Subtopic: Specific Heat Capacities of Diatomic Gases
Question Statement:
One kg of diatomic gas is heated and its temperature increases from 100 K to 600 K. The energy added at constant pressure during this process is 500 kJ. The specific heat at constant volume for the gas is _________ kJ/(kg K) (round off to 2 decimal places).
Official Answer Key:
0.71
Step-by-Step Solution & Explanation
### **Detailed Step-by-Step Solution (GATE AE 2019 Q.31 / AE Q.21)**
**1. Intuitive Physical Setup & Given Parameters**:
- Mass of gas: $m = 1\text{ kg}$
- Initial temperature: $T_1 = 100\text{ K}$
- Final temperature: $T_2 = 600\text{ K}$
- Temperature rise: $\Delta T = T_2 - T_1 = 600 - 100 = 500\text{ K}$
- Heat added at constant pressure: $Q_p = 500\text{ kJ} = 500 \times 10^3\text{ J}$
- Diatomic gas: Ratio of specific heats $\gamma = \frac{C_p}{C_v} = 1.4$ (or $7/5$).
**2. Step-by-Step Mathematical Derivation**:
1. **Specific Heat at Constant Pressure ($C_p$)**:
$Q_p = m C_p \Delta T \implies C_p = \frac{Q_p}{m \Delta T} = \frac{500\text{ kJ}}{1\text{ kg} \times 500\text{ K}} = 1.00\text{ kJ/(kg K)}$
2. **Specific Heat at Constant Volume ($C_v$)**:
For a diatomic ideal gas:
$\gamma = \frac{C_p}{C_v} = 1.4 = \frac{7}{5}$
$C_v = \frac{C_p}{\gamma} = \frac{1.00\text{ kJ/(kg K)}}{1.4} = \frac{5}{7} \approx 0.71428\text{ kJ/(kg K)}$
3. Rounding to two decimal places:
$C_v \approx 0.71\text{ kJ/(kg K)}$
This falls right in the official key range **[0.70, 0.72]**.
**3. Comprehensive Options & Distractor Analysis**:
- **Question Type**: Numerical Answer Type (NAT).
- **Official Range**: **0.70 to 0.72 kJ/(kg K)**.
**4. Official Master Key & Exam Takeaway**:
- **Official Answer**: **0.71** (Range: **0.70 to 0.72**)
- **Key Formula**: $C_p = \frac{Q_p}{m\Delta T} = 1.0$, $C_v = \frac{C_p}{\gamma} = \frac{1.0}{1.4} \approx 0.71$.
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