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PHXI12:THERMODYNAMICS

371263 The work done by the gas in the cyclic process is
supporting img

1

- 75 π k J

2

- 50 π k J

3

- 40 π k J

4

- 20 π k J

Explanation:

$| W_{cycle} | =$ Area of the cycle $= π r^{2} = π r_{V} r_{P}$
Pressure radius $(r_{P}) = \frac{(20 - 10)}{2} k P a = 5 k P a$
Volume radius $(r_{V}) = \frac{(40 - 10)}{2} m^{3} = 15 m^{3}$
$W_{cycle} = - 75 π K J$
$W = (+) v e$ (clock wise direction when $P$ is on $y -$ axis)
$W = (-)$ ve (clock wise direction when $P$ is on $x$ - axis)

PHXI12:THERMODYNAMICS

371264 An ideal gas after going through a series of four thermodynamic states in order, reaches the initial state against (cyclic process). The amounts of heat and work involved in these states are
$Q_{1} = 6000 J, Q_{2} = - 5500 J$
$Q_{3} = - 3000 J, Q_{2} = 3500 J$
$W_{1} = 2500 J, W_{2} = - 1000 J$
$W_{3} = - 1200 J, W_{4} = x J$
The ratio of net work done by the gas to the total heat absorbed by the gas is $η$ . The value of $x$ and $η$ are nearly

1

500, 7.5 %

2

700, 10.5 %

3

1000, 21 %

4

1500, 15 %

Explanation:

$Δ Q = Q_{1} + Q_{2} + Q_{3} + Q_{4}$
$= 6000 - 5500 - 3000 + 3500$
$Δ Q = 1000 J$
Total work done by ideal gas, i.e.,
$\begin{aligned} W & = W_{1} + W_{2} + W_{3} + W_{4} \\ = 2500 - 1000 - 1200 + x = 300 + x \end{aligned}$
For cyclic process, $Δ U = U_{f} - U_{i} = 0$
So, from first law of thermodynamics,
i.e., $Δ Q = Δ U + W$
$Δ Q = W (∴ Δ U = 0)$
We have, $1000 = (300 + x)$
i.e., $x = 1000 - 300 = 700 J$
Efficiency of the cycle is defined as
$η = \frac{Work done}{heat Input}$
$= \frac{W}{Q_{1} + Q_{4}} = \frac{Δ Q}{(Q_{1} + Q_{4})}$
$= \frac{1000}{(6000 + 3500)} = \frac{1000}{9500} \times 100$
$= 0.105 \times 100 = 10.5 %$
Correct option is (2).

PHXI12:THERMODYNAMICS

371265 The heat energy absorbed by a system in going through a cyclic process shown in figure is
supporting img

1

10^{3} π J

2

10^{2} π J

3

10^{4} π J

4

10^{7} π J

Explanation:

From $P - V$ diagram work done by system in cyclic process is equal to area of the loop then Heat absorbed = Workdone $=$ area of the loop.
$π r^{2} = π {(\frac{30 - 10}{2})}^{2} = π \times 10^{2} J$ .

PHXI12:THERMODYNAMICS

371266 One mole of a diatomic ideal gas undergoes a cyclic process $A B C$ as shown in figure. The process $B C$ is adiabatic. The temperatures at $A, B$ and $C$ are $400 K, 800 K$ and $600 K$ respectively. Choose the correct statement:
supporting img

1 The change in internal energy in whole cyclic process in

250 R

.

2 The change in internal energy in the process

C A

is

700 R

.

3 The change in internal energy in the process

A B

is

350 R

.

4 The change in internal energy in the process

B C

is

500 R

.

