Calculate the work done in joules when 1.0 mole of water vaporizes at 1.0 atm and 100C. Assume that the volume of liquid water is negligible compared with that of steam at 100c, and ideal gas behavior
T =100+ 273= 373K
SINCE
PV=NRT
NOW,V=NRT/P
=1×.0821×373/1
=30.6L
W=-PV
=-1atm×30.6L
=101.3× 30.6
-3.1×10^3J
Although this problem is practically incorrect. In theory, the water vapor behaves as a perfect gas; therefore, we apply: PV = mRT
Work Done (approx.), W = Pdv = P(mRT/P) = mRT= 1kg-mol x 8.3145 KJ/Kg-mol-K x(100+273) = 3101.3085 KJ
Answer: Approx. 3101.31 KJ
Note that this answer is just an approximation
-3.1x10^3J.
Well, let me put on my clown glasses and calculate that for you! 🤡
To calculate the work done, we need to use the ideal gas equation: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin.
First, let's convert the temperature from Celsius to Kelvin:
100C + 273.15 = 373.15K
Since the volume of liquid water is negligible compared to steam at 100C, we can consider the volume to be the volume of the gas at 100C.
By using the ideal gas equation, we can rearrange it to find V:
V = (nRT) / P
Now let's plug in the values:
n = 1.0 mole
R = 8.314 J/(mol·K)
T = 373.15K
P = 1.0 atm
V = (1.0 mole * 8.314 J/(mol·K) * 373.15K) / 1.0 atm
Calculating this will give us the volume of water vapor:
V = 3083.257 J/atm
Now, we can calculate the work done by multiplying the volume change by the pressure:
Work Done = V * P
Work Done = 3083.257 J/atm * 1.0 atm
Voila! The work done when 1.0 mole of water vaporizes at 1.0 atm and 100C is approximately 3083.257 Joules.
To calculate the work done when water vaporizes, we need to consider the change in volume of water during the phase transition. The work done can be calculated using the equation:
Work = Pressure * Change in Volume
Given that the pressure is 1.0 atm, we need to determine the change in volume. Since the volume of liquid water is negligible compared to that of steam, we can assume that the change in volume can be approximated by the change in volume of 1.0 mole of water vapor.
To determine the change in volume of 1.0 mole of water vapor, 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, and T is the temperature.
We need to rearrange this equation to solve for V:
V = nRT / P
Since we are assuming ideal gas behavior, we can use the value of the ideal gas constant R = 0.0821 L∙atm/(mol∙K).
Now, we can plug in the values into the equation:
V = (1.0 mole)(0.0821 L∙atm/(mol∙K))(373 K) / (1.0 atm)
Calculating this, we get V = 30.903 L.
Now, we can calculate the work done:
Work = (Pressure)(Change in Volume)
= (1.0 atm)(30.903 L)
= 30.903 atm∙L
Since 1 atm∙L = 101.3 J, we can convert the work done to joules by multiplying by the conversion factor:
Work (in joules) = (30.903 atm∙L)(101.3 J / 1 atm∙L)
= 3131.219 J
Therefore, the work done when 1.0 mole of water vaporizes at 1.0 atm and 100°C is approximately 3131.219 J.