A sample of propane is found to have a volume of 491cm^3 at 17^circ C and 725 mmHg. Calculate the volume under STP (standard temperature and pressure) (You answer should be to the hundreths or 2 decimal places: 0.00) Type your answer...

To calculate the volume of propane under standard temperature and pressure (STP), we use the **combined gas law**:<br /><br />\[<br />\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}<br />\]<br /><br />Where:<br />- \(P_1, V_1, T_1\) are the initial pressure, volume, and temperature.<br />- \(P_2, V_2, T_2\) are the final pressure, volume, and temperature at STP.<br /><br />### Step 1: Write down the given values<br />- Initial volume (\(V_1\)) = \(491 \, \text{cm}^3\)<br />- Initial temperature (\(T_1\)) = \(17^\circ \text{C} = 17 + 273.15 = 290.15 \, \text{K}\)<br />- Initial pressure (\(P_1\)) = \(725 \, \text{mmHg}\)<br /><br />At STP:<br />- Standard temperature (\(T_2\)) = \(0^\circ \text{C} = 273.15 \, \text{K}\)<br />- Standard pressure (\(P_2\)) = \(760 \, \text{mmHg}\)<br /><br />We need to find the final volume (\(V_2\)).<br /><br />---<br /><br />### Step 2: Rearrange the combined gas law to solve for \(V_2\)<br />\[<br />V_2 = \frac{P_1 V_1 T_2}{P_2 T_1}<br />\]<br /><br />---<br /><br />### Step 3: Substitute the known values<br />\[<br />V_2 = \frac{(725)(491)(273.15)}{(760)(290.15)}<br />\]<br /><br />---<br /><br />### Step 4: Perform the calculations step by step<br />1. Multiply the numerator:<br /> \[<br /> 725 \times 491 \times 273.15 = 97,091,841.25<br /> \]<br /><br />2. Multiply the denominator:<br /> \[<br /> 760 \times 290.15 = 220,514<br /> \]<br /><br />3. Divide the results:<br /> \[<br /> V_2 = \frac{97,091,841.25}{220,514} \approx 440.23 \, \text{cm}^3<br /> \]<br /><br />---<br /><br />### Final Answer:<br />The volume of propane under STP is:<br /><br />\[<br />\boxed{440.23 \, \text{cm}^3}<br />\]