1.278 mL of water at 10°C was placed in a 3.0 L container that was sealed and evacuated. Then, 3440 J of heat was added to the water. Determine the pressure inside the container after heating. (Assume all of the heat goes into the water.)
d (H2O(s)) 1.00 g/cm3
Csp(H2O(s)) 2.08 J/g•°C
Csp(H2O(l)) 4.184J/g•°C
Csp(H2O(g)) 2.01 J/g•°C
Delta Hfus 6.01 kJ/mol
Delta Hvap 40.7 kJ/mol
melting pt 273 K (760 mm Hg)
boiling pt 373 K (760 mm Hg)
thanks
To determine the pressure inside the container after heating, we need to use the ideal gas law equation: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.
Step 1: Calculate the number of moles of water using the given mass and molar mass.
The molar mass of water (H2O) is 18 g/mol.
Given mass of water = 1.278 mL = 1.278 g (since density = mass/volume and the density of water is approximately 1.00 g/cm3)
Number of moles = mass / molar mass = 1.278 g / 18 g/mol = 0.071 moles
Step 2: Convert the given temperature from °C to Kelvin.
The given temperature is 10°C.
Temperature in Kelvin = 10°C + 273.15 = 283.15 K
Step 3: Calculate the pressure using the ideal gas law equation.
Rearranging the equation PV = nRT, we can solve for the pressure (P):
P = (nRT) / V
Given volume of the container = 3.0 L = 3000 mL
Ideal gas constant R = 0.0821 L·atm/mol·K (or any unit consistent with the other units being used in the equation)
P = (0.071 moles * 0.0821 L·atm/mol·K * 283.15 K) / 3000 mL
P = 0.0174 atm
Therefore, the pressure inside the container after heating is approximately 0.0174 atm.