Question

You lower the temperature of a sample of liquid carbon disulfide from 90.3 ∘ C until its volume contracts by 0.507 % 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

Therefore final temperature = 85.89 °C

Step-by-step explanation:

Coefficient of volume expansion: This is defined as an increase in volume, per unit volume per degree rise in temperature. The SI unit is 1/k. mathematically,

γ = ΔV/(V₁ΔT)......................... equation 1

Making ΔT the subject of formula in equation 1

ΔT = ΔV/(V₁γ)......................... equation 2

Where γ = coefficient of volume expansion, ΔV = increase in volume, ΔT = change in temperature, V₁ = Initial volume.

Where γ = 1.15 × 10⁻³ C⁻¹, V₁ = X ΔV = 0.00507X

Substituting this values into equation 2,

ΔT = 0.00507X/(X × 1.15 × 10⁻³ )

ΔT = 0.00507/0.00115

ΔT = 4.41 °C.

For contraction,

ΔT = T₁ - T₂

∴ T₂ = T₁ - ΔT

Where T₁ = 90.3 °C

T₂ = 90.3 - 4.41 = 85.89 °C

T₂ = 85.89 °C

Therefore final temperature = 85.89 °C

Answer 2

`T_(f)` = 85.89 ° C

Step-by-step explanation:

The linear thermal expansion process is given by

ΔL = L α ΔT

For the three-dimensional case, the expression takes the form

ΔV = V β ΔT

Let's apply this equation to our case

ΔV / V = ​​-0.507% = -0.507 10-2

ΔT = (ΔV / V) 1 /β

ΔT = -0.507 10⁻² 1 / 1.15 10⁻³

ΔT = -4.409

`T_(f)` –T₀ = 4,409

`T_(f)` = T₀ - 4,409

`T_(f)` = 90.3-4409

`T_(f)` = 85.89 ° C