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.
What is the pressure (in atmospheres) of the gas inside the container connected to an open-end, mercury-filled manometer as shown in the picture? The atmospheric pressure is 0.95 \rm atm.
1.19 atm
deltaE=-37.6kJ w=3.10
Can someone show a step by step solution?
bcjs
To calculate ΔE (the change in energy) for the condensation of one mole of water at 1.00 atm and 100°C, we can use the equation:
ΔE = q + w
Where:
- ΔE is the change in energy
- q is the heat transferred
- w is the work done
Given that 40.66 kJ of heat is released during the condensation, we can substitute q = -40.66 kJ.
Now, let's determine the work done:
w = -PΔV
Where:
- P is the pressure
- ΔV is the change in volume
Since the water changes from gas to liquid, the change in volume is equal to the volume of gas minus the volume of liquid.
ΔV = Vgas - Vliquid
Given that the volume of gas is 30.6 L and the density of liquid water is 0.996 g/cm³, we can find the volume of the liquid water:
Vliquid = (mass of liquid water) / (density of liquid water)
Since we have one mole of water, the mass of liquid water is equal to the molar mass of water.
mass of water = 1 mol × molar mass of water
The molar mass of water is approximately 18 g/mol.
mass of water = 1 mol × 18 g/mol = 18 g
Now we can calculate the volume of liquid water using the given density:
Vliquid = (18 g) / (0.996 g/cm³)
Since the density is given in g/cm³, we need to convert the volume to liters by dividing by 1000 cm³/L:
Vliquid = (18 g) / (0.996 g/cm³) / (1000 cm³/L) = 18.1 L
Now we can substitute the values into the equation for ΔV:
ΔV = Vgas - Vliquid = 30.6 L - 18.1 L = 12.5 L
Finally, we can calculate the work done:
w = -PΔV = -(1.00 atm) × (12.5 L)
Since 1 atm is equivalent to 101.3 J/L, we can convert the units:
w = -(1.00 atm) × (12.5 L) × (101.3 J/L) / (1 atm) = -1.28 kJ
Now we can substitute the values into the equation for ΔE:
ΔE = q + w = -40.66 kJ + (-1.28 kJ) = -41.94 kJ
Therefore, the change in energy (ΔE) for the condensation of one mole of water at 1.00 atm and 100°C is approximately -41.94 kJ.