When 1 g of water at 0°C and 1 × 10^5 N⁄m^2 pressure is converted into ice of volume 1. 091 cm^3, the external work done will be

Q: When 1 g of water at 0°C and 1 × 10⁵ N⁄m² pressure is converted into ice of volume 1. 091 cm³ , the external work done will be

(a) 0.0091 joule

(b) 0.0182 joule

(c) -0.0091 joule

(d) -0.0 joule

Sol:(a) Volume of water V_w = (1 g)/(1 g/cc) = 1 cm³

Volume of ice V_i = 1.091 cm³

∆W = P ∆ V = P (V_i – V_w)

= 10⁵ (1. 091 – 1) × 10^-6

= 0.0091 J