One mole of H2O(g) at 1.00 atm and 100.°C occupies a volume of 30.6 L. When one mole of H2O(g) is condensed to one mole of H2O(l) at 1.00 atm and 100.°C, 40.66 kJ of heat is released. If the density of H2O(l) at this temperature and pressure is 0.996 g/cm3, calculate ΔE for the condensation of one mole of water at 1.00 atm and 100.°C.

To calculate the change in internal energy (ΔE) for the condensation of one mole of water at 1.00 atm and 100.°C, we can use the equation:

ΔE = q + PΔV

where ΔE is the change in internal energy, q is the heat transferred, P is the pressure, and ΔV is the change in volume.

Given information:
- The heat released during condensation, q = -40.66 kJ
- The pressure, P = 1.00 atm
- The initial volume of gas, V1 = 30.6 L
- The final volume of liquid, V2 = (density of liquid) / (molar mass of water)

First, let's find the final volume of the liquid state, V2:
- Density of water (at 100.°C and 1.00 atm) = 0.996 g/cm^3 = 0.996 g/mL
- Molar mass of water (H2O) = 18.015 g/mol

To convert the density from g/mL to L, divide by 1000:
Density of water = 0.996 g/mL / 1000 = 0.000996 g/mL = 0.000996 g/cm^3

Now, we can calculate V2:
V2 = (0.000996 g/cm^3) / (18.015 g/mol) = 0.0552 cm^3

Since 1 cm^3 is equivalent to 1 mL, we can convert the volume to liters:
V2 = 0.0552 cm^3 = 0.0552 mL = 0.0552 L

Using the values we have obtained, we can now calculate ΔE:
ΔE = q + PΔV
ΔE = -40.66 kJ + (1.00 atm)(30.6 L - 0.0552 L)

Remember to convert the pressure from atm to J/L:
1 atm = 101.325 J/L

ΔE = -40.66 kJ + (101.325 J/L)(30.6 L - 0.0552 L)
ΔE = -40.66 kJ + (101.325 J/L)(30.5448 L)
ΔE = -40.66 kJ + 3093.22 J
ΔE = -40,660 J + 3093.22 J
ΔE = -37,566.78 J or -37.56678 kJ

Therefore, the change in internal energy (ΔE) for the condensation of one mole of water at 1.00 atm and 100.°C is approximately -37.56678 kJ.

To calculate the change in energy (ΔE) for the condensation of one mole of water at 1.00 atm and 100.°C, we need to consider the changes in both volume and heat.

Step 1: Calculate the initial volume of 1 mole of H2O(g) at 1.00 atm and 100.°C.
Given:
Pressure (P) = 1.00 atm
Temperature (T) = 100.°C (we need to convert it to Kelvin scale)
Volume (V) = 30.6 L

Convert temperature to Kelvin:
T(K) = 100.°C + 273.15 = 373.15 K

Step 2: Calculate the final volume of 1 mole of H2O(l) at 1.00 atm and 100.°C.
Given:
Density of H2O(l) = 0.996 g/cm^3

We need to calculate the mass first:
Density = Mass / Volume
Mass = Density × Volume

Convert volume from L to cm^3:
Volume(cm^3) = 30.6 L × 1000 cm^3/L = 30600 cm^3

Mass = 0.996 g/cm^3 × 30600 cm^3 = 30477.6 g

Since 1 mole of H2O is equivalent to 18.015 g, we need to calculate the moles:
Moles = 30477.6 g / 18.015 g/mol ≈ 1695.39 mol

Step 3: Calculate the change in energy (ΔE) using the heat released.
Given:
Heat released = 40.66 kJ

Since 1 kJ = 1000 J, we need to convert kJ to J:
Heat released = 40.66 kJ × 1000 J/1 kJ ≈ 40660 J

ΔE = -q (change in energy = negative heat released)
ΔE = -40660 J

Therefore, the change in energy (ΔE) for the condensation of one mole of water at 1.00 atm and 100.°C is approximately -40660 J.