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

Step-by-step solution

Given, mass of water, m = lg Volume of 1 ~g of water =1 ~cm ^3=10⁻⁶ ~m ^3 Volume of 1 ~g of steam =1671 ~cm ^3=1671 10⁻⁶ ~m ^3 Pressure, p =1 10^5 ~Pa Latent heat of vaporization of water, L =2256 ~J / g Change in volume, V=(1671-1) 10⁻⁶ ~m ^3 =1670 10⁻⁶ ~m ^3 Heat supplied, Q = mL =1 2256=2256 ~J As the steam expands, so the work done in expansion is array rlr W & = p V =1 10^5 1670 10⁻⁶ [from Eq. (i)] & =167 ~J & .(iii) array According to first law of thermodynamics, array rlrl Q & = U + W U = Q - W & =2256-167 &