Subtopic: Stress and Strain Tensors & Generalized Hooke's Law
Question Statement:
𝛿Q and 𝛿W are the heat and work interactions of a system with its surroundings,
and dU is the change in the internal energy of the system. For an adiabatic process
in a closed, constant pressure combustor, which one of the following options is
correct?
Options:
(A) |𝛿Q| = |dU| ≠ 0 and |𝛿W| = 0
(B) |𝛿Q| = |𝛿W| = 0 and |dU| ≠ 0
(C) |𝛿Q| = |𝛿W| = |dU| = 0
(D) |𝛿W| = |dU| ≠ 0 and |𝛿Q| = 0
Official Answer Key:
D
Step-by-Step Solution & Explanation
### **Detailed Step-by-Step Solution (GATE AE 2025 Q.31)**
**1. Intuitive Physical Setup & Given Parameters**:
Consider a closed, constant pressure combustor:
- Process is **adiabatic**: no heat interaction with surroundings ($\delta Q = 0$).
- Constant pressure process: $p = \text{constant}$.
- Chemical combustion reaction occurs inside the closed system, changing the chemical internal energy and temperature, and doing boundary work on surroundings as gas expands.
We need to determine the relations between heat $\delta Q$, work $\delta W$, and internal energy change $dU$.
**2. Step-by-Step Thermodynamic Derivation**:
1. **First Law of Thermodynamics**:
For any closed system undergoing a process:
$dU = \delta Q - \delta W$
2. **Apply Adiabatic Condition**:
Since the combustor is perfectly adiabatic:
$\delta Q = 0 \implies |\delta Q| = 0$
3. **Work Interaction ($\delta W$)**:
- The combustion reaction releases chemical energy, raising temperature and causing substantial thermal expansion of the gases against the constant ambient pressure $p$.
- Boundary work done is:
$\delta W = p dV \neq 0$
- Hence, $|\delta W| \neq 0$.
4. **Internal Energy Change ($dU$)**:
- From the first law with $\delta Q = 0$:
$dU = -\delta W$
- Since $\delta W \neq 0$:
$|dU| = |\delta W| \neq 0$
5. Combining conditions:
$|\delta W| = |dU| \neq 0 \quad \text{and} \quad |\delta Q| = 0$
**3. Comprehensive Options & Distractor Analysis**:
- **(A) |𝛿Q| = |dU| ≠ 0 and |𝛿W| = 0**: ❌ **INCORRECT**. Violates the adiabatic condition ($\delta Q = 0$).
- **(B) |𝛿Q| = |𝛿W| = 0 and |dU| ≠ 0**: ❌ **INCORRECT**. Violates the first law ($dU = \delta Q - \delta W = 0 - 0 = 0$).
- **(C) |𝛿Q| = |𝛿W| = |dU| = 0**: ❌ **INCORRECT**. Combustion produces non-zero expansion work.
- **(D) |𝛿W| = |dU| ≠ 0 and |𝛿Q| = 0**: ✅ **CORRECT**. Directly satisfies adiabatic boundary ($\delta Q = 0$) and first law $|dU| = |\delta W| \neq 0$.
**4. Official Master Key & Exam Takeaway**:
- **Official Answer Key**: **D**
- **Exam Takeaway**: For any adiabatic closed process, the first law demands $dU = -\delta W$. When gases expand against constant pressure, boundary work $p\Delta V$ is drawn directly from internal energy.
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