Question

You lower the temperature of a sample of liquid carbon disulfide from 87.7°C until its volume contracts by 0.585% of its initial value. What is the final temperature of the substance? The coefficient of volume expansion for carbon disulfide is 1.15 × 10-3 (C°)–1.

Answer 1

Let initial volume be V₀

Given,

final volume, V_f = V₀ - 0.585%*V₀ = 0.99415*V₀

Initial temperature, T_i = 87.7 C

Let final temperature be T_f

Coefficient of volume expansion, = 1.15 * 10^-3C

Now,

as we know

V_f = V₀(1 + (T) )

=> 0.99415*V₀ = V₀(1 + ( 1.15 * 10^-3)*(T_f - 87.7) )

=> 0.99415 = 1 + ( 1.15 * 10^-3)*(T_f - 87.7)

=> ( 1.15 * 10^-3)*(T_f - 87.7) = 0.99415 - 1 = -0.00585

=> (T_f - 87.7) = -0.00585 / ( 1.15 * 10^-3)

=> (T_f - 87.7) = -5.086

=> T_f = 87.7 - 5.086 = 82.614

Hence final temperature is 82.614 C