In a system the mass is kept constant. The initial temperature of the system is \( 31.1^{\circ} \mathrm{C} \), the pressure is \( 0.61 \mathrm{~atm} \), and the volume is \( 3 \mathrm{~L} \). If the temperature changes to \( 145.7^{\circ} \mathrm{C} \) and the pressure changes to 2.3 atm what is the volume?
Answer
\( V_2 = \frac{P_1V_1T_2}{P_2T_1} = \frac{0.61 \mathrm{~atm} \cdot 3 \mathrm{~L} \cdot 418.85 \mathrm{~K}}{2.3 \mathrm{~atm} \cdot 304.25 \mathrm{~K}} = 1.3225 \mathrm{~L} \)
Step 1 ：\( T_1 = 31.1 + 273.15 = 304.25 \mathrm{~K} \)
Step 2 ：\( T_2 = 145.7 + 273.15 = 418.85 \mathrm{~K} \)
Step 3 ：\( V_2 = \frac{P_1V_1T_2}{P_2T_1} = \frac{0.61 \mathrm{~atm} \cdot 3 \mathrm{~L} \cdot 418.85 \mathrm{~K}}{2.3 \mathrm{~atm} \cdot 304.25 \mathrm{~K}} = 1.3225 \mathrm{~L} \)
