Math Snap

STEP 1

1. The container is rigid with a volume of 2.00 L.
2. The initial pressure of argon gas ($\mathrm{Ar(g)}$) is 752 mmHg.
3. The temperature is $35^{\circ} \mathrm{C}$.
4. A 0.728 g sample of benzene ($\mathrm{C}_{6} \mathrm{H}_{6}$) vapor is added.
5. Ideal gas law applies to both gases.

STEP 2

1. Convert temperature to Kelvin.
2. Calculate the number of moles of $\mathrm{C}_{6} \mathrm{H}_{6}$.
3. Use the ideal gas law to find the partial pressure of $\mathrm{C}_{6} \mathrm{H}_{6}$.
4. Calculate the total pressure in the container.
5. Determine the partial pressures of $\mathrm{Ar}$ and $\mathrm{C}_{6} \mathrm{H}_{6}$.

STEP 3

Convert the temperature from Celsius to Kelvin.
$$T(K) = 35^{\circ} \mathrm{C} + 273.15 = 308.15 \, \mathrm{K}$$

STEP 4

Calculate the number of moles of $\mathrm{C}_{6} \mathrm{H}_{6}$.
The molar mass of $\mathrm{C}_{6} \mathrm{H}_{6}$ is approximately $78.11 \, \mathrm{g/mol}$.
$$n = \frac{\text{mass}}{\text{molar mass}} = \frac{0.728 \, \mathrm{g}}{78.11 \, \mathrm{g/mol}} \approx 0.00932 \, \mathrm{mol}$$

STEP 5

Use the ideal gas law to find the partial pressure of $\mathrm{C}_{6} \mathrm{H}_{6}$.
The ideal gas law is $PV = nRT$.
$$P_{\mathrm{C}_{6} \mathrm{H}_{6}} = \frac{nRT}{V}$$ Where:
- $R = 0.0821 \, \mathrm{L \cdot atm/(mol \cdot K)}$
- $n = 0.00932 \, \mathrm{mol}$
- $T = 308.15 \, \mathrm{K}$
- $V = 2.00 \, \mathrm{L}$
$$P_{\mathrm{C}_{6} \mathrm{H}_{6}} = \frac{0.00932 \, \mathrm{mol} \times 0.0821 \, \mathrm{L \cdot atm/(mol \cdot K)} \times 308.15 \, \mathrm{K}}{2.00 \, \mathrm{L}}$$ $$P_{\mathrm{C}_{6} \mathrm{H}_{6}} \approx 0.118 \, \mathrm{atm}$$ Convert to mmHg: $0.118 \, \mathrm{atm} \times 760 \, \mathrm{mmHg/atm} \approx 89.68 \, \mathrm{mmHg}$

STEP 6

Calculate the total pressure in the container.
The total pressure is the sum of the initial pressure of $\mathrm{Ar}$ and the partial pressure of $\mathrm{C}_{6} \mathrm{H}_{6}$.
$$P_{\text{total}} = P_{\mathrm{Ar}} + P_{\mathrm{C}_{6} \mathrm{H}_{6}}$$ $$P_{\text{total}} = 752 \, \mathrm{mmHg} + 89.68 \, \mathrm{mmHg} \approx 841.68 \, \mathrm{mmHg}$$

Determine the partial pressures of $\mathrm{Ar}$ and $\mathrm{C}_{6} \mathrm{H}_{6}$.
- Partial pressure of $\mathrm{Ar}$ is given as $752 \, \mathrm{mmHg}$.
- Partial pressure of $\mathrm{C}_{6} \mathrm{H}_{6}$ is calculated as $89.68 \, \mathrm{mmHg}$.
The total pressure in the container is approximately $841.68 \, \mathrm{mmHg}$.
The partial pressure of $\mathrm{Ar}$ is $752 \, \mathrm{mmHg}$ and the partial pressure of $\mathrm{C}_{6} \mathrm{H}_{6}$ is $89.68 \, \mathrm{mmHg}$.