Respuesta :
Answer:To find the new temperature at which the sulfur hexafluoride gas will occupy a volume of 2.86 L while the pressure remains constant, we can use the combined gas law equation: \( \frac{P_1 \cdot V_1}{T_1} = \frac{P_2 \cdot V_2}{T_2} \) Given: \( V_1 = 6.89 L \) (initial volume) \( T_1 = 212°C \) (initial temperature) \( V_2 = 2.86 L \) (final volume) We need to solve for \( T_2 \) (final temperature). 1. Convert the temperatures to Kelvin: \( T_{1(K)} = T_{1(°C)} + 273.15 \) \( T_{1(K)} = 212 + 273.15 = 485.15 K \) 2. Substitute the values into the equation: \( \frac{P_1 \cdot 6.89}{485.15} = \frac{P_2 \cdot 2.86}{T_2} \) 3. Since the pressure remains constant, we can simplify the equation: \( 6.89 = \frac{2.86}{T_2} \) 4. Rearrange the equation to solve for \( T_2 \): \( T_2 = \frac{2.86}{6.89} \) \( T_2 ≈ 0.4154 \) 5. Finally, convert the temperature back to Celsius: \( T_2(°C) = T_2 - 273.15 \) \( T_2(°C) ≈ -272.7346 \) Therefore, the temperature needed to reduce the volume to 2.86 L while the pressure remains constant is approximately -272.7°C.
