In a system the pressure and mass are kept constant. The initial temperature of the system is \( -48.5^{\circ} \mathrm{C} \). If the temperature changes to \( 119.8^{\circ} \mathrm{C} \) as the volume becomes \( 6.43 \mathrm{~L} \), what was the initial volume?
Answer
Solve for initial volume \(V_1\): \(V_1 = \frac{V_2 \times T_1}{T_2} = \frac{6.43 L \times 224.65 K}{392.95 K} = 3.6984 L\).
Step 1 ：Convert initial and final temperatures to Kelvin: \(T_1 = -48.5 + 273.15 = 224.65 K\), \(T_2 = 119.8 + 273.15 = 392.95 K\).
Step 2 ：Apply the formula for constant pressure temperature-volume relation: \(\frac{V_1}{T_1} = \frac{V_2}{T_2}\).
Step 3 ：Solve for initial volume \(V_1\): \(V_1 = \frac{V_2 \times T_1}{T_2} = \frac{6.43 L \times 224.65 K}{392.95 K} = 3.6984 L\).
