Question

1 g of water of volume 1 cm^3 at 100° c is converted into steam at same temperature under normal pressure (= 1 × 10^5 pa) . The volume of steam formed equals 1671 cm^3. If specific latent heat of vaporisation of water is 2256 J/g. the change in internal energy is



a) 2256 J
b)2423 J
c)2089 J
d)167 J​

Answer 1

Answer:

The answer is 2086.829 J

Explanation:

According to the problem the amount of water is converted into steam at the presence of temperature of 100° c.

Now mass of the water, m = 1 g = 10^(-3) kg

Now it is given that the latent of heat , l = 2256J/g = 2256 x 10^3 J/kg

It is also given that the atmospheric pressure , p = 1 x 10^5 pa

Now the volume of steam , vs = 1671 cm^3 = 1671 x 10^(-6) m^3

Now we know that the volume of water is, vw 1 cm^3 = 1 x 10^(-6) m^3

Now according to the first law of thermodynamics,

dq = workdone +internal energy

dq = dv +pdv

=> m x l = dv + p(vs - vw )

=> dv = ml - p(vs - vw)

          =  10^(-3)  x 2256 x 10^3 - 1 x 10^5(1671-1) x 10^(-6)

            = 2086.829 J

Answer 2

Answer:

ans option (3) is correct