Part A A multipatient hyperbaric chamber has a volume of 3900 L . At a temperature of \( 23^{\circ} \mathrm{C} \), how many grams of oxygen are needed to give a pressure of 1.9 atm ? Express the mass to two significant figures and include the appropriate units. \( m= \) Value ? Units

To solve this problem, we can use the Ideal Gas Law, which states \( PV = nRT \). Here, we want to find the mass of oxygen, so we need to rearrange the equation to find \( n \) (the number of moles) first, and then convert that to grams. **Step 1: Rearranging the Ideal Gas Law** \[ n = \frac{PV}{RT} \] **Step 2: Plugging in values** - \( P = 1.9 \, \text{atm} \) - \( V = 3900 \, \text{L} \) (which is 3900 L) - \( R = 0.0821 \, \text{L atm/(K mol)} \) - \( T = 23^{\circ}C = 23 + 273.15 = 296.15 \, \text{K} \) **Step 3: Calculating \( n \)** Now, substitute the values into the equation: \[ n = \frac{(1.9 \, \text{atm})(3900 \, \text{L})}{(0.0821 \, \text{L atm/(K mol)})(296.15 \, \text{K})} \] Calculating this gives: \[ n \approx \frac{7410}{24.475} \approx 303.1 \, \text{mol} \] **Step 4: Converting moles to grams** Oxygen (\( O_2 \)) has a molar mass of approximately 32.00 g/mol: \[ m = n \times \text{Molar Mass} = 303.1 \, \text{mol} \times 32.00 \, \text{g/mol} \approx 9699.2 \, \text{g} \] **Step 5: Rounding to two significant figures** Thus, we round the mass to two significant figures: \[ m = 9700 \, \text{g} \] Therefore, the final answer is: \( m = 9700 \, \text{g} \)