Explanation:

In cyclic process, change in total internal energy is zero.
$\begin{aligned} Δ U_{cyclic} = 0 \\ Δ U_{B C} = n C_{V} Δ T = 1 \times \frac{5 R}{2} Δ T \end{aligned}$
Where, $C_{V} =$ molar specific heat at constant volume.
For $B C, Δ T = - 200 K$
$∴ Δ U_{B C} = - 500 R$

PHXI12:THERMODYNAMICS

371263 The work done by the gas in the cyclic process is
supporting img

1

- 75 π k J

2

- 50 π k J

3

- 40 π k J

4

- 20 π k J

Explanation:

$| W_{cycle} | =$ Area of the cycle $= π r^{2} = π r_{V} r_{P}$
Pressure radius $(r_{P}) = \frac{(20 - 10)}{2} k P a = 5 k P a$
Volume radius $(r_{V}) = \frac{(40 - 10)}{2} m^{3} = 15 m^{3}$
$W_{cycle} = - 75 π K J$
$W = (+) v e$ (clock wise direction when $P$ is on $y -$ axis)
$W = (-)$ ve (clock wise direction when $P$ is on $x$ - axis)

PHXI12:THERMODYNAMICS

371264 An ideal gas after going through a series of four thermodynamic states in order, reaches the initial state against (cyclic process). The amounts of heat and work involved in these states are
$Q_{1} = 6000 J, Q_{2} = - 5500 J$
$Q_{3} = - 3000 J, Q_{2} = 3500 J$
$W_{1} = 2500 J, W_{2} = - 1000 J$
$W_{3} = - 1200 J, W_{4} = x J$
The ratio of net work done by the gas to the total heat absorbed by the gas is $η$ . The value of $x$ and $η$ are nearly

1

500, 7.5 %

2

700, 10.5 %

3

1000, 21 %

4

1500, 15 %

Explanation:

$Δ Q = Q_{1} + Q_{2} + Q_{3} + Q_{4}$
$= 6000 - 5500 - 3000 + 3500$
$Δ Q = 1000 J$
Total work done by ideal gas, i.e.,
$\begin{aligned} W & = W_{1} + W_{2} + W_{3} + W_{4} \\ = 2500 - 1000 - 1200 + x = 300 + x \end{aligned}$
For cyclic process, $Δ U = U_{f} - U_{i} = 0$
So, from first law of thermodynamics,
i.e., $Δ Q = Δ U + W$
$Δ Q = W (∴ Δ U = 0)$
We have, $1000 = (300 + x)$
i.e., $x = 1000 - 300 = 700 J$
Efficiency of the cycle is defined as
$η = \frac{Work done}{heat Input}$
$= \frac{W}{Q_{1} + Q_{4}} = \frac{Δ Q}{(Q_{1} + Q_{4})}$
$= \frac{1000}{(6000 + 3500)} = \frac{1000}{9500} \times 100$
$= 0.105 \times 100 = 10.5 %$
Correct option is (2).

PHXI12:THERMODYNAMICS

371265 The heat energy absorbed by a system in going through a cyclic process shown in figure is
supporting img

1

10^{3} π J

2

10^{2} π J

3

10^{4} π J

4

10^{7} π J

Explanation:

From $P - V$ diagram work done by system in cyclic process is equal to area of the loop then Heat absorbed = Workdone $=$ area of the loop.
$π r^{2} = π {(\frac{30 - 10}{2})}^{2} = π \times 10^{2} J$ .

PHXI12:THERMODYNAMICS

371266 One mole of a diatomic ideal gas undergoes a cyclic process $A B C$ as shown in figure. The process $B C$ is adiabatic. The temperatures at $A, B$ and $C$ are $400 K, 800 K$ and $600 K$ respectively. Choose the correct statement:
supporting img

1 The change in internal energy in whole cyclic process in

250 R

.

2 The change in internal energy in the process

C A

is

700 R

.

3 The change in internal energy in the process

A B

is

350 R

.

4 The change in internal energy in the process

B C

is

500 R

.

