What is w when 3.37 kg of H2O(l), initially at 25.0oC, is converted into water vapour at 111oC against a constant external pressure of 1.00 atm? Assume that the vapour behaves ideally and that the density of liquid water is 1.00 g/mL. (Remember to include a "+" or "-" sign as appropriate.)
To find the value of w, we can use the equation:
w = ΔH - P ΔV
where:
ΔH is the change in enthalpy
P is the external pressure
ΔV is the change in volume
First, let's find the change in volume.
The initial volume of liquid water, V1, can be calculated using the mass and density:
V1 = m / ρ
= (3.37 kg) / (1.00 g/mL)
= 3.37 L
Since water doesn't compress significantly, the change in volume, ΔV, is equal to the change in the volume of water vapor, V2:
ΔV = V2 - V1
To find V2, we can use the ideal gas law:
PV = nRT
where:
P is the pressure
V is the volume
n is the number of moles
R is the ideal gas constant (0.0821 L·atm/mol·K)
T is the temperature in Kelvin
First, let's convert the temperatures to Kelvin:
T1 = 25.0 + 273.15 = 298.15 K
T2 = 111 + 273.15 = 384.15 K
Next, we need to find the number of moles of water present.
n = m / M
where:
m is the mass of water
M is the molar mass of water (18.015 g/mol)
n = (3.37 kg) / (18.015 g/mol)
= 186.84 mol
Now, let's calculate the initial volume of water vapor, V2:
V2 = (nRT2) / P
= [(186.84 mol) * (0.0821 L·atm/mol·K) * (384.15 K)] / (1.00 atm)
= 6059.92 L
Now we can find the change in volume:
ΔV = V2 - V1
= 6059.92 L - 3.37 L
= 6056.55 L
Next, let's calculate the change in enthalpy, ΔH.
ΔH = q
where q is the heat absorbed or released during the process.
q = m * c * ΔT
where:
m is the mass of water
c is the specific heat capacity of water (4.18 J/g·K)
ΔT is the change in temperature
ΔT = T2 - T1
= 384.15 K - 298.15 K
= 86 K
q = (3.37 kg) * (4.18 J/g·K) * (86 K)
= 1201.77 kJ
Now we can find w:
w = ΔH - P ΔV
= (1201.77 kJ) - (1.00 atm * 6056.55 L)
Note: It's important to convert the units of pressure and volume to match the energy unit (kJ) before performing the calculation.
To find the work (w) done in this process, we can use the formula:
w = -PΔV
Where P is the external pressure and ΔV is the change in volume.
To calculate ΔV, we need to find the initial and final volumes of water.
First, let's find the initial volume (Vi) of water. We are given the mass of water (m) and the density of liquid water (ρ).
Using the formula:
density = mass / volume
We can rearrange it to solve for volume:
volume = mass / density
Given that mass (m) of water = 3.37 kg and density (ρ) of water = 1.00 g/mL, we need to convert the mass to grams:
mass (m) = 3.37 kg * 1000 g/kg = 3370 g
Now we can find the initial volume:
Vi = 3370 g / 1.00 g/mL = 3370 mL
Next, let's find the final volume (Vf) of water vapor.
Since the water has been converted into water vapor, its volume will increase. The change in volume (ΔV) is therefore:
ΔV = Vf - Vi
We can rewrite this as:
Vf = ΔV + Vi
To find ΔV, we can use the ideal gas law:
PV = nRT
Where P is the pressure, V is the volume, n is the number of moles of gas, R is the ideal gas constant, and T is the temperature in Kelvin.
Given that the pressure (P) is 1.00 atm, the external pressure remains constant throughout the process.
The temperature (T) also changes from the initial temperature (25.0°C) to the final temperature (111°C). To convert these temperatures to Kelvin, we add 273.15:
Initial temperature (Ti) = 25.0°C + 273.15 = 298.15 K
Final temperature (Tf) = 111°C + 273.15 = 384.15 K
Since water vapor behaves ideally, we can assume it follows the ideal gas law. We need to find the number of moles (n) to determine the change in volume (ΔV).
To find the number of moles (n), we can use the formula:
n = mass / molar mass
The molar mass of water (H2O) is the sum of the atomic masses of hydrogen (H) and oxygen (O):
Atomic mass of H = 1.008 g/mol
Atomic mass of O = 16.00 g/mol
Molar mass of water (H2O) = 2 * 1.008 g/mol + 16.00 g/mol = 18.02 g/mol
Now we can find the number of moles:
n = 3370 g / 18.02 g/mol ≈ 186.97 mol
Since the number of moles (n) remains constant during the process, we can calculate the final volume (Vf) using the ideal gas law:
Vf = (nRTf) / P
Substituting the values:
Vf = (186.97 mol * 0.0821 L·atm/mol·K * 384.15 K) / 1.00 atm
Vf ≈ 7248 L
Now we have all the values to calculate the change in volume (ΔV):
ΔV = Vf - Vi
ΔV = 7248 L - 3370 mL
To convert mL to L, we divide by 1000:
ΔV ≈ 7248 L - 3.370 L
ΔV ≈ 7244.63 L
Finally, we can calculate the work (w) using the formula:
w = -PΔV
Substituting the values:
w = -(1.00 atm * 7244.63 L)
w ≈ -7244.63 atm∙L
Therefore, w is approximately -7244.63 atm∙L.
dE = q + w
q1 = mass H2O x specific heat H2O x (Tfinal-Tinitial) to get from 25 C to 100C.
q2 = mass H2O x heat vaporization to get from liquid water at 100 to steam at 100
q3 = mass steam x specific heat steam x (Tfinal-Tinitial) to get from steam at 100 C to steam at 111 C.
Add q1+q2+q3 to obtain total q and that is delta H. The sign is + because you ar adding heat for q1 and q2 and q3.
For work:
Use PV = nRT. You can calculate n, you know P and R and T. Solve for V. I assume you are expected to ignore the volume of the liquid but if you aren't you should calculate the volume of liquid water at 100 C. Then p*delta V will be the work. Since this is expanding into the atmosphere it will be -. Then substitute into dE = dH+w. Remember dH will be + and work will be -; dE can come out with either sign.