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.
