Question

A balloon filled with 2.00 L of helium initially at 1.15 atm of pressure rises into the atmosphere. When the surrounding pressure reaches 500. mmHg , the balloon will burst. If 1 atm = 760. mmHg , what volume will the balloon occupy in the instant before it bursts?

Answer 1

Answer: The balloon will occupy volume of 3.53 L before getting burst

Explanation:

Initial Volume of the balloon=$V_1=2.00 L$

Initial pressure of the atmosphere=$P_1=1.15 atm$

Final pressure of the atmosphere =$P_2=500 mmHg=0.65 atm$ (1atm =760 mmHg)

Final Volume of the ballon =$V_2$

According Boyle's Law:

$Pressure\propto \frac{1}{Volume}$ (At constant Temperature in Kelvins)

$P_1V_1=P_2V_2$

$V_2=\frac{P_1V_1}{P_2}=\frac{1.15 atm\times 2.00 L}{0.65 atm}=3.53 L$

The balloon will occupy volume of 3.53 L before getting burst.

Answer 2

Answer:

The correct answer is 3.53 volume

Explanation:

The equation used in the question is boyle's equation.

$P_{1} V_{1}= P_{2}V_{2}$.

The data given is as follows:-

$P_{1}$ is 1.15atm, $P_{2}$ is 0.65atm, and $V_{1}$ is 2L

Therefore the $V_{2}$ is

$V_{2} =\frac{1.15 X 2.00}{0.65}$

$V_{2}$ is 3.53L

Hence, 3.53 volume is required to burst the ballon.

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