What volume does 15.6 g of H2O(g) occupy at 36.2 degrees Celsius and 1.25 atm?
Calculation of Volume for 15.6 g of H2O(g)
Final answer:
The volume occupied by 15.6 g of H2O(g) at 36.2 degrees Celsius and 1.25 atm can be calculated using the Ideal Gas Law. The volume is approximately 18.13 liters.
Explanation:
To find the volume occupied by 15.6 g of H2O(g) at 36.2 degrees Celsius and 1.25 atm, we can use the Ideal Gas Law, which is PV = nRT. Here, P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin.
First, we convert the mass of water to moles:
- Molar mass of H2O = 18.015 g/mol
- Moles of H2O = 15.6 g / 18.015 g/mol = 0.866 moles (approximately)
Next, we convert the temperature from degrees Celsius to Kelvin:
- Kelvin temperature = 36.2 + 273.15 = 309.35 K
Now we can plug these values into the Ideal Gas Law:
- PV = nRT
- V = nRT / P
- V = (0.866 moles) * (0.0821 L*atm/(K*mol)) * (309.35 K) / (1.25 atm)
- V ≈ 18.13 L
So, the volume occupied by 15.6 g of H2O(g) at 36.2 degrees Celsius and 1.25 atm is approximately 18.13 liters.
How can the volume occupied by 15.6 g of H2O(g) at 36.2 degrees Celsius and 1.25 atm be calculated? The volume occupied by 15.6 g of H2O(g) at 36.2 degrees Celsius and 1.25 atm can be calculated using the Ideal Gas Law formula, PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin.