QBManager

Thermodynamics

148114 The internal energy change in a system that has a absorbed 2 kcal of heat and done $500 J$ of work is

1

6400 J

2

5400 J

3

7900 J

4

8900 J

Explanation:

C Given, $W = 500 J, dQ = 2 kcal = 2 \times 10^{3} \times$ $4.2 = 8.4 \times 10^{3} J$
We know that,
$dQ = dU + W$
$dU = dQ - W$
$dU = 8.4 \times 10^{3} - 500$
$dU = 8400 - 500$
$dU = 7900 J$

BHU 2000

Thermodynamics

148115 Two moles of a monoatomic ideal gas is confined in a container and is heated such that its temperature increases by $10^{\circ} C$ . The approximate change in its internal energy is $(R = 8.31 J / mole - K)$

1 +250 joules

2 +350 joules

3 -250 joules

4 +450 joules

Explanation:

A Given, $n = 2 moles, Δ T = 10^{\circ} C = 10 K$
$R = 8.31 J / mol . K$ , For monoatomic gas, $C_{v} = \frac{3}{2} R$
$∵ dU = {nC}_{v} Δ T$
$= n \times \frac{3}{2} R Δ T$
$= 2 \times \frac{3}{2} \times 8.31 \times 10$
$= 249.3 J \approx 250 J$

AIIMS-2011

Thermodynamics

148117 The latent heat of vaporization of water is 2240 $J / g$ . If the work done in the process of vaporization of $1 g$ is $168 J$ , then increase in internal energy is

1

1940 J

2

2072 J

3

2240 J

4

2408 J

Explanation:

B Given that, $L_{V} = 2240 J / g$ , mass, $m = 1 g$ , $Δ W = 168 J$
$∵ Δ Q = {mL}_{V}$
$= 1 \times 2240$
$= 2240 J$
According to the first law of thermodynamics-
$Δ Q = Δ U + Δ W$
$Δ U = Δ Q - Δ W$
$= 2240 - 168 = 2072 J$

Punjab PET 1998

Thermodynamics

148118 A system is given 300 calories of heat and it does 600 joules of work. How much does the internal energy of the system change in this process? $(J = 4.18 Joules / cal)$

1 654 Joule

2 156.5 Joule

3 -300 Joule

4 -528.2 Joule

Explanation:

A Given, $Δ Q = 300 cal = 300 \times 4.18 J = 1254$ $J, Δ W = 600 J$
According to first law of thermodynamics-
$Δ Q = Δ U + Δ W$
$Δ U = Δ Q - Δ W$
$= 1254 - 600$
$= 654 J$

MP PET 1991

Thermodynamics

148114 The internal energy change in a system that has a absorbed 2 kcal of heat and done $500 J$ of work is

1

6400 J

2

5400 J

3

7900 J

4

8900 J

Explanation:

C Given, $W = 500 J, dQ = 2 kcal = 2 \times 10^{3} \times$ $4.2 = 8.4 \times 10^{3} J$
We know that,
$dQ = dU + W$
$dU = dQ - W$
$dU = 8.4 \times 10^{3} - 500$
$dU = 8400 - 500$
$dU = 7900 J$

BHU 2000

Thermodynamics

148115 Two moles of a monoatomic ideal gas is confined in a container and is heated such that its temperature increases by $10^{\circ} C$ . The approximate change in its internal energy is $(R = 8.31 J / mole - K)$

1 +250 joules

2 +350 joules

3 -250 joules

4 +450 joules

Explanation:

A Given, $n = 2 moles, Δ T = 10^{\circ} C = 10 K$
$R = 8.31 J / mol . K$ , For monoatomic gas, $C_{v} = \frac{3}{2} R$
$∵ dU = {nC}_{v} Δ T$
$= n \times \frac{3}{2} R Δ T$
$= 2 \times \frac{3}{2} \times 8.31 \times 10$
$= 249.3 J \approx 250 J$

AIIMS-2011

Thermodynamics

148117 The latent heat of vaporization of water is 2240 $J / g$ . If the work done in the process of vaporization of $1 g$ is $168 J$ , then increase in internal energy is

1

1940 J

2

2072 J

3

2240 J

4

2408 J

Explanation:

B Given that, $L_{V} = 2240 J / g$ , mass, $m = 1 g$ , $Δ W = 168 J$
$∵ Δ Q = {mL}_{V}$
$= 1 \times 2240$
$= 2240 J$
According to the first law of thermodynamics-
$Δ Q = Δ U + Δ W$
$Δ U = Δ Q - Δ W$
$= 2240 - 168 = 2072 J$

Punjab PET 1998

Thermodynamics

148118 A system is given 300 calories of heat and it does 600 joules of work. How much does the internal energy of the system change in this process? $(J = 4.18 Joules / cal)$

1 654 Joule

2 156.5 Joule

3 -300 Joule

4 -528.2 Joule

Explanation:

A Given, $Δ Q = 300 cal = 300 \times 4.18 J = 1254$ $J, Δ W = 600 J$
According to first law of thermodynamics-
$Δ Q = Δ U + Δ W$
$Δ U = Δ Q - Δ W$
$= 1254 - 600$
$= 654 J$

