40 Numeric 1 point 6.00 L of Neon gas is being cooled from \( 30.0^{\circ} \mathrm{C} \) to \( 0.0^{\circ} \mathrm{C} \). What is the resulting volume of the gas assuming pressure do not change? Round your answer to the nearest hundredth. Type your answer...

To determine the resulting volume of the neon gas when cooled from \( 30.0^{\circ} \mathrm{C} \) to \( 0.0^{\circ} \mathrm{C} \) at constant pressure, we can use **Charles's Law**: \[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \] Where: - \( V_1 = 6.00 \, \text{L} \) (initial volume) - \( T_1 = 30.0^{\circ} \mathrm{C} = 303.15 \, \text{K} \) (initial temperature in Kelvin) - \( T_2 = 0.0^{\circ} \mathrm{C} = 273.15 \, \text{K} \) (final temperature in Kelvin) - \( V_2 \) is the final volume we need to find. Rearranging the equation to solve for \( V_2 \): \[ V_2 = V_1 \times \frac{T_2}{T_1} = 6.00 \, \text{L} \times \frac{273.15 \, \text{K}}{303.15 \, \text{K}} \] \[ V_2 \approx 6.00 \times 0.901 \approx 5.406 \, \text{L} \] Rounding to the nearest hundredth: \[ V_2 \approx 5.41 \, \text{L} \] **Answer:** 5.41