# Question #c13ab

Oct 29, 2016

$\textsf{1.96 \textcolor{w h i t e}{x} \text{cm}}$

#### Explanation:

Because the pressure remains constant we can use:

$\textsf{{V}_{1} / {T}_{1} = {V}_{2} / {T}_{2}}$

The volume of a sphere is given by:

$\textsf{V = \frac{4}{3} \pi {r}^{3}}$

So the expression becomes:

$\textsf{\frac{\cancel{\frac{4}{3} \pi} {r}_{1}^{3}}{T} _ 1}$=$\textsf{\frac{\cancel{\frac{4}{3} \pi} {r}_{2}^{3}}{T} _ 2}$

$\therefore$$\textsf{{r}_{2}^{3} = {r}_{1}^{3} / {T}_{1} \times {T}_{2}}$

We need to convert to absolute temperature so:

$\textsf{{T}_{1} = 8.50 + 273 = 281.5 \textcolor{w h i t e}{x} \text{K}}$

$\textsf{{T}_{2} = 93.7 + 273 = 366.7 \textcolor{w h i t e}{x} \text{K}}$

Putting in the numbers:

$\textsf{{r}_{2}^{3} = {\left(1.80\right)}^{3} \times \frac{366.7}{281.5}}$

$\textsf{{r}_{2}^{3} = 7.579}$

$\therefore$$\textsf{r = \text{^3sqrt(7.579)=1.96color(white)(x)"cm}}$

This seems a reasonable answer as you would expect the radius to increase as the temperature is raised.