Question

A balloon blown up has a volume of 500 mL at $5^{o}C$. The balloon is distended to 7/8th of its maximum capacity.
(a) Will it burst at ${30}^{o}C$?
(b) Calculate the minimum temperature above which it will burst.

Answer 1

Maximum capacity of balloon $=\frac{8}{7}\times 480=548.57mL$

Let $T$ be the temperature at which the volume of balloon becomes $548.57mL$ and the balloon will burst.

Given:-

$V_1=480mL$

$V_2=548.57mL$

$T_1=5℃=(273+5)K=278K$

$T_2=T\left(\text{say}\right)=?$

Now, using Charle's Law,

$V\propto T$

$\Rightarrow \frac{V_1}{V_2}=\frac{T_1}{T_2}$

$\Rightarrow \frac{480}{548.57}=\frac{278}{T}$

$\Rightarrow T=\frac{278\times 548.57}{480}$

$\Rightarrow T=317.71K=44.71℃$

Hence the balloon will burst at $44.71℃$.

(a) The balloon will not burst at $30℃$.

(b) The minimum temperature above which it will burst is $44.71℃$.