Thermodynamics - 7 feb Contact Number: 9667591930 / 8527521718

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The variation of pressure versus temperature of an ideal gas is shown in the given diagram. From this diagram, one can conclude that

1. Volume increases continuously

2. Volume decreases continuously

3. Volume first increases then decreases

4. Volume first decreases, then increase

A carnot engine having an efficiency of $\frac{1}{10}$th of heat engine, is used as a refrigerator. If then work done on the system is 10 J, the amount of energy absorbed from the reservoir at lower temperature is:

1. 1 J

2. 90 J

3. 99 J

4. 100 J

The temperature inside a refrigerator is ${{t}_{2}}^{\circ}C$and the room temperature is ${{t}_{1}}^{\circ}C$. The amount of heat delivered to the room for each joule of electrical energy consumed ideally will be:

1. $\frac{{t}_{1}}{{t}_{1}-{t}_{2}}$

2. $\frac{{t}_{1}+273}{{t}_{1}-{t}_{2}}$

3. $\frac{{t}_{2}+273}{{t}_{1}+{t}_{2}}$

4. $\frac{{t}_{1}+{t}_{2}}{{t}_{1}+273}$

The coefficient of performance of a refrigerator is 5. If the temperature inside freezer is -20°C, the temperature of the surroundings to which it rejects heat is -

1. 31°C

2. 41°C

3. 11°C

4. 21°C

One mole of an ideal gas at an initial temperature of *T K* does 6*R* joules of work adiabatically. If the ratio of specific heats of this gas at constant pressure and at constant volume is 5/3, the final temperature of the gas will be:

1. | (T + 2.4)K |
2. | (T – 2.4)K |

3. | (T + 4)K |
4. | (T – 4)K |

The molar heat capacity in case of a diatomic gas if it does a work of $\frac{Q}{4}$ when heat *Q* is supplied to it is:

1. $\frac{2}{5}R$

2. $\frac{5}{2}R$

3. $\frac{10}{3}R$

4. $\frac{6}{7}R$

An insulator container contains 4 moles of an ideal diatomic gas at temperature *T*. Heat *Q* is supplied to this gas, due to which 2 moles of the gas are dissociated into atoms but temperature of the gas remains constant. Then

1. *Q* = 2*RT*

2. *Q* = *RT*

3. *Q* = 3*RT*

4. *Q* = 4*RT*

*P-V* diagram of a diatomic gas is a straight line passing through origin. The molar heat capacity of the gas in the process will be -

1. 4 *R*

2. 2.5 *R *

3. 3 *R*

4. $\frac{4R}{3}$

The temperature-entropy diagram of a reversible engine cycle is given in the figure. Its efficiency is

1. 1/3

2. 2/3

3. 1/2

4. 1/4

When a system is taken from state *i* to a state *f* along path *iaf*, *Q* = 50 *J* and *W* = 20 *J*.

If *W* = –13 *J* for the curved return path *fi*, *Q* for this path is -

1. 33 *J*

2. 23 *J*

3. – 7 *J*

4. – 43 *J*

An ideal gas is taken from point *A* to the point *B, *as shown in the *P-V* diagram. The work done in the process is -

1. $({P}_{A}-{P}_{B})({V}_{B}-{V}_{A})$

2. $\frac{1}{2}({P}_{B}-{P}_{A})({V}_{B}+{V}_{A})$

3. $\frac{1}{2}({P}_{B}-{P}_{A})({V}_{B}-{V}_{A})$

4. $\frac{1}{2}({P}_{B}+{P}_{A})({V}_{B}-{V}_{A})$

Thermodynamic processes are indicated in the following diagram:

Match the following:

Column-I |
Column-II |
||

P. | Process I | a. | Adiabatic |

Q. | Process II | b. | Isobaric |

R. | Process III | c. | Isochoric |

S. | Process IV | d. | Isothermal |

P | Q | R | S | |

1. | c | a | d | b |

2. | c | d | b | a |

3. | d | b | a | c |

4. | a | c | d | b |

One mole of an ideal monatomic gas undergoes a process described by the equation \(PV^3=\mathrm{constant}.\) The heat capacity of the gas during this process is:

1. \(\frac{3}{2}R\)

2. \(\frac{5}{2}R\)

3. \(2R\)

4. \(R\)

The volume (\(V\)) of a monatomic gas varies with its temperature (\(T\)), as shown in the graph. The ratio of work done by the gas to the heat absorbed by it when it undergoes a change from state \(\mathrm{A}\) to state \(\mathrm{B}\) will be:

1. | \(2 \over 5\) | 2. | \(2 \over 3\) |

3. | \(1 \over 3\) | 4. | \(2 \over 7\) |

An engine has an efficiency of $\frac{1}{6}$. When the temperature of the sink is reduced by ${62}^{0}C$, its efficiency is doubled. the temperature of the source is:

1. 124^{o}C

2. 37^{o}C

3. 62^{o}C

4. 99^{o}C

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