To test the effectiveness of a gunpowder mixture 1 gram was exploded under controlled STP conditions, and the reaction chamber was found to expand to 310cm^3 What volume would these gases occupy at the temperature produced by an uncontrolled explosion (2,200^circ C and 2.1 atmospheres of pressure? 1,000cm^3 1,300cm^3 7.7times 10^4cm^3 5,900cm^3

To solve this problem, we can use the ideal gas law and the combined gas law. The combined gas law relates the pressure, volume, and temperature of a gas before and after a change in conditions:<br /><br />\[<br />\frac{P_1 \times V_1}{T_1} = \frac{P_2 \times V_2}{T_2}<br />\]<br /><br />where:<br />- \( P_1 \) and \( V_1 \) are the initial pressure and volume,<br />- \( T_1 \) is the initial temperature in Kelvin,<br />- \( P_2 \) and \( V_2 \) are the final pressure and volume,<br />- \( T_2 \) is the final temperature in Kelvin.<br /><br />Given:<br />- Initial conditions (STP): \( P_1 = 1 \) atm, \( V_1 = 310 \, \text{cm}^3 \), \( T_1 = 273 \, \text{K} \)<br />- Final conditions: \( P_2 = 2.1 \) atm, \( T_2 = 2200^\circ \text{C} = 2473 \, \text{K} \)<br /><br />We need to find \( V_2 \).<br /><br />Rearranging the combined gas law to solve for \( V_2 \):<br /><br />\[<br />V_2 = \frac{P_1 \times V_1 \times T_2}{T_1 \times P_2}<br />\]<br /><br />Substitute the known values:<br /><br />\[<br />V_2 = \frac{1 \times 310 \times 2473}{273 \times 2.1}<br />\]<br /><br />Calculate:<br /><br />\[<br />V_2 = \frac{310 \times 2473}{573.3}<br />\]<br /><br />\[<br />V_2 \approx \frac{766630}{573.3} \approx 1337.5<br />\]<br /><br />Thus, the volume that these gases would occupy at the temperature produced by an uncontrolled explosion is approximately \( 1337.5 \, \text{cm}^3 \). Rounding to the nearest option, the closest answer is \( 1,300 \, \text{cm}^3 \).