Let's break down each part step by step. --- ## Part a **Given:** - Initial concentration of chloroform, \( C_ = 10 \) μg/L - First order rate constant, \( k = .167 \) min\(^{-1} \) - Time, \( t = 30 \) min **Asked:** Final concentration of chloroform in water after one wash cycle (30 min). ### **Step 1: First Order Kinetics Equation** For a first-order reaction: \[ C = C_ e^{-kt} \] ### **Step 2: Plug in the Values** \[ C = 10\,\mu g/L \times e^{-.167 \times 30} \] Calculate the exponent: \[ .167 \times 30 = 5.01 \] \[ e^{-5.01} \approx .00667 \] ### **Step 3: Calculate Final Concentration** \[ C = 10\,\mu g/L \times .00667 = .0667\,\mu g/L \] --- **Final answer (a):** \[ \boxed{.067 \ \mu g/L} \quad \text{(rounded to 2 decimal places)} \] --- ## Part b **Given:** - 5 μg/min chloroform released at a steady rate for 1 hour (60 min) - Room size: \( 12\,m^3 \) - Ventilation rate: \( 6\,m^3/\text{hour} \) - Time: 1 hour = 60 min - Temperature: 30°C - Pressure: 1 atm - Assume well-mixed air, no chloroform in incoming air. **Asked:** What is the concentration of chloroform in the air (μg/m³) after 1 hour? --- ### **Step 1: Mass Balance Equation for Steady-State** Since the system is ventilated, use the following mass balance for a well-mixed room with a continuous source and outflow: \[ \text{Accumulation} = \text{Input (generation)} - \text{Output (ventilation)} \] Let: - \( V = 12\,m^3 \) (room volume) - \( Q = 6\,m^3/\text{hr} \) (ventilation rate) - \( G = 5\,\mu g/\text{min} = 300\,\mu g/\text{hr} \) (source rate) - \( C(t) \) = Chloroform concentration in air at time \( t \) The differential equation: \[ V \frac{dC}{dt} = G - Q C \] --- ### **Step 2: Solve the Differential Equation** Rearrange and solve: \[ \frac{dC}{dt} + \frac{Q}{V}C = \frac{G}{V} \] This is a first-order linear ODE. Its solution with \( C() = \) is: \[ C(t) = \frac{G}{Q} (1 - e^{-Qt/V}) \] Plug in the values: - \( G = 300 \, \mu g/hr \) - \( Q = 6 \, m^3/hr \) - \( V = 12 \, m^3 \) - \( t = 1 \) hr \[ C(1) = \frac{300}{6} (1 - e^{-6 \times 1 / 12}) \] Calculate exponent: \[ \frac{6}{12} = .5 \] \[ e^{-.5} \approx .6065 \] \[ C(1) = 50 \times (1 - .6065) = 50 \times .3935 = 19.675 \, \mu g/m^3 \] --- **Final answer (b):** \[ \boxed{20 \ \mu g/m^3} \quad \text{(rounded to 2 significant figures)} \] --- ## **Summary of Answers** a. Final chloroform concentration in water after one wash cycle: \[ \boxed{.067\ \mu g/L} \] b. Chloroform concentration in room air after 1 hour: \[ \boxed{20\ \mu g/m^3} \]