For carbon dioxide the numerical value of a is 3.592 and the
numerical value of b is 0.0429.
Use the van der Waals equation to calculate the pressure of a
sample of carbon dioxide at 50.0 °C with a molar volume of
0.515 L/mol. Round your answer to 3 significant digits.



Answer :

Alright! Let's solve this using the van der Waals equation, which takes into account the non-ideal behavior of a gas. The equation is: \[ (P + \frac{a}{V_m^2})(V_m - b) = RT \] where: - \( P \) is the pressure, - \( a \) and \( b \) are empirical constants specific to each gas, - \( V_m \) is the molar volume of the gas, - \( R \) is the ideal gas constant, and - \( T \) is the temperature in Kelvin. Given values for carbon dioxide (\( CO_2 \)): - \( a = 3.592 \) (L^2 atm/mol^2), - \( b = 0.0429 \) L/mol, - Temperature \( T = 50.0 \) °C, which we need to convert to Kelvin by adding 273.15 to it, - Molar volume \( V_m = 0.515 \) L/mol. Let's solve the equation step-by-step. First, convert temperature to Kelvin: \[ T = 50.0 + 273.15 = 323.15 \text{ K} \] Now, use the ideal gas constant for the units we're working in: \[ R = 0.0821 \text{ L atm} / \text{(K mol)} \] Now, plug in all the values into the van der Waals equation: \[ (P + \frac{3.592}{(0.515)^2})(0.515 - 0.0429) = (0.0821)(323.15) \] Calculate the denominators and simplify the equation: \[ (P + \frac{3.592}{0.265225})(0.4721) = 26.612715 \] Next, evaluate the term \( \frac{3.592}{0.265225} \): \[ (P + 13.5419)(0.4721) = 26.612715 \] Distribute the left side of the equation: \[ 0.4721P + 6.39406739 = 26.612715 \] Now, solve for \( P \) by first subtracting the constant from both sides: \[ 0.4721P = 26.612715 - 6.39406739 \] \[ 0.4721P = 20.21864761 \] Finally, divide by the coefficient of \( P \): \[ P = \frac{20.21864761}{0.4721} \] \[ P \approx 42.848 \text{ atm} \] Thus, the pressure of the sample of carbon dioxide at 50.0 °C with a molar volume of 0.515 L/mol is approximately 42.848 atm, rounded to three significant digits.