Explanation:

In cyclic process, change in total internal energy is zero.
$\begin{aligned} Δ U_{cyclic} = 0 \\ Δ U_{B C} = n C_{V} Δ T = 1 \times \frac{5 R}{2} Δ T \end{aligned}$
Where, $C_{V} =$ molar specific heat at constant volume.
For $B C, Δ T = - 200 K$
$∴ Δ U_{B C} = - 500 R$

PHXI12:THERMODYNAMICS

371263 The work done by the gas in the cyclic process is
supporting img

1

- 75 π k J

2

- 50 π k J

3

- 40 π k J

4

- 20 π k J

Explanation:

$| W_{cycle} | =$ Area of the cycle $= π r^{2} = π r_{V} r_{P}$
Pressure radius $(r_{P}) = \frac{(20 - 10)}{2} k P a = 5 k P a$
Volume radius $(r_{V}) = \frac{(40 - 10)}{2} m^{3} = 15 m^{3}$
$W_{cycle} = - 75 π K J$
$W = (+) v e$ (clock wise direction when $P$ is on $y -$ axis)
$W = (-)$ ve (clock wise direction when $P$ is on $x$ - axis)

PHXI12:THERMODYNAMICS

371264 An ideal gas after going through a series of four thermodynamic states in order, reaches the initial state against (cyclic process). The amounts of heat and work involved in these states are
$Q_{1} = 6000 J, Q_{2} = - 5500 J$
$Q_{3} = - 3000 J, Q_{2} = 3500 J$
$W_{1} = 2500 J, W_{2} = - 1000 J$
$W_{3} = - 1200 J, W_{4} = x J$
The ratio of net work done by the gas to the total heat absorbed by the gas is $η$ . The value of $x$ and $η$ are nearly

1

500, 7.5 %

2

700, 10.5 %

3

1000, 21 %

4

1500, 15 %

Explanation:

$Δ Q = Q_{1} + Q_{2} + Q_{3} + Q_{4}$
$= 6000 - 5500 - 3000 + 3500$
$Δ Q = 1000 J$
Total work done by ideal gas, i.e.,
$\begin{aligned} W & = W_{1} + W_{2} + W_{3} + W_{4} \\ = 2500 - 1000 - 1200 + x = 300 + x \end{aligned}$
For cyclic process, $Δ U = U_{f} - U_{i} = 0$
So, from first law of thermodynamics,
i.e., $Δ Q = Δ U + W$
$Δ Q = W (∴ Δ U = 0)$
We have, $1000 = (300 + x)$
i.e., $x = 1000 - 300 = 700 J$
Efficiency of the cycle is defined as
$η = \frac{Work done}{heat Input}$
$= \frac{W}{Q_{1} + Q_{4}} = \frac{Δ Q}{(Q_{1} + Q_{4})}$
$= \frac{1000}{(6000 + 3500)} = \frac{1000}{9500} \times 100$
$= 0.105 \times 100 = 10.5 %$
Correct option is (2).

PHXI12:THERMODYNAMICS

371265 The heat energy absorbed by a system in going through a cyclic process shown in figure is
supporting img

1

10^{3} π J

2

10^{2} π J

3

10^{4} π J

4

10^{7} π J

Explanation:

From $P - V$ diagram work done by system in cyclic process is equal to area of the loop then Heat absorbed = Workdone $=$ area of the loop.
$π r^{2} = π {(\frac{30 - 10}{2})}^{2} = π \times 10^{2} J$ .

PHXI12:THERMODYNAMICS

371266 One mole of a diatomic ideal gas undergoes a cyclic process $A B C$ as shown in figure. The process $B C$ is adiabatic. The temperatures at $A, B$ and $C$ are $400 K, 800 K$ and $600 K$ respectively. Choose the correct statement:
supporting img

1 The change in internal energy in whole cyclic process in

250 R

.

2 The change in internal energy in the process

C A

is

700 R

.

3 The change in internal energy in the process

A B

is

350 R

.

4 The change in internal energy in the process

B C

is

500 R

.

