Calculate the difference between H and U when 1 mole of water is boiled at 373 K and 1 atm. The volume of 1 mole of perfect gas at 373 K is 0.03 m3 and the volume of liquid water may be neglected. (1 atm = 105 N m-2)
what equations should be used ?
Im confused with the content
no idea
To calculate the difference between ∆H (enthalpy change) and ∆U (internal energy change) when 1 mole of water is boiled at 373 K and 1 atm, you can use the following equation:
∆H = ∆U + P∆V
Where:
∆H is the enthalpy change,
∆U is the internal energy change,
P is the pressure, and
∆V is the change in volume.
In this case, the volume of liquid water can be neglected, so the change in volume (∆V) is equal to the volume of 1 mole of perfect gas at 373 K, which is given as 0.03 m^3.
Now, you can substitute the known values into the equation to find the difference between ∆H and ∆U.
Note: Since water is being boiled at a constant pressure of 1 atm, the pressure (P) is 1 atm or 105 N/m^2.
So, the equation becomes:
∆H = ∆U + (P∆V)
∆H = ∆U + (1 atm)(0.03 m^3)
∆H = ∆U + 0.03 atm m^3
Therefore, the equation that should be used is ∆H = ∆U + 0.03 atm m^3.
To calculate the difference between ΔH (enthalpy change) and ΔU (internal energy change) when 1 mole of water is boiled at 373 K and 1 atm, you can use the equation:
ΔH = q + PΔV
Where:
ΔH is the enthalpy change
q is the heat added or released during the process
P is the pressure
ΔV is the change in volume
First, you need to calculate the value of q, which represents the heat added or released during the boiling process. You can use the equation:
q = m × ΔHvap
Where:
q is the heat added or released
m is the mass (in moles) of water
ΔHvap is the enthalpy of vaporization of water
Since 1 mole of water is being boiled, the value of m is 1 mole.
Now, you need to find the value of ΔHvap from a reliable source or table. The enthalpy of vaporization of water at 373 K is approximately 40.7 kJ/mol.
Next, you can calculate the value of q by substituting the values into the equation:
q = 1 mol × 40.7 kJ/mol = 40.7 kJ
Now, you need to calculate the work done during the process using the equation:
PΔV = nRT
Where:
P is the pressure (1 atm = 105 N m-2)
ΔV is the change in volume (from liquid to gas state)
n is the number of moles (1 mole of water)
R is the gas constant (8.314 J/(mol·K))
T is the temperature (373 K)
Substituting the values into the equation:
PΔV = (105 N/m^2) × (0.03 m^3) = 3.15 J
Now that you have both the values for q and PΔV, you can calculate the difference between ΔH and ΔU using the equation:
ΔH - ΔU = q - PΔV
Substituting the values:
ΔH - ΔU = 40.7 kJ - 3.15 J
Note that the units of q, PΔV, and ΔU are different. You need to convert the units to be consistent (either all in joules or all in kilojoules) before calculating the difference.
Therefore, in this case, the equations you should use are:
- q = m × ΔHvap
- PΔV = nRT
- ΔH - ΔU = q - PΔV