Question

You lower the temperature of a sample of liquid carbon disulfide from 94.1°C until its volume contracts by 0.491% 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

We know that,

$V=V_{0}(1+\alpha \Delta T)$

Where is the final volume, $V_{0}$ is the original volume, $\alpha$ is the coefficient of volume expansion and $\Delta T$ = (Final Temperature - Initial Temperature)

The given data shows that,

$\frac{(V-V_{0})}{V_{0}}=-0.491/100$

The negative sign indicated contraction

$\frac{V}{V_{0}}=1- \frac{0.491}{100}$

$\frac{V}{V_{0}}=0.99509$

$1+\alpha \Delta T=0.99509$

$\alpha \Delta T=0.99509-1=-0.00491$

Final Temperature - Initial Temperature = $-0.00491/\alpha$

Initial Temperature = 94.1 °C

$\alpha$ = 1.15*10^-3 (°C^-1)

Final Temperature = $\frac{-0.00491}{0.00115}+94.1$ = 89.83 °C