Question

A balloon blown up with $1$ mole of a gas, has a volume of $480$ ml at $5^{\circ }C$. If the balloon is filled to ${\frac{7}{8}}^{th}$ of its maximum capacity, then which of the following options is/are correct?

• A. The balloon will burst at ${30}^{\circ }C$
• B. The pressure of the gas inside the balloon at $5^{\circ }C$ is $47.5atm$
• C. The minimum temperature at which the balloon will burst is ${44.71}^{\circ }C$
• D. The pressure of gas when balloon burst at minimum temperature is $50atm$

Answer 1

Answer: (B), (C)

The correct options are
B The pressure of the gas inside the balloon at $5^{\circ }C$ is $47.5atm$
C The minimum temperature at which the balloon will burst is ${44.71}^{\circ }C$

Maximum capacity or volumes of balloon $=\frac{8}{7}\times 480=548.57ml$

Also, $V_1=480ml,T_1=278K,n=1mole$

$\left(A\right)$ The balloon will burst at a minimum temperature $\left(T_2\right)$ when volume becomes $548.57ml$.

Using Charle's law,

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

$\frac{480}{278}=\frac{548.57}{T_2}$

$\therefore T_2=317.71K$

or $T_2={44.71}^{\circ }C$Hence $\left(A\right)$ is incorrect but $\left(C\right)$ is correct.

$\left(B\right)$ Pressure of the gas at $5^{\circ }C$ having $1moleandV=480ml$

$PV=nRT$

$\therefore P\times \frac{480}{1000}=1\times 0.0821\times 278$

$\therefore P=47.5atm$

Hence $\left(B\right)$ is also correct.

$\left(D\right)$ Since $V$ increase in temperature from $5^{\circ }Cto{44.71}^{\circ }C$, at which balloon bursts and therefore pressure remains constant.

Thus the pressure of the gas inside the balloon when it bursts is $47.5atm$. Hence $\left(D\right)$ is incorrect.

Hence $(B,C)$ are correct.