1gram of liquid water is vapourised at 1atm pressure and 373k ,calculate Q,W,∆U∆H,assume water vapour behave as ideal gas ,latent heat of vapourisation of water vapour is 540cal per gram?

To tackle the problem of vaporizing 1 gram of liquid water at 1 atm pressure and 373 K, we need to calculate several thermodynamic quantities: the heat absorbed (Q), work done (W), change in internal energy (∆U), and change in enthalpy (∆H). Given that the latent heat of vaporization of water is 540 cal/g, we can use this information to derive our answers.

Calculating Heat Absorbed (Q)

The heat absorbed during the phase change from liquid to vapor can be calculated using the formula:

Q = m * L

Where:

• m = mass of water (1 g)
• L = latent heat of vaporization (540 cal/g)

Substituting the values:

Q = 1 g * 540 cal/g = 540 cal

Work Done (W)

When vaporizing water, work is done against the atmospheric pressure. The work done can be calculated using the formula:

W = -P * ∆V

Where:

• P = pressure (1 atm = 101.3 J)
• ∆V = change in volume

To find ∆V, we can use the ideal gas law. The volume of 1 gram of water vapor can be calculated as:

V = nRT/P

First, we need to find the number of moles (n) of water:

n = m/M

Where:

• M = molar mass of water (approximately 18 g/mol)

Thus, for 1 g of water:

n = 1 g / 18 g/mol ≈ 0.0556 mol

Now, substituting into the volume equation:

V = (0.0556 mol) * (8.314 J/(mol·K)) * (373 K) / (101.3 J) ≈ 0.0132 m³

Now, we can find ∆V. The initial volume of 1 g of liquid water is negligible compared to the volume of vapor, so:

∆V ≈ 0.0132 m³

Now substituting into the work equation:

W = - (101.3 J) * (0.0132 m³) ≈ -1.34 J

Change in Internal Energy (∆U)

Using the first law of thermodynamics, we can relate the heat absorbed, work done, and change in internal energy:

∆U = Q + W

Substituting the values we calculated:

∆U = 540 cal + (-1.34 J)

To convert calories to joules (1 cal = 4.184 J):

540 cal = 540 * 4.184 J ≈ 2260.56 J

Now, substituting into the equation:

∆U = 2260.56 J - 1.34 J ≈ 2260.22 J

Change in Enthalpy (∆H)

The change in enthalpy during a phase change can be calculated using:

∆H = Q

Since the heat absorbed is equal to the change in enthalpy during the phase transition:

∆H = 540 cal ≈ 2260.56 J

Summary of Results

• Q (Heat absorbed): 540 cal (2260.56 J)
• W (Work done): -1.34 J
• ∆U (Change in internal energy): 2260.22 J
• ∆H (Change in enthalpy): 540 cal (2260.56 J)

In summary, we have calculated the thermodynamic properties associated with the vaporization of 1 gram of water at 373 K and 1 atm. Each step illustrates the relationships between heat, work, internal energy, and enthalpy during this phase transition.