What is w when 1.27 kg of H2O(l), initially at 25.0oC, is converted into water vapour at 133oC 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.)
I think I've answered this before.
H2O(l) ==> H2O(steam)
Convert kg to mols H2O at 25 and calculate the volume occupied. Do the same for volume at 133 C using PV = nRT.
w = -p*delta V.
To find the change in enthalpy, we can use the formula:
ΔH = m * c * ΔT
where ΔH is the change in enthalpy, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.
First, let's find the change in temperature:
ΔT = final temperature - initial temperature
= 133°C - 25°C
= 108°C
Next, let's calculate the mass of water:
mass of water = volume of water * density of water
= 1.27 kg / (1.00 g/mL)
= 1270 mL
Now, let's convert mL to grams:
mass of water = 1270 mL * 1.00 g/mL
= 1270 g
Finally, let's calculate the change in enthalpy:
ΔH = (mass of water) * c * ΔT
The specific heat capacity of water is approximately 4.18 J/g·°C. Therefore:
ΔH = 1270 g * 4.18 J/g·°C * 108°C
= 582,505.6 J
Since the water is being converted into water vapor, it is an endothermic process. Therefore, the change in enthalpy (ΔH) is positive.
So, the value of ΔH for the given process is +582,505.6 J.
To find the work (w) done in this process, we need to use the equation:
w = -PΔV
where P is the external pressure and ΔV is the change in volume.
In this case, we have water (H2O) initially at 25.0oC and it is being converted into water vapor at 133oC against a constant external pressure of 1.00 atm.
To find ΔV, we need to calculate the difference in volume between the liquid water and water vapor. We can use the ideal gas law and the density of liquid water to find the initial volume (Vinitial) of the water.
Vinitial = m/ρ
where m is the mass of the water (1.27 kg) and ρ is the density of liquid water (1.00 g/mL).
Converting the units, we have:
m = 1.27 kg
ρ = 1.00 g/mL = 1000 g/L
Vinitial = 1.27 kg / (1000 g/L) = 1.27 L
Next, we need to determine the final volume (Vfinal) of the water vapor. We can use the ideal gas law with the given conditions: Tfinal = 133oC (converted to Kelvin) and P = 1.00 atm.
Vfinal = (nRT) / P
where n is the number of moles of the water vapor, R is the ideal gas constant, and T is the temperature in Kelvin.
First, we need to calculate the number of moles of water vapor (n). We can use the molar mass of water (18.015 g/mol) to convert the mass of the water (1.27 kg) to moles.
n = mass / molar mass
n = 1.27 kg / 18.015 g/mol = 70.536 mol
Next, we need to convert the temperature from Celsius to Kelvin.
Tfinal = 133oC + 273.15 = 406.15 K
Now, we can substitute these values into the equation for Vfinal:
Vfinal = (70.536 mol * 0.0821 L·atm/mol·K * 406.15 K) / 1.00 atm
Vfinal = 2408.49 L·atm
Finally, we can calculate the change in volume (ΔV) by subtracting Vinitial from Vfinal:
ΔV = Vfinal - Vinitial
ΔV = 2408.49 L·atm - 1.27 L
ΔV = 2407.22 L·atm
Finally, we can calculate the work (w) done in the process using the equation:
w = -PΔV
Substituting the values:
w = -(1.00 atm) * (2407.22 L·atm)
w ≈ -2407 J
Therefore, the work done (w) when converting 1.27 kg of water into water vapor at the given conditions is approximately -2407 J, where the negative sign indicates work done by the system.