MP PET 1991

Thermodynamics

148114 The internal energy change in a system that has a absorbed 2 kcal of heat and done $500 J$ of work is

1

6400 J

2

5400 J

3

7900 J

4

8900 J

Explanation:

C Given, $W = 500 J, dQ = 2 kcal = 2 \times 10^{3} \times$ $4.2 = 8.4 \times 10^{3} J$
We know that,
$dQ = dU + W$
$dU = dQ - W$
$dU = 8.4 \times 10^{3} - 500$
$dU = 8400 - 500$
$dU = 7900 J$

BHU 2000

Thermodynamics

148115 Two moles of a monoatomic ideal gas is confined in a container and is heated such that its temperature increases by $10^{\circ} C$ . The approximate change in its internal energy is $(R = 8.31 J / mole - K)$

1 +250 joules

2 +350 joules

3 -250 joules

4 +450 joules

Explanation:

A Given, $n = 2 moles, Δ T = 10^{\circ} C = 10 K$
$R = 8.31 J / mol . K$ , For monoatomic gas, $C_{v} = \frac{3}{2} R$
$∵ dU = {nC}_{v} Δ T$
$= n \times \frac{3}{2} R Δ T$
$= 2 \times \frac{3}{2} \times 8.31 \times 10$
$= 249.3 J \approx 250 J$

AIIMS-2011

Thermodynamics

148117 The latent heat of vaporization of water is 2240 $J / g$ . If the work done in the process of vaporization of $1 g$ is $168 J$ , then increase in internal energy is

1

1940 J

2

2072 J

3

2240 J

4

2408 J

Explanation:

B Given that, $L_{V} = 2240 J / g$ , mass, $m = 1 g$ , $Δ W = 168 J$
$∵ Δ Q = {mL}_{V}$
$= 1 \times 2240$
$= 2240 J$
According to the first law of thermodynamics-
$Δ Q = Δ U + Δ W$
$Δ U = Δ Q - Δ W$
$= 2240 - 168 = 2072 J$

Punjab PET 1998

Thermodynamics

148118 A system is given 300 calories of heat and it does 600 joules of work. How much does the internal energy of the system change in this process? $(J = 4.18 Joules / cal)$

1 654 Joule

2 156.5 Joule

3 -300 Joule

4 -528.2 Joule

Explanation:

A Given, $Δ Q = 300 cal = 300 \times 4.18 J = 1254$ $J, Δ W = 600 J$
According to first law of thermodynamics-
$Δ Q = Δ U + Δ W$
$Δ U = Δ Q - Δ W$
$= 1254 - 600$
$= 654 J$

MP PET 1991

Thermodynamics

148114 The internal energy change in a system that has a absorbed 2 kcal of heat and done $500 J$ of work is

1

6400 J

2

5400 J

3

7900 J

4

8900 J

Explanation:

C Given, $W = 500 J, dQ = 2 kcal = 2 \times 10^{3} \times$ $4.2 = 8.4 \times 10^{3} J$
We know that,
$dQ = dU + W$
$dU = dQ - W$
$dU = 8.4 \times 10^{3} - 500$
$dU = 8400 - 500$
$dU = 7900 J$

BHU 2000

Thermodynamics

148115 Two moles of a monoatomic ideal gas is confined in a container and is heated such that its temperature increases by $10^{\circ} C$ . The approximate change in its internal energy is $(R = 8.31 J / mole - K)$

1 +250 joules

2 +350 joules

3 -250 joules

4 +450 joules

Explanation:

A Given, $n = 2 moles, Δ T = 10^{\circ} C = 10 K$
$R = 8.31 J / mol . K$ , For monoatomic gas, $C_{v} = \frac{3}{2} R$
$∵ dU = {nC}_{v} Δ T$
$= n \times \frac{3}{2} R Δ T$
$= 2 \times \frac{3}{2} \times 8.31 \times 10$
$= 249.3 J \approx 250 J$

AIIMS-2011

Thermodynamics

148117 The latent heat of vaporization of water is 2240 $J / g$ . If the work done in the process of vaporization of $1 g$ is $168 J$ , then increase in internal energy is

1

1940 J

2

2072 J

3

2240 J

4

2408 J

Explanation:

B Given that, $L_{V} = 2240 J / g$ , mass, $m = 1 g$ , $Δ W = 168 J$
$∵ Δ Q = {mL}_{V}$
$= 1 \times 2240$
$= 2240 J$
According to the first law of thermodynamics-
$Δ Q = Δ U + Δ W$
$Δ U = Δ Q - Δ W$
$= 2240 - 168 = 2072 J$

Punjab PET 1998

Thermodynamics

148118 A system is given 300 calories of heat and it does 600 joules of work. How much does the internal energy of the system change in this process? $(J = 4.18 Joules / cal)$

1 654 Joule

2 156.5 Joule

3 -300 Joule

4 -528.2 Joule

Explanation:

A Given, $Δ Q = 300 cal = 300 \times 4.18 J = 1254$ $J, Δ W = 600 J$
According to first law of thermodynamics-
$Δ Q = Δ U + Δ W$
$Δ U = Δ Q - Δ W$
$= 1254 - 600$
$= 654 J$

MP PET 1991

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