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.)
w = +2.25 kJ
To find the value of w (work) when 1.27 kg of H2O(l) is converted into water vapor at a constant external pressure of 1.00 atm, we can use the equation:
w = -PΔV
where P is the pressure and ΔV is the change in volume.
Given:
Initial pressure (P1) = 1.00 atm
Final pressure (P2) = 1.00 atm
Density of liquid water = 1.00 g/mL
Mass of H2O(l) = 1.27 kg
Temperature T1 = 25.0°C = 298.15 K
Final temperature T2 = 133°C = 406.15 K
First, we need to calculate the change in volume (ΔV) using the formula:
ΔV = V2 - V1
where V1 and V2 are the initial and final volumes of water, respectively.
The initial volume V1 can be calculated using the density and mass of water:
V1 = m/d
where m is the mass of water and d is the density of liquid water. Converting the mass of water to grams:
m = 1.27 kg = 1270 g
Now, substituting the values:
V1 = 1270 g / (1.00 g/mL) = 1270 mL
Next, we need to calculate the final volume V2 using 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)), and T is the temperature in Kelvin.
First, we need to find the number of moles of water vapor. We can use the equation:
n = m/M
where m is the mass of water and M is the molar mass of water.
The molar mass of water (H2O) is:
M = 2(1.01 g/mol for hydrogen) + 16.00 g/mol for oxygen = 18.02 g/mol
Now, substituting the values:
n = 1270 g / 18.02 g/mol = 70.49 mol
Now, we can calculate the final volume V2:
V2 = nRT2 / P2
Substituting the values:
V2 = (70.49 mol) * (0.0821 L·atm/(mol·K)) * (406.15 K) / (1.00 atm)
V2 = 2385.62 L
Finally, we can calculate the change in volume (ΔV):
ΔV = V2 - V1
ΔV = 2385.62 L - 1270 mL
ΔV = 1115.62 L
Now, we can calculate the work w:
w = -PΔV
w = -(1.00 atm) * (1115.62 L)
w = -1115.62 atm·L
So, the value of w (work) is -1115.62 atm·L.
To find the value of w in this scenario, we can use the equation:
w = ΔH - ΔE
Where:
w is the work done
ΔH is the change in enthalpy
ΔE is the change in internal energy
To calculate ΔH, we can use the equation:
ΔH = q + PΔV
Where:
q is the heat transferred
P is the pressure
ΔV is the change in volume
Since the pressure is constant, the equation simplifies to:
ΔH = q + PΔV
Now, we need to find the values for q and ΔV:
1. First, let's calculate q. We can use the equation:
q = mcΔT
Where:
m is the mass of the substance
c is the specific heat capacity of water
ΔT is the change in temperature
Given:
m = 1.27 kg
c = 4.18 J/g°C (specific heat capacity of water at constant pressure)
ΔT = 133°C - 25°C = 108°C
Substituting the values into the equation:
q = (1.27 kg) * (4.18 J/g°C) * (108°C) = 573.7344 J
2. Next, let's calculate ΔV. We can use the equation:
ΔV = Vf - Vi
Where:
Vf is the final volume (volume of water vapor)
Vi is the initial volume (volume of liquid water)
We can find Vi using the density of water, which is given as 1.00 g/mL.
Given:
m = 1.27 kg (mass of liquid water)
density = 1.00 g/mL
Using the equation:
density = mass / volume
Volume of liquid water:
Vi = m / density = (1.27 kg) / (1.00 g/mL) = 1270 mL
To find the final volume, we need to find the volume of water vapor.
To do that, we can assume the ideal gas law, which is:
PV = nRT
Where:
P is the pressure
V is the volume
n is the number of moles
R is the ideal gas constant
T is the temperature in Kelvin
Given:
P = 1.00 atm
T = 133°C = 406.15 K (converted to Kelvin)
R = 0.0821 L·atm/(K·mol) (ideal gas constant)
We know that n (number of moles) is equal to the moles of water present initially. To find this, we divide the mass of water by its molar mass.
The molar mass of water (H2O) is:
(1.01 g/mol for hydrogen * 2) + (16.00 g/mol for oxygen) = 18.02 g/mol
Moles of water:
n = mass / molar mass = (1.27 kg) / (18.02 g/mol) = 70.39 mol
Now we can calculate the final volume:
(1.00 atm) * V = (70.39 mol) * (0.0821 L·atm/(K·mol)) * (406.15 K)
V = [(70.39 mol) * (0.0821 L·atm/(K·mol)) * (406.15 K)] / (1.00 atm) = 2367.727 L
Finally, we can calculate ΔV:
ΔV = Vf - Vi
ΔV = (2367.727 L) - (1270 mL) = 1097.727 L
Now we have obtained the values for q and ΔV, we can substitute them into the equation for ΔH:
ΔH = q + PΔV
ΔH = 573.7344 J + (1.00 atm) * (1097.727 L) * (101.325 J/L) = -390.641 J
Since w = ΔH - ΔE, and we assume there is no change in internal energy (ΔE = 0) in this case, the value of w is:
w = ΔH - ΔE = -390.641 J (rounded to three decimal places)
So, the value of w is -390.641 J.