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If \(\Delta Q\) is the heat flowing out of a system, \(\Delta W\) is the work done by the system on its surroundings, and \(\Delta U\) is the decrease in internal energy of the system, then the first law of thermodynamics can be stated as:

1. | \(\Delta Q=\Delta U+\Delta W\) |

2. | \(\Delta U=\Delta Q+\Delta W\) |

3. | \(\Delta U=\Delta Q-\Delta W\) |

4. | \(\Delta U+\Delta Q+\Delta W=0\) |

Subtopic: First Law of Thermodynamics |

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A thermodynamic process involving a system takes place with \(100\) J of heat being rejected to the environment and an increase of \(20\) J of internal energy. Then,

1. | work done by the system is \(120\) J. |

2. | work done on the system is \(120\) J. |

3. | work done by the system is \(80\) J. |

4. | work done on the system is \(80\) J. |

Subtopic: First Law of Thermodynamics |

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If an average person jogs, he produces \(14.5 \times10^3\) cal/min. This is removed by the evaporation of sweat. The amount of sweat evaporated per minute (assuming \(1\) kg requires \(580 \times10^3\) cal for evaporation) is:

1. | \(0.25\) kg | 2. | \(0.50\) kg |

3. | \(0.025\) kg | 4. | \(0.20\) kg |

Subtopic: First Law of Thermodynamics |

66%

From NCERT

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\(1\) g of water of volume \(1\) cm^{3} at \(100^\circ \text{C}\) is converted into steam at the same temperature under normal atmospheric pressure \(\approx 1\times10^{5} \) Pa. The volume of steam formed equals \(1671\) cm^{3}. If the specific latent heat of vaporization of water is \(2256\) J/g, the change in internal energy is:

1. \(2423\) J

2. \(2089\) J

3. \(167\) J

4. \(2256\) J

Subtopic: First Law of Thermodynamics |

66%

From NCERT

NEET - 2019

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The internal energy of a gas is given by \(U=\frac32PV.\) The gas expands in such a way that its internal energy (initially \(U_0\)) remains constant throughout the process, but its volume changes from \(V_0\) to \(2V_0.\) The heat supplied to the gas equals:

1. \(U_0\mathrm{ln}(2)\)

2. \(\frac12U_0~\mathrm{ln}(2)\)

3. \(\frac13U_0~\mathrm{ln}(2)\)

4. \(\frac23U_0~\mathrm{ln}(2)\)

1. \(U_0\mathrm{ln}(2)\)

2. \(\frac12U_0~\mathrm{ln}(2)\)

3. \(\frac13U_0~\mathrm{ln}(2)\)

4. \(\frac23U_0~\mathrm{ln}(2)\)

Subtopic: First Law of Thermodynamics |

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From NCERT

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The first law of thermodynamics is a statement of:

1. | conservation of heat |

2. | conservation of work |

3. | conservation of momentum |

4. | conservation of energy |

Subtopic: First Law of Thermodynamics |

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Consider the process on a system shown in the figure. During the process, the work done by the system:

1. | continuously increases |

2. | continuously decreases |

3. | first increases then decreases |

4. | first decreases then increases |

Subtopic: Work Done by a Gas |

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An ideal gas goes from the state \(i\) to the state \(f\) as shown in figure given below. The work done by the gas during the process,

1. | is positive |

2. | is negative |

3. | is zero |

4. | cannot be obtained from this information |

Subtopic: Work Done by a Gas |

65%

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For the given cycle, the work done during the isobaric process is:

1. \(200\) J

2. zero

3. \(400\) J

4. \(600\) J

1. \(200\) J

2. zero

3. \(400\) J

4. \(600\) J

Subtopic: Work Done by a Gas |

52%

From NCERT

NEET - 2023

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The molar specific heat at a constant pressure of an ideal gas is \(\dfrac{7}{2}R.\) The ratio of specific heat at constant pressure to that at constant volume is:

1. | \(\dfrac{7}{5}\) | 2. | \(\dfrac{8}{7}\) |

3. | \(\dfrac{5}{7}\) | 4. | \(\dfrac{9}{7}\) |

Subtopic: Molar Specific Heat |

77%

From NCERT

AIPMT - 2006

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