To solve the problem of finding the new temperature of a gas when its pressure is reduced at constant volume, we can use the ideal gas law and the relationship between pressure and temperature. Let's break this down step-by-step. ### Step 1: Understand the Ideal Gas Law The ideal gas law is given by the equation: \[ PV = nRT \] Where: - \( P \) = pressure (in atm) - \( V \) = volume (in liters) - \( n \) = number of moles of gas - \( R \) = ideal gas constant (0.0821 L·atm/(K·mol)) - \( T \) = temperature (in Kelvin) ### Step 2: Use the Relationship Between Pressure and Temperature Since the volume and the number of moles of gas remain constant, we can use the relationship derived from the ideal gas law that relates pressure and temperature: \[ \frac{P_1}{T_1} = \frac{P_2}{T_2} \] Where: - \( P_1 \) = initial pressure - \( T_1 \) = initial temperature - \( P_2 \) = final pressure - \( T_2 \) = final temperature ### Step 3: Convert Initial Temperature to Kelvin The initial temperature is given as 0°C. To convert this to Kelvin, we use the formula: \[ T(K) = T(°C) + 273.15 \] So, \[ T_1 = 0 + 273.15 = 273.15 \, K \] ### Step 4: Substitute Known Values into the Equation Now we can substitute the known values into the equation: - \( P_1 = 9 \, atm \) - \( T_1 = 273.15 \, K \) - \( P_2 = 3 \, atm \) - \( T_2 = ? \) Substituting these values into the equation: \[ \frac{9 \, atm}{273.15 \, K} = \frac{3 \, atm}{T_2} \] ### Step 5: Solve for \( T_2 \) Cross-multiplying gives us: \[ 9 \, atm \cdot T_2 = 3 \, atm \cdot 273.15 \, K \] Now, divide both sides by \( 9 \, atm \): \[ T_2 = \frac{3 \cdot 273.15}{9} \] Calculating the right side: \[ T_2 = \frac{819.45}{9} = 91.05 \, K \] ### Step 6: Round to the Appropriate Significant Figures Since the options provided are in whole numbers, we can round \( 91.05 \, K \) to \( 91 \, K \). ### Conclusion: Correct Answer Thus, the final answer is: **A. 91K** ### Explanation of Other Options - **B. 182K**: This value is incorrect because it does not follow the direct proportionality of pressure and temperature as derived from the ideal gas law. - **C. 273K**: This is the initial temperature and does not account for the change in pressure. - **D. 819K**: This value is too high and does not reflect the decrease in pressure from 9 atm to 3 atm. ### Revision Summary - Use the ideal gas law to relate pressure and temperature. - Convert Celsius to Kelvin for calculations. - Apply the formula \( \frac{P_1}{T_1} = \frac{P_2}{T_2} \) for constant volume scenarios. - Carefully perform calculations and round appropriately based on significant figures.