Question

1gm of water at a pressure of 1.01×105 Pa is converted into steam without any change of temperature. The volume of 1 g of steam is 1671 cc and the latent heat of evaporation is 540cal. The change in internal energy due to evaporation of 1gm of water is

A. ≈167cal

B. ≈500cal
C. 540cal

D.581cal

Answer 1

dW = PΔV = 1.01 × 10⁵ [ 1671 − 1 ] × 10⁻⁶ joule 

= (1.01 × 167)/4.2 cal 

= 40 cal. nearly 

ΔQ = m L = 1 × 540 , 

Δ Q = Δ W + Δ U 

or Δ U = 540 − 40 

= 500 cal .

Comments

Thank You sir