Absolutely, let's determine the volume of 231 grams of water vapor at 652 K and 0.785 atm.
First, here are the steps you'll need to solve this problem:
1. Calculate Moles of Water:
- The molar mass of water (H[tex]\(_2\)[/tex]O) is approximately 18.01528 g/mol.
- Use the formula: [tex]\( \text{moles} = \frac{\text{mass}}{\text{molar mass}} \)[/tex]
[tex]\[
\text{moles of water} = \frac{231 \text{ g}}{18.01528 \text{ g/mol}} = 12.823 \text{ moles}
\][/tex]
2. Convert Pressure to Pascals:
- 1 atm = 101325 Pa.
- Convert pressure from atm to Pa:
[tex]\[
\text{pressure in Pa} = 0.785 \text{ atm} \times 101325 \text{ Pa/atm} = 79540.125 \text{ Pa}
\][/tex]
3. Use Ideal Gas Law to Find Volume:
- The ideal gas law equation is [tex]\( PV = nRT \)[/tex], where [tex]\( P \)[/tex] is the pressure, [tex]\( V \)[/tex] is the volume, [tex]\( n \)[/tex] is the number of moles, [tex]\( R \)[/tex] is the ideal gas constant (8.3144621 J/(mol·K)), and [tex]\( T \)[/tex] is the temperature.
- Solve for [tex]\( V \)[/tex]:
[tex]\[
V = \frac{nRT}{P}
\][/tex]
- Plug in the values:
[tex]\[
V = \frac{12.823 \text{ moles} \times 8.3144621 \text{ J/(mol·K)} \times 652 \text{ K}}{79540.125 \text{ Pa}}
\][/tex]
4. Calculate the Volume:
[tex]\[
V = \frac{12.823 \times 8.3144621 \times 652}{79540.125} = \frac{69454.192 \text{ J}}{79540.125 \text{ Pa}} = 0.873 \text{ m}^3
\][/tex]
- Convert cubic meters to liters (1 m³ = 1000 liters):
[tex]\[
V = 0.873 \text{ m}^3 \times 1000 \frac{\text{liters}}{\text{m}^3} = 873 \text{ liters}
\][/tex]
5. Round the Answer:
- The volume of 231 g of water vapor at 652 K and 0.785 atm is approximately [tex]\(873\)[/tex] liters when rounded to the nearest whole number.
Thus, the volume is 873 liters.
