**Calculation of work (W):**

The work done during the conversion of water into steam can be calculated using the equation:

W = P∆V

Where:
W = work done
P = pressure
∆V = change in volume

In this case, we are given that the pressure is 1 atm. To calculate the change in volume, we need to consider the difference in volume between 1 mole of water and 1 mole of steam.

The molar volume of water at 373K and 1 atm pressure is approximately 18 cm³/mol. The molar volume of steam at the same temperature and pressure is approximately 30.5 cm³/mol.

Therefore, the change in volume (∆V) is:

∆V = (30.5 - 18) cm³/mol = 12.5 cm³/mol

Now we can calculate the work done (W):

W = (1 atm) * (12.5 cm³/mol) = 12.5 atm·cm³/mol

**Calculation of ∆E (Change in Energy):**

The change in energy (∆E) during the conversion of water into steam can be calculated using the equation:

∆E = q + W

Where:
∆E = change in energy
q = heat transferred
W = work done

In this case, we are given that the latent heat of vaporization of water is 9720 cal/mol. The heat transferred (q) can be calculated by multiplying the latent heat by the number of moles of water:

q = (9720 cal/mol) * (1 mol) = 9720 cal

Now we can calculate the change in energy (∆E):

∆E = (9720 cal) + (12.5 atm·cm³/mol)

Since atm·cm³ is not a standard unit for energy, we need to convert it to cal. 1 atm·cm³ is equivalent to 101.325 J. And 1 J is equivalent to 0.24 cal.

Therefore:

∆E = (9720 cal) + (12.5 atm·cm³/mol) * (101.325 J) * (0.24 cal/J)

∆E = (9720 cal) + (303.66 cal)

∆E ≈ 10023.66 cal

Therefore, the change in energy (∆E) during the conversion of 1 mole of water into 1 mole of steam at a temperature of 373K and a pressure of 1 atm is approximately 10023.66 cal.

Q= W+ ∆E

∆E =0 because during phase change of liquid into vapour temp. remains constant.
so Q=W
Q=m×L mass of one mole water will be equal to it's molecular wt. that is 18g.

also L=9720 cal/mol
1 mol=18g
for 18g L is 9720 so for 1g it would be 9720/18=540 cal/g
so now Q=18×540=9720 cal.