- 1(current)
Q. No.An ideal gas is made to undergo a cycle depicted by the \((P\text-V)\) diagram alongside. If the curved line from \(A\) to \(B\) is adiabatic, then:
| 1. | The efficiency of this cycle is given by unity, as no heat is released during the cycle. |
| 2. | Heat is absorbed in the upper part of the straight-line path and released in the lower. |
| 3. | If \(T_1\) and \(T_2\) are the maximum and minimum temperatures reached during the cycle, then the efficiency is given by, \(\left(1-\dfrac{T_2}{T_1}\right).\) |
| 4. | The cycle can only be carried out in the reverse direction as shown in the figure. |
| 1. | \(4\) | 2. | \(1\) |
| 3. | \(2\) | 4. | \(3\) |
Match the thermodynamic processes taking place in a system with the correct conditions. In the table, \(\Delta Q\) is the heat supplied, \(\Delta W\) is the work done and \(\Delta U\) is the change in internal energy of the system.
| Process | Condition | ||
| (I) | Adiabatic | (A) | \(\Delta W=0\) |
| (II) | Isothermal | (B) | \(\Delta Q=0\) |
| (III) | Isochoric | (C) | \(\Delta U\neq0, \Delta W\neq0,\Delta Q\neq0\) |
| (IV) | Isobaric | (D) | \(\Delta U=0\) |
Codes:
| 1. | (I) – (B), (II) – (A), (III) – (D), (IV) – (C) |
| 2. | (I) – (A), (II) – (A), (III) – (B), (IV) – (C) |
| 3. | (I) – (A), (II) – (B), (III) – (D), (IV) – (D) |
| 4. | (I) – (B), (II) – (D), (III) – (A), (IV) – (C) |
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 |
| 1. | P → c, Q → a, R → d, S→ b |
| 2. | P→ c, Q → d, R → b, S → a |
| 3. | P → d, Q → b, R → b, S → c |
| 4. | P → a, Q → c, R → d, S → b |
Figure shows P-T diagram for given mass of an ideal gas for the process A→B. During this process, density of the gas is
1. Decreasing
2. Increasing
3. Constant
4. First decreasing then increasing
- 1(current)
Q. No.
© 2025 GoodEd Technologies Pvt. Ltd.