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PHXI12:THERMODYNAMICS

371228 A thermodynamical system is changed from state $(p_{1} V_{1})$ to $(p_{2} V_{2})$ by two different process. The quantity which will remain same is

1

Δ W

2

Δ Q

3

Δ Q - Δ W

4

Δ Q + Δ W

Explanation:

Change in internal energy does not depend upon path so $Δ U = Δ Q - Δ W$ remains constant.

PHXI12:THERMODYNAMICS

371229 In thermodynamic process, $200 J o u l e s$ of heat is given to a gas and $100 J o u l e s$ of work is also done on it. The change in internal energy of the gas is

1

100 J

2

300 J

3

419 J

4

24 J

Explanation:

$Δ Q = Δ U + Δ W; Δ Q = 200 J a n d Δ W = - 100 J$
$\Rightarrow Δ U = Δ Q - Δ W = 2000 - (100) = 300 J .$

PHXI12:THERMODYNAMICS

371230 An electric heater supplies heat to a system at a rate of $120 W$ . If system performs work at a rate of $80 J s^{- 1}$

1

30 J s^{- 1}

2

40 J s^{- 1}

3

50 J s^{- 1}

4

60 J s^{- 1}

Explanation:

According to first Law of thermodynamic $Δ Q = Δ U + Δ W$
$∴ \frac{Δ Q}{Δ t} = \frac{Δ U}{Δ t} + \frac{Δ W}{Δ t}$
Here, $\frac{Δ Q}{Δ t} = 120 W, \frac{Δ W}{Δ t} = 80 J s^{- 1}$
$∴ \frac{Δ U}{Δ t} = 120 - 80 = 40 J s^{- 1}$

PHXI12:THERMODYNAMICS

371231 The change in internal energy of $3 m o l$ of helium gas when its temperature is increased by $2 K$

1

74.8 c a l

2

37.4 c a l

3

37.4 J

4

74.8 J

Explanation:

$Δ U = μ C_{ν} d T = μ \times \frac{3}{2} R d T$
$= 3 \times \frac{3}{2} \times 8.31 \times 2 = 74.8 J$

PHXI12:THERMODYNAMICS

371228 A thermodynamical system is changed from state $(p_{1} V_{1})$ to $(p_{2} V_{2})$ by two different process. The quantity which will remain same is

1

Δ W

2

Δ Q

3

Δ Q - Δ W

4

Δ Q + Δ W

Explanation:

Change in internal energy does not depend upon path so $Δ U = Δ Q - Δ W$ remains constant.

PHXI12:THERMODYNAMICS

371229 In thermodynamic process, $200 J o u l e s$ of heat is given to a gas and $100 J o u l e s$ of work is also done on it. The change in internal energy of the gas is

1

100 J

2

300 J

3

419 J

4

24 J

Explanation:

$Δ Q = Δ U + Δ W; Δ Q = 200 J a n d Δ W = - 100 J$
$\Rightarrow Δ U = Δ Q - Δ W = 2000 - (100) = 300 J .$

PHXI12:THERMODYNAMICS

371230 An electric heater supplies heat to a system at a rate of $120 W$ . If system performs work at a rate of $80 J s^{- 1}$

1

30 J s^{- 1}

2

40 J s^{- 1}

3

50 J s^{- 1}

4

60 J s^{- 1}

Explanation:

According to first Law of thermodynamic $Δ Q = Δ U + Δ W$
$∴ \frac{Δ Q}{Δ t} = \frac{Δ U}{Δ t} + \frac{Δ W}{Δ t}$
Here, $\frac{Δ Q}{Δ t} = 120 W, \frac{Δ W}{Δ t} = 80 J s^{- 1}$
$∴ \frac{Δ U}{Δ t} = 120 - 80 = 40 J s^{- 1}$

PHXI12:THERMODYNAMICS

371231 The change in internal energy of $3 m o l$ of helium gas when its temperature is increased by $2 K$

1

74.8 c a l

2

37.4 c a l

3

37.4 J

4

74.8 J

Explanation:

$Δ U = μ C_{ν} d T = μ \times \frac{3}{2} R d T$
$= 3 \times \frac{3}{2} \times 8.31 \times 2 = 74.8 J$

PHXI12:THERMODYNAMICS

371228 A thermodynamical system is changed from state $(p_{1} V_{1})$ to $(p_{2} V_{2})$ by two different process. The quantity which will remain same is

1

Δ W

2

Δ Q

3

Δ Q - Δ W

4

Δ Q + Δ W

Explanation:

Change in internal energy does not depend upon path so $Δ U = Δ Q - Δ W$ remains constant.

