GATE AE 2019 • Q31 NAT • 1 Mark Propulsion

Propulsion

Thermodynamic Cycles of Gas Turbines

Subtopic: Turbojet, Turbofan & Turboprop Engine Performance

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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