W_{d \to a})$
$W = W_{a \to b} - W_{c \to d}$
$[as W_{b \to c} = W_{d \to a}]$
i.e. $W_{a \to b} + W_{c \to d} > 0$
$W_{a \to b} + W_{b \to c} + W_{c \to d} + W_{d \to a} > 0$
$W_{b \to c} = W_{d \to a}$
$∴ W_{b \to c} + W_{d \to a} = 0$

PHXI12:THERMODYNAMICS

371260 A thermodynamic system is taken through the cycle $P Q R S P$ process. The net work done by the system is
supporting img

1

20 J

2

- 40 J

3

400 J

4

- 374 J

Explanation:

Work done $=$ area of the cycle.
$W = - 200 c c \times 200 K P a ((-)$ sign indicates the path is in anticlockwise direction). $= - 200 \times 10^{- 6} \times 200 \times 10^{3} = - 40 J$

PHXI12:THERMODYNAMICS

371261 A thermodynamic system is taken through the cycle $A B C D$ as shown in the figure. Heat rejected by the gas during the cycle is
supporting img

1

P V

2

2 P V

3

4 P V

4

1 / 2 P V

Explanation:

From the first law of thermodynamics $Q = Δ U + Δ W$
For cyclic process $Δ U = 0$
Hence $Q = W$
Work done $W =$ area under $P - V$ diagram
$\Rightarrow Q = - 2 P V .$

PHXI12:THERMODYNAMICS

371262 In the following indicator diagram, the net amount of work done will be :
supporting img

1 positive

2 Negative

3 zero

4 Infinity

PHXI12:THERMODYNAMICS

371258 A gas can be taken from $A$ to $B$ via two different processes $A C B$ and $A D B$ . When path $A C B$ is used $60 J$ of heat flows into the system and $30 J$ of work is done by the system. If path $A D B$ is used work done by the system is $10 J$ . The heat flow into the system in path $A D B$ is
supporting img

1

100 J

2

80 J

3

40 J

4

20 J

Explanation:

$Δ Q = Δ U + W$
$Δ U = Δ Q - W$
$(Δ U)_{A C B} = (Δ U)_{A D B}$
$\Rightarrow 60 - 30 = Δ Q - 10$
$\Rightarrow Δ Q = 40 J$

JEE - 2019

PHXI12:THERMODYNAMICS

371259 A cyclic Process of an ideal gas is shown in the figure. Let $W_{a \to b}, W_{b \to c}, W_{c \to d}$ and $W_{d \to a}$
represents the work done by the system during the processes $a \to b, b \to c, c \to d$ and $d \to a$ respectively. Consider the following relations
supporting img

I. $W_{a \to b} + W_{b \to c} + W_{c \to d} + W_{d \to a} > 0$
II. $W_{a \to b} + W_{b \to c} + W_{c \to a} + W_{d \to a} < 0$
III. $W_{a \to b} + W_{c \to d} > 0$
IV. $W_{b \to c} + W_{d \to a} = 0$
Which of the above relations is/are true?

1 I and III

2 III and IV

3 I, III and IV

4 II, III and IV

Explanation:

In Carnot cycle, in order to obtain continuous supply of work, net work done
As, $W_{a \to b}$ is isothermal
$W_{b \to c}$ is adiabatic
$W_{c \to d}$ is isothermal (negative)
$W_{d \to a}$ is adiabatic (negative)
Net work done during the complete cycle,
$W = W_{a \to b} + W_{b \to c} + (- W_{c \to d}) + (- W_{d \to a})$
$W = W_{a \to b} - W_{c \to d}$
$[as W_{b \to c} = W_{d \to a}]$
i.e. $W_{a \to b} + W_{c \to d} > 0$
$W_{a \to b} + W_{b \to c} + W_{c \to d} + W_{d \to a} > 0$
$W_{b \to c} = W_{d \to a}$
$∴ W_{b \to c} + W_{d \to a} = 0$

PHXI12:THERMODYNAMICS

371260 A thermodynamic system is taken through the cycle $P Q R S P$ process. The net work done by the system is
supporting img

1

20 J

2

- 40 J

3

400 J

4

- 374 J

Explanation:

Work done $=$ area of the cycle.
$W = - 200 c c \times 200 K P a ((-)$ sign indicates the path is in anticlockwise direction). $= - 200 \times 10^{- 6} \times 200 \times 10^{3} = - 40 J$

PHXI12:THERMODYNAMICS

371261 A thermodynamic system is taken through the cycle $A B C D$ as shown in the figure. Heat rejected by the gas during the cycle is
supporting img

1

P V

2

2 P V

3

4 P V

4

1 / 2 P V

Explanation:

From the first law of thermodynamics $Q = Δ U + Δ W$
For cyclic process $Δ U = 0$
Hence $Q = W$
Work done $W =$ area under $P - V$ diagram
$\Rightarrow Q = - 2 P V .$

PHXI12:THERMODYNAMICS

371262 In the following indicator diagram, the net amount of work done will be :
supporting img

1 positive

2 Negative

3 zero

4 Infinity

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