Question

On the surface of the earth at $1\text{atm}$ pressure, a balloon filled with $H_2$ gas occupies $500\text{mL}$, the volume being $5/6$ of its maximum capacity. The balloon is left in air. It starts rising. The height above which the balloon will burst if temperature of the atmosphere remains constant and the pressure decreases by $1\text{mm}$ of Hg for every $100\text{cm}$ rise of height is?

• A. $120\text{m}$
• B. $136.67\text{m}$
• C. $126.67\text{m}$
• D. $100\text{m}$

Answer 1

The correct option is C $126.67\text{m}$

According to Boyle's Law:
$P_1{V}_1=P_2{V}_2$
$P_1=1atm=760\text{mm},V_1=500\text{mL}$
$P_2=?,V_2=?$
Now, let $x$ be maximum volume
$500=\frac{5}{6}x,x=600\text{mL}$
Since balloon will burst at $600\text{mL}(\because V_{max})$
$V_2=600\text{mL}$
Now, as Boyle's law: $760\times 500=P_2\times 600$
$P_2=633.33\text{mm}$
Pressure decreased
$=760-633.33=126.67\text{mm}$
Now, $1\text{mm dec.}\to 100\text{cm}$
$126.67\text{mm dec.}\to 100\times 126.67=12667\text{cm}=126.67\text{m}$