1. **State the problem**: We have two containers A and B connected, each containing an ideal gas at the same pressure but at different volumes and temperatures. 2. **Convert volumes to cubic meters**: - Volume in container A: $$V_A = 3.1 \times 10^3 \text{ cm}^3 = 3.1 \times 10^{-3} \text{ m}^3$$ - Volume in container B: $$V_B = 4.6 \times 10^3 \text{ cm}^3 = 4.6 \times 10^{-3} \text{ m}^3$$ 3. **Convert temperatures to Kelvin**: - Temperature in container A: $$T_A = 17 + 273.15 = 290.15 \text{ K}$$ - Temperature in container B: $$T_B = 30 + 273.15 = 303.15 \text{ K}$$ 4. **Given data**: - Pressure $$P = 2.3 \times 10^5 \text{ Pa}$$ - Ideal gas constant $$R = 8.314 \text{ J mol}^{-1} \text{K}^{-1}$$ 5. **Use the ideal gas law for each container**: $$n = \frac{PV}{RT}$$ 6. **Calculate moles of gas in container A**: $$n_A = \frac{(2.3 \times 10^5)(3.1 \times 10^{-3})}{8.314 \times 290.15}$$ $$n_A = \frac{713}{2409.8} \approx 0.296 \text{ mol}$$ 7. **Calculate moles of gas in container B**: $$n_B = \frac{(2.3 \times 10^5)(4.6 \times 10^{-3})}{8.314 \times 303.15}$$ $$n_B = \frac{1058}{2520.7} \approx 0.420 \text{ mol}$$ 8. **Calculate total moles of gas**: $$n_{total} = n_A + n_B = 0.296 + 0.420 = 0.716 \text{ mol}$$ **Final answer:** There are approximately **0.72 moles** of gas in the two containers combined.