Explanation:

In cyclic process, change in total internal energy is zero.
$\begin{aligned} Δ U_{cyclic} = 0 \\ Δ U_{B C} = n C_{V} Δ T = 1 \times \frac{5 R}{2} Δ T \end{aligned}$
Where, $C_{V} =$ molar specific heat at constant volume.
For $B C, Δ T = - 200 K$
$∴ Δ U_{B C} = - 500 R$

PHXI12:THERMODYNAMICS

371263 The work done by the gas in the cyclic process is
supporting img

1

- 75 π k J

2

- 50 π k J

3

- 40 π k J

4

- 20 π k J

Explanation:

$| W_{cycle} | =$ Area of the cycle $= π r^{2} = π r_{V} r_{P}$
Pressure radius $(r_{P}) = \frac{(20 - 10)}{2} k P a = 5 k P a$
Volume radius $(r_{V}) = \frac{(40 - 10)}{2} m^{3} = 15 m^{3}$
$W_{cycle} = - 75 π K J$
$W = (+) v e$ (clock wise direction when $P$ is on $y -$ axis)
$W = (-)$ ve (clock wise direction when $P$ is on $x$ - axis)

PHXI12:THERMODYNAMICS

371264 An ideal gas after going through a series of four thermodynamic states in order, reaches the initial state against (cyclic process). The amounts of heat and work involved in these states are
$Q_{1} = 6000 J, Q_{2} = - 5500 J$
$Q_{3} = - 3000 J, Q_{2} = 3500 J$
$W_{1} = 2500 J, W_{2} = - 1000 J$
$W_{3} = - 1200 J, W_{4} = x J$
The ratio of net work done by the gas to the total heat absorbed by the gas is $η$ . The value of $x$ and $η$ are nearly

1

500, 7.5 %

2

700, 10.5 %

3

1000, 21 %

4

1500, 15 %

Explanation:

$Δ Q = Q_{1} + Q_{2} + Q_{3} + Q_{4}$
$= 6000 - 5500 - 3000 + 3500$
$Δ Q = 1000 J$
Total work done by ideal gas, i.e.,
$\begin{aligned} W & = W_{1} + W_{2} + W_{3} + W_{4} \\ = 2500 - 1000 - 1200 + x = 300 + x \end{aligned}$
For cyclic process, $Δ U = U_{f} - U_{i} = 0$
So, from first law of thermodynamics,
i.e., $Δ Q = Δ U + W$
$Δ Q = W (∴ Δ U = 0)$
We have, $1000 = (300 + x)$
i.e., $x = 1000 - 300 = 700 J$
Efficiency of the cycle is defined as
$η = \frac{Work done}{heat Input}$
$= \frac{W}{Q_{1} + Q_{4}} = \frac{Δ Q}{(Q_{1} + Q_{4})}$
$= \frac{1000}{(6000 + 3500)} = \frac{1000}{9500} \times 100$
$= 0.105 \times 100 = 10.5 %$
Correct option is (2).

PHXI12:THERMODYNAMICS

371265 The heat energy absorbed by a system in going through a cyclic process shown in figure is
supporting img

1

10^{3} π J

2

10^{2} π J

3

10^{4} π J

4

10^{7} π J

Explanation:

From $P - V$ diagram work done by system in cyclic process is equal to area of the loop then Heat absorbed = Workdone $=$ area of the loop.
$π r^{2} = π {(\frac{30 - 10}{2})}^{2} = π \times 10^{2} J$ .

PHXI12:THERMODYNAMICS

371266 One mole of a diatomic ideal gas undergoes a cyclic process $A B C$ as shown in figure. The process $B C$ is adiabatic. The temperatures at $A, B$ and $C$ are $400 K, 800 K$ and $600 K$ respectively. Choose the correct statement:
supporting img

1 The change in internal energy in whole cyclic process in

250 R

.

2 The change in internal energy in the process

C A

is

700 R

.

3 The change in internal energy in the process

A B

is

350 R

.

4 The change in internal energy in the process

B C

is

500 R

.

Explanation:

In cyclic process, change in total internal energy is zero.
$\begin{aligned} Δ U_{cyclic} = 0 \\ Δ U_{B C} = n C_{V} Δ T = 1 \times \frac{5 R}{2} Δ T \end{aligned}$
Where, $C_{V} =$ molar specific heat at constant volume.
For $B C, Δ T = - 200 K$
$∴ Δ U_{B C} = - 500 R$

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