Question

A balloon contains 30.0 L of helium gas at 1.00 atm. What is the volume of the helium when the balloon rises to an altitude where the pressure is only 0.250 atm? Assume that the temperature remains constant.

Answer 1

Answer:

The volume of that ballon would be approximately $1.20 \times 10^2\; \rm L$ at a pressure of $0.250\; \rm atm$.

Explanation:

Let $V_1$ and $P_1$ be the volume and pressure of the gas before the change.

By Boyle's Law, the volume of an ideal gas is inversely proportional to its pressure. In other words, if $V_2$ and $P_2$ represent the volume after the change, then $V_1 \cdot P_1 = V_2 \cdot P_2$.

In this case,

• $V_1 = 30.0\; \rm L$.
• $P_1 = \rm 1.00\; \rm atm$.
• $P_2 = 0.250\; \rm atm$.

The value of $V_2$ needs to be found. Rearrange the equation $V_1 \cdot P_1 = V_2 \cdot P_2$:

$\begin{aligned}& V_1\cdot P_1 = V_2 \cdot P_2 \\ & \implies V_2 = \frac{V_1 \cdot P_1}{P_2} = \frac{P_1}{P_2} \cdot V_1\end{aligned}$.

Therefore,

$\begin{aligned}\displaystyle V_2 &= \frac{P_1}{P_2}\cdot V_1 \\ &= \frac{1.00\; \rm atm}{0.250\; \rm atm} \times 30.0\; \rm L = 1.20 \times 10^2 \; \rm L\end{aligned}$.