what pressure, in atmosphere, is exerted by 0.52 mol of water vapor at its boiling point of 100 degrees celcius in a 3L teakettle?
The vapor pressure of a fluid is 1 atmosphere at its boiling point.
To calculate the pressure exerted by the water vapor, we can use the ideal gas law, which states:
PV = nRT
Where:
P is the pressure in atmospheres (atm)
V is the volume in liters (L)
n is the number of moles
R is the ideal gas constant (0.0821 L·atm/(mol·K))
T is the temperature in Kelvin (K)
First, we need to convert the temperature from Celsius to Kelvin:
T(K) = T(°C) + 273.15
T(K) = 100°C + 273.15 = 373.15K
Next, we have the values:
n = 0.52 mol
V = 3 L
R = 0.0821 L·atm/(mol·K)
T = 373.15K
Now we can substitute these values into the ideal gas law equation and solve for P:
PV = nRT
P * 3 = 0.52 * 0.0821 * 373.15
P * 3 = 15.968
P = 15.968 / 3
P ≈ 5.33 atm
Therefore, the pressure exerted by 0.52 mol of water vapor at its boiling point of 100 degrees Celsius in a 3L teakettle is approximately 5.33 atmospheres.