where you live?(11.8) KN A 41. 4.000times 10^3 L of propane gas is held in a tank at 25^circ C The tank has a moveable diaphragm to keep the pressure constant at 200 kPa. If the temperature falls to -5^circ C on a cold winter day,what volume will the gas occupy? (11.8)T/I

Here's how to solve this problem using the combined gas law:<br /><br />**Understanding the Combined Gas Law**<br /><br />The combined gas law relates the pressure, volume, and temperature of a gas under two different sets of conditions, assuming the amount of gas (number of moles) remains constant. The formula is:<br /><br />(P₁V₁)/T₁ = (P₂V₂)/T₂<br /><br />Where:<br /><br />* P₁ and P₂ are the initial and final pressures.<br />* V₁ and V₂ are the initial and final volumes.<br />* T₁ and T₂ are the initial and final temperatures (in Kelvin).<br /><br />**Important Note:** Temperatures *must* be converted to Kelvin for this equation to work correctly.<br /><br />**1. Convert Temperatures to Kelvin:**<br /><br />* T₁ = 25°C + 273.15 = 298.15 K<br />* T₂ = -5°C + 273.15 = 268.15 K<br /><br />**2. Identify Known Values:**<br /><br />* P₁ = 200 kPa<br />* V₁ = 4.000 x 10³ L<br />* T₁ = 298.15 K<br />* P₂ = 200 kPa (pressure is constant)<br />* T₂ = 268.15 K<br /><br />**3. Solve for V₂:**<br /><br />We want to find V₂, the final volume. Rearrange the combined gas law equation to solve for V₂:<br /><br />V₂ = (P₁V₁T₂)/(T₁P₂)<br /><br />**4. Plug in the Values and Calculate:**<br /><br />V₂ = (200 kPa * 4.000 x 10³ L * 268.15 K) / (298.15 K * 200 kPa)<br /><br />V₂ = 3600 L (rounded to two significant figures, matching the initial volume's precision)<br /><br />**Answer:** The propane gas will occupy a volume of approximately 3600 L at -5°C.<br />