1 g of water, of volume 1 cm3 at 100°C, is converted into steam at same temperature under normal atmospheric pressure (≃1 × 105 Pa). The volume of steam formed equals 1671 cm3. If the specific latent heat of vaporisation of water is 2256 J/g, the change in internal energy is,

CONCEPT:

According to the first law of thermodynamics we have;

Q = ΔU + W

Here we have Q is heated, ΔU is the change in internal energy and W is the work done.

CALCULATION:

Given: mass, m = 1 g

Latent heat of vaporization, L = 2256 J/g

Pressure, P = 1 × 105 Pa

The volume, \((V_2-V_1)\) = 1671 cm³

As we know that

Q = ΔU + W   ----(1)

The work done, W = PdV

and heat Q = mL

Now, on putting the values in equation (1) we have;

mL = ΔU + P(V2 - V1)   ----(2)

Now, on putting all the given values in (2) we have;

\(1 \times 2256 = Δ U +1 \times 10^5 (1670 \times 10^{-6})\)

⇒ΔU = 2089 J

Hence, option 4) is the correct answer.