A balloon blown up has a volume of 500 mL at `5^@C`. The balloon is distended to 7/8th of its maximum capacity.
Will it burst at `30^@C` ?

Explanation

`480 mL` represents `7//8` of the balloon's maximum capacity. Therefore, maximum volume capacity of the balloon is
`480xx(8)/(7)=548.6 mL`
Thus, if volume exceeds `548.6 mL`m the balloon will burst.
(`a`) Calculation of the volume of balloon at `30^(@)C`:
By Charles's law, we get
`(V_(1))/(T_(1))=(V_(2))/(T_(2))`
Here, `V_(1)=480 mL`, `V_(2)=? T_(1)=278 K`, `T_(2)=303 K`
Thus, `V_(2)=(V_(1)T_(2))/(T_(1))=(480xx303)/(278)=523.16mL`
Therefore, the balloon will not burst at `30^(@)C`.
(`b`) Calculation of the temperature at which balloon will burst:
Here, `V_(1)=480 mL`, `T_(2)=278 K`, `V_(2)=548.6 mL`, `T_(2)=?`
`(V_(1))/(T_(1))=(V_(2))/(T_(2))`
`:. T_(2)=(V_(2)xxT_(1))/(V_(1))=(548.6xx278)/(480)=317.7 K`
`=317.7-273=44.7C^(@)`
Thus the ballon will burst as soon as temperature exceeds `44.7^(@)C`.