deremine (a) the work done and (b) the change in internal energy of 1.00 kg of water when it is all boiled to steam at 100c. Assume a constant pressure of 1.00 atm and heat of vaporisation = 2260000 J.

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To determine the work done and the change in internal energy of water when it is boiled to steam at 100°C, we can use the first law of thermodynamics, which states that the change in internal energy (ΔU) of a system is equal to the heat (Q) added to the system minus the work (W) done by the system:

ΔU = Q - W

(a) To find the work done (W), we need to consider the expansion of the water as it is boiled to steam. Since the pressure is constant at 1.00 atm, we can use the formula:

W = PΔV

Here, P is the pressure, and ΔV is the change in volume. In this case, we know that the water is boiling, which means it is transforming from a liquid to a gas (steam). Since gases occupy more volume than liquids, there is an expansion of the water during boiling.

To calculate the change in volume (ΔV), 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 in Kelvin. Since we are given the pressure (1.00 atm) and know that the boiling point is 100°C, we can convert this to Kelvin as follows:

T = 100 + 273.15

Given that we have 1.00 kg of water, we can calculate the number of moles using the molar mass of water (18.015 g/mol):

n = mass / molar mass
= 1000 g / 18.015 g/mol
= 55.49 mol

Now, we can rearrange the ideal gas law equation to solve for volume:

V = (nRT) / P

Substituting the known values:

V = (55.49 mol)(0.0821 L·atm/(mol·K))(373.15 K) / 1.00 atm
= 1791.93 L

The change in volume (ΔV) during the boiling process is equal to the final volume (V) minus the initial volume (V initial), which is the volume of 1.00 kg of water:

ΔV = V - V initial
= 1791.93 L - 1.00 L
= 1790.93 L

Now that we have the change in volume (ΔV) and the pressure (P), we can calculate the work done (W):

W = PΔV
= (1.00 atm)(1790.93 L)
= 1790.93 atm·L

(b) To find the change in internal energy (ΔU), we can now use the first law of thermodynamics equation:

ΔU = Q - W

In this case, we know that we are adding heat to the system to convert water to steam. The heat of vaporization (Q) is given as 2,260,000 J. Therefore:

ΔU = Q - W
= 2,260,000 J - 1790.93 atm·L

Note: We need to convert atm·L to J to have consistent units. Since 1 atm·L is equal to 101.325 J, we can convert the work done:

W = 1790.93 atm·L × 101.325 J/atm·L
= 181,632.50 J

Substituting the values:

ΔU = 2,260,000 J - 181,632.50 J
= 2,078,367.50 J

Therefore, the work done (W) is equal to 181,632.50 J, and the change in internal energy (ΔU) is equal to 2,078,367.50 J.