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:

  1. PV = nRT
  2. V = nRT / P
  3. V = (0.866 moles) * (0.0821 L*atm/(K*mol)) * (309.35 K) / (1.25 atm)
  4. 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.
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