What is w when 3.93 kg of H2O(l), initially at 25.0 ºC, is converted into water vapour at 157 ºC 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.
w is work
3930 g H2O = 3930/18 = approx 2 mol but this an estimate. You need to do a better job when you calculate this an all calculations that follow.
Ideal gas law is obeyed (from the problem) so V = nRT/P or
V (in liters) = 2*0.08206*(273.15+157)/1 atm = ? about 71 L
If you assume the volume of the liquid is negligible (I don't think I would do that), then work done is -P(Vvapor -Vliquid) = -1*(71-0) = about 71 L*atm.
Multiply by 101.325 to convert to J
If you want to include volume of the liquid and a number and not 0, then
3930 grams H2O is volume = mass/density = 3930/1.00 = 3930 cc or 3.93 L = about 4 L. The problem doesn't say to ignore the volume of the liquid and I would not. Post your work if you get stuck.
for the first part, how did you get 2 mol? 3930/18 = ~218?
Note that I said APPROX 2 mols. 2.18 is correct but I'm trying to show you HOW to do it and leave the calculations to you. I almost always say about or approximate. I expect you to follow up, calculate a more exact answer for each time I've estimated a number.
i meant, how did you get 2.18, 3930/18 is literally 218. where did you get the decimal from? sorry for the misunderstanding!
I get it and thank you for persisting. 3930/18 is 218 and I, of all people, should have recognized that 2.18 is an unrealistic answer. How did that happen? I have an old calculator and some of the pixels don't work anymore. A few dark dots (spots if you will) show up now and then. Those look like decimals to me. You guessed it. One of those sporadic spots showed up and I read the answer as 2.18 and not 218. My bad. Sorry about that. Also, I failed to note that your answer explicitly showed 218 and I mentally added a decimal to your answer to make it read 2.18 but there was no decimal there. By the way, with that being 218 and not 2.18, the answer will be 100 times greater so the volume of the liquid may very well be negligible when compared to V2. Thanks for catching that.
To determine the value of work (w) in this scenario, we need to use the equation:
w = -PΔV
where w represents work, P is the external pressure, and ΔV is the change in volume.
To calculate the change in volume, we need to consider the different states of water: liquid and vapor.
1. First, let's determine the initial volume (Vi) of the water. We are given the mass of water, 3.93 kg, and the density of liquid water, 1.00 g/mL. To convert the mass to volume, we use the formula:
Vi = (mass of water) / (density of water)
Vi = (3.93 kg) / (1.00 g/mL)
Note that we need to convert the mass from kg to g in order to match the units.
Vi = (3.93 kg) / (1000 g/kg) = 3930 g
Since 1 mL = 1 cm^3, the volume of water in mL is equal to the mass in grams. Therefore, Vi = 3930 mL.
2. Next, we need to determine the final volume (Vf) of the water in the vapor state. To do this, we need to consider the ideal gas law, which relates pressure, volume, and temperature:
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 in Kelvin.
3. We are given the temperature of the water initially (25.0 ºC) and the final temperature when converted into water vapor (157 ºC). We need to convert both temperatures to Kelvin:
Initial temperature (Ti) = 25.0 ºC + 273.15 = 298.15 K
Final temperature (Tf) = 157 ºC + 273.15 = 430.15 K
4. Since the external pressure is constant at 1.00 atm, the pressure in the vapor state (Pf) is also 1.00 atm.
5. To find the final volume (Vf), we need to determine the number of moles of water vapor. This can be calculated using the ideal gas law:
Vf = (n * R * Tf) / Pf
6. To find the number of moles (n) of water vapor, we use the mass and molar mass of water (H2O):
n = (mass of water) / (molar mass of water)
The molar mass of water (H2O) is the sum of the atomic masses of hydrogen (H) and oxygen (O):
Molar mass of water = (2 * atomic mass of hydrogen) + atomic mass of oxygen
Atomic mass of hydrogen (H) = 1.01 g/mol
Atomic mass of oxygen (O) = 16.00 g/mol
Molar mass of water = (2 * 1.01 g/mol) + 16.00 g/mol = 18.02 g/mol
n = (3.93 kg) / (18.02 g/mol)
Note: We need to convert the mass from kg to g and divide by the molar mass to obtain the number of moles.
n ≈ (3.93 kg) / (18.02 g/mol) = 218 mol
7. Now that we have the number of moles (n), we can calculate the final volume (Vf):
Vf = (n * R * Tf) / Pf
Using R = 0.0821 (atm * L / mol * K) as the ideal gas constant:
Vf = (218 mol * 0.0821 * 430.15 K) / 1.00 atm
8. Lastly, we can calculate the change in volume (ΔV) by taking the difference between the final volume (Vf) and the initial volume (Vi):
ΔV = Vf - Vi
Finally, we can substitute the values of ΔV and the external pressure (P) into the equation w = -PΔV to calculate the work (w) done.
Note: Remember to convert units accordingly to ensure consistency throughout the calculations.