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Q. No.An ideal monoatomic gas at a temperature of \(300\) K and a pressure of \(10\) atm is suddenly allowed to expand into vacuum so that its volume is doubled. No exchange of heat is allowed to take place between the gas and its surroundings during the process. After equilibrium is reached, the final temperature is:
1.
\(300\) K
2.
\(\dfrac{300}{2^{5/3}}\) K
3.
\(\dfrac{300}{2^{2/3}}\) K
4.
\(600\) K
Subtopic: Types of Processes |
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The internal energy of a gas is given by \(U=\dfrac32PV.\) 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. | \(\dfrac12U_0~\mathrm{ln}(2)\) |
| 3. | \(\dfrac13U_0~\mathrm{ln}(2)\) | 4. | \(\dfrac23U_0~\mathrm{ln}(2)\) |
Subtopic: First Law of Thermodynamics |
67%
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An ideal gas obeys the law \(P^2V\) = constant during a reversible thermodynamic process. The relationship between volume \((V)\) and absolute temperature \((T)\) during this process is:
| 1. | \(\dfrac {V}{T}\) = constant | 2. | \(\dfrac {V^2}{T}\) = constant |
| 3. | \(\dfrac {T^2}{V}\) = constant | 4. | \(TV^2\) = constant |
Subtopic: Types of Processes |
79%
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The pressure of an ideal gas \(\left(\gamma=\dfrac32\right)\) is increased by \(1\%\) in an adiabatic process. The temperature of the gas:
| 1. | increases by \(1.5\%\) |
| 2. | decreases by \(1.5\%\) |
| 3. | increases by \(\frac13\%\) |
| 4. | increases by \(\frac23\%\) |
Subtopic: Types of Processes |
61%
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An ideal gas forms the working substance of a Carnot engine, and is taken around the Carnot cycle. We form the integral: \(I=\int\dfrac{dQ}{T},\)
where \(dQ\) is the heat supplied to the gas and \(T\) is the temperature of the gas. The integral is evaluated over the entire cycle. The value of the integral \(I\) is:
where \(dQ\) is the heat supplied to the gas and \(T\) is the temperature of the gas. The integral is evaluated over the entire cycle. The value of the integral \(I\) is:
| 1. | zero |
| 2. | negative |
| 3. | positive |
| 4. | non-negative(positive or zero) |
Subtopic: Carnot Engine |
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A gas \((\gamma = 1.5)\) undergoes a process in which its volume is doubled, but the speed of sound in the gas remains unchanged. Then,
| 1. | the pressure is halved |
| 2. | the pressure decreases by a factor of \(2\sqrt 2\) |
| 3. | the temperature is halved |
| 4. | the temperature decreases by a factor of \(2 \sqrt 2\) |
Subtopic: Types of Processes |
50%
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An ideal monatomic gas and a diatomic gas, both undergo adiabatic expansion starting from the same point on the \(P\)-\(V\) (indicator) diagram. The gases also undergo isothermal expansion. The curves are given by \(a,b,c.\) Which of the following is correct?
| 1. | \(a\)–isothermal, \(b\)–monatomic adiabatic, \(c\)–diatomic adiabatic |
| 2. | \(a\)–monatomic adiabatic, \(b\)–diatomic adiabatic, \(c\)–isothermal |
| 3. | \(a\)–diatomic adiabatic, \(b\)–monatomic adiabatic, \(c\)–isothermal |
| 4. | \(a\)–isothermal, \(b\)–diatomic adiabatic, \(c\)–monatomic adiabatic |
Subtopic: Types of Processes |
57%
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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 |
59%
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Given below are two statements:
| Statement I: | The efficiency of any thermodynamic engine can approach \(100\%\) if friction and all dissipative processes are reduced. |
| Statement II: | The first law of thermodynamics is applicable only to non-living systems. |
| 1. | Statement I is incorrect and Statement II is correct. |
| 2. | Both Statement I and Statement II are correct. |
| 3. | Both Statement I and Statement II are incorrect. |
| 4. | Statement I is correct and Statement II is incorrect. |
Subtopic: Second Law of Thermodynamics |
54%
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If the volume of an ideal gas is doubled, isothermally, the final pressure is \(1\) atm. If the process was adiabatic, the pressure would have been:
1. \(1\) atm
2. \(0.5\) atm
3. less than \(1\) atm
4. greater than \(1\) atm
1. \(1\) atm
2. \(0.5\) atm
3. less than \(1\) atm
4. greater than \(1\) atm
Subtopic: Types of Processes |
60%
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