4. In the chamber below, the volume is 20.5 mathrm(~L) . What would the volume be if the temperature is increased to 500 mathrm(~K) , the moles are reduced to 0.5 , and the pressure is increased to 2.0 atm? 41.0 mathrm(~L) 61.5 mathrm(~L) 10.25 mathrm(~L) 20.5 mathrm(~L)

To solve this problem, we can use the ideal gas law equation, which is PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature.Step 1: First, we need to find the number of moles in the original setup. We know the pressure (P1 = 1.0 atm), volume (V1 = 20.5 L), and temperature (T1 = 250 K). We also know that R = 0.0821 L.atm/(mol.K) for these units. So, we can rearrange the equation to solve for n:n1 = PV / RTn1 = (1.0 atm * 20.5 L) / (0.0821 L.atm/(mol.K) * 250 K)n1 = 1 molStep 2: Now, we can use the ideal gas law again to find the new volume (V2) when the temperature is increased to 500 K, the moles are reduced to 0.5, and the pressure is increased to 2.0 atm. We rearrange the equation to solve for V2:V2 = nRT / PV2 = (0.5 mol * 0.0821 L.atm/(mol.K) * 500 K) / 2.0 atmV2 = 10.25 LSo, the volume would be 10.25 L if the temperature is increased to 500 K, the moles are reduced to 0.5, and the pressure is increased to 2.0 atm. The correct answer is .