In a system the mass is kept constant. The initial temperature of the system is \( 91.2^{\circ} \mathrm{C} \), the pressure is \( 2.93 \mathrm{~atm} \), and the volume is \( 6.7 \mathrm{~L} \). If the temperature changes to \( 18.5^{\circ} \mathrm{C} \) and the volume changes to \( 20.2 \mathrm{~L} \) what is the pressure?
Answer
3. Solve for \(P_2\): \(P_2 = \frac{P_1V_1T_2}{T_1V_2} = \frac{2.93 \mathrm{~atm} \cdot 6.7 \mathrm{~L} \cdot 291.65 \mathrm{K}}{364.35 \mathrm{K} \cdot 20.2 \mathrm{~L}} = 1.0511 \mathrm{~atm} \)
Step 1 ：1. Convert Celsius to Kelvin: \(T_1 = 91.2^{\circ} \mathrm{C} + 273.15 = 364.35 \mathrm{K} \), \(T_2 = 18.5^{\circ} \mathrm{C} + 273.15 = 291.65 \mathrm{K} \)
Step 2 ：2. Use the combined gas law: \(\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} \)
Step 3 ：3. Solve for \(P_2\): \(P_2 = \frac{P_1V_1T_2}{T_1V_2} = \frac{2.93 \mathrm{~atm} \cdot 6.7 \mathrm{~L} \cdot 291.65 \mathrm{K}}{364.35 \mathrm{K} \cdot 20.2 \mathrm{~L}} = 1.0511 \mathrm{~atm} \)