PHXI12:THERMODYNAMICS

371229 In thermodynamic process, $200 J o u l e s$ of heat is given to a gas and $100 J o u l e s$ of work is also done on it. The change in internal energy of the gas is

1

100 J

2

300 J

3

419 J

4

24 J

Explanation:

$Δ Q = Δ U + Δ W; Δ Q = 200 J a n d Δ W = - 100 J$
$\Rightarrow Δ U = Δ Q - Δ W = 2000 - (100) = 300 J .$

PHXI12:THERMODYNAMICS

371230 An electric heater supplies heat to a system at a rate of $120 W$ . If system performs work at a rate of $80 J s^{- 1}$

1

30 J s^{- 1}

2

40 J s^{- 1}

3

50 J s^{- 1}

4

60 J s^{- 1}

Explanation:

According to first Law of thermodynamic $Δ Q = Δ U + Δ W$
$∴ \frac{Δ Q}{Δ t} = \frac{Δ U}{Δ t} + \frac{Δ W}{Δ t}$
Here, $\frac{Δ Q}{Δ t} = 120 W, \frac{Δ W}{Δ t} = 80 J s^{- 1}$
$∴ \frac{Δ U}{Δ t} = 120 - 80 = 40 J s^{- 1}$

PHXI12:THERMODYNAMICS

371231 The change in internal energy of $3 m o l$ of helium gas when its temperature is increased by $2 K$

1

74.8 c a l

2

37.4 c a l

3

37.4 J

4

74.8 J

Explanation:

$Δ U = μ C_{ν} d T = μ \times \frac{3}{2} R d T$
$= 3 \times \frac{3}{2} \times 8.31 \times 2 = 74.8 J$

PHXI12:THERMODYNAMICS

371228 A thermodynamical system is changed from state $(p_{1} V_{1})$ to $(p_{2} V_{2})$ by two different process. The quantity which will remain same is

1

Δ W

2

Δ Q

3

Δ Q - Δ W

4

Δ Q + Δ W

Explanation:

Change in internal energy does not depend upon path so $Δ U = Δ Q - Δ W$ remains constant.

PHXI12:THERMODYNAMICS

371229 In thermodynamic process, $200 J o u l e s$ of heat is given to a gas and $100 J o u l e s$ of work is also done on it. The change in internal energy of the gas is

1

100 J

2

300 J

3

419 J

4

24 J

Explanation:

$Δ Q = Δ U + Δ W; Δ Q = 200 J a n d Δ W = - 100 J$
$\Rightarrow Δ U = Δ Q - Δ W = 2000 - (100) = 300 J .$

PHXI12:THERMODYNAMICS

371230 An electric heater supplies heat to a system at a rate of $120 W$ . If system performs work at a rate of $80 J s^{- 1}$

1

30 J s^{- 1}

2

40 J s^{- 1}

3

50 J s^{- 1}

4

60 J s^{- 1}

Explanation:

According to first Law of thermodynamic $Δ Q = Δ U + Δ W$
$∴ \frac{Δ Q}{Δ t} = \frac{Δ U}{Δ t} + \frac{Δ W}{Δ t}$
Here, $\frac{Δ Q}{Δ t} = 120 W, \frac{Δ W}{Δ t} = 80 J s^{- 1}$
$∴ \frac{Δ U}{Δ t} = 120 - 80 = 40 J s^{- 1}$

PHXI12:THERMODYNAMICS

371231 The change in internal energy of $3 m o l$ of helium gas when its temperature is increased by $2 K$

1

74.8 c a l

2

37.4 c a l

3

37.4 J

4

74.8 J

Explanation:

$Δ U = μ C_{ν} d T = μ \times \frac{3}{2} R d T$
$= 3 \times \frac{3}{2} \times 8.31 \times 2 = 74.8 J$