An engineer is designing an engine. Each cycle, it ignites a negligible amount of fuel, releasing $ 2000 \, \text{J} $ of energy into the cubic decimeter of air, which we assume here is gaseous nitrogen at $ 20^\circ \, \text{C} $ at $ 1 \, \text{atm} $ in the engine in a process which we can regard as instantaneous and isochoric. It then expands adiabatically until its pressure once again reaches $ 1 \, \text{atm} $, and shrinks isobarically until it reaches its initial state. What is the efficiency of this engine?

[i]Problem proposed by B. Dejean[/i]
1. **Convert to SI Units:**
   \[
   P_1 = 1 \, \text{atm} = 101325 \, \text{Pa}
   \]
   \[
   T_1 = 20^\circ \, \text{C} = 293.15 \, \text{K}
   \]
   \[
   V_1 = 1 \, \text{dm}^3 = 10^{-3} \, \text{m}^3
   \]

2. **Calculate the number of moles \( n \):**
   Using the ideal gas law \( P_1 V_1 = n R T_1 \):
   \[
   n = \frac{P_1 V_1}{R T_1} = \frac{101325 \, \text{Pa} \times 10^{-3} \, \text{m}^3}{8.314 \, \text{J/(mol K)} \times 293.15 \, \text{K}} \approx 0.041592 \, \text{mol}
   \]

3. **Isochoric Heating:**
   Since nitrogen is a diatomic gas, the change in internal energy is:
   \[
   \Delta U = \frac{5}{2} n R \Delta T
   \]
   Given \( \Delta U = 2000 \, \text{J} \):
   \[
   2000 \, \text{J} = \frac{5}{2} \times 0.041592 \, \text{mol} \times 8.314 \, \text{J/(mol K)} \times \Delta T
   \]
   Solving for \( \Delta T \):
   \[
   \Delta T = \frac{2000 \, \text{J}}{\frac{5}{2} \times 0.041592 \, \text{mol} \times 8.314 \, \text{J/(mol K)}} \approx 2314 \, \text{K}
   \]
   Therefore:
   \[
   T_2 = T_1 + \Delta T = 293.15 \, \text{K} + 2314 \, \text{K} = 2607.15 \, \text{K}
   \]
   Using the ideal gas law to find \( P_2 \):
   \[
   P_2 = \frac{n R T_2}{V_2} = \frac{0.041592 \, \text{mol} \times 8.314 \, \text{J/(mol K)} \times 2607.15 \, \text{K}}{10^{-3} \, \text{m}^3} \approx 901325 \, \text{Pa}
   \]

4. **Adiabatic Expansion:**
   For an adiabatic process, \( P V^\gamma = \text{constant} \) where \( \gamma = \frac{7}{5} \) for diatomic gases:
   \[
   P_2 V_2^{\gamma} = P_3 V_3^{\gamma}
   \]
   Solving for \( V_3 \):
   \[
   V_3 = V_2 \left( \frac{P_2}{P_3} \right)^{\frac{1}{\gamma}} = 10^{-3} \, \text{m}^3 \left( \frac{901325 \, \text{Pa}}{101325 \, \text{Pa}} \right)^{\frac{5}{7}} \approx 4.764 \times 10^{-3} \, \text{m}^3
   \]

5. **Work Done:**
   - **Adiabatic Process:**
     \[
     W_{\text{adiabatic}} = \frac{P_2 V_2 - P_3 V_3}{\gamma - 1} = \frac{901325 \, \text{Pa} \times 10^{-3} \, \text{m}^3 - 101325 \, \text{Pa} \times 4.764 \times 10^{-3} \, \text{m}^3}{\frac{7}{5} - 1} \approx 1046.52 \, \text{J}
     \]
   - **Isobaric Compression:**
     \[
     W_{\text{isobaric}} = P_3 (V_1 - V_3) = 101325 \, \text{Pa} \times (10^{-3} \, \text{m}^3 - 4.764 \times 10^{-3} \, \text{m}^3) \approx -381.39 \, \text{J}
     \]
   - **Net Work Done:**
     \[
     W_{\text{net}} = 1046.52 \, \text{J} - 381.39 \, \text{J} = 665.13 \, \text{J}
     \]

6. **Efficiency Calculation:**
   \[
   \eta = \frac{W_{\text{net}}}{Q_H} = \frac{665.13 \, \text{J}}{2000 \, \text{J}} \approx 0.33
   \]

The final answer is \(\boxed{0.33}\)