Question

A weather balloon is filled with helium that occupies a volume of 5.57 104 L at 0.995 atm and 32.0°C. After it is released, it rises to a location where the pressure is 0.720 atm and the temperature is -14.5°C. What is the volume of the balloon at that new location?

Answer 1

6.52 × 10⁴ L. (3 sig. fig.)

Explanation

Helium is a noble gas. The interaction between two helium molecules is rather weak, which makes the gas rather "ideal."

Consider the ideal gas law:

$P\cdot V = n\cdot R\cdot T$,

where

• $P$ is the pressure of the gas,
• $V$ is the volume of the gas,
• $n$ is the number of gas particles in the gas,
• $R$ is the ideal gas constant, and
• $T$ is the absolute temperature of the gas in degrees Kelvins.

The question is asking for the final volume $V$ of the gas. Rearrange the ideal gas equation for volume:

$V = \dfrac{n \cdot R \cdot T}{P}$.

Both the temperature of the gas, $T$, and the pressure on the gas changed in this process. To find the new volume of the gas, change one variable at a time.

Start with the absolute temperature of the gas:

• $T_0 = (32.0 + 273.15) \;\text{K} = 305.15\;\text{K}$,
• $T_1 = (-14.5 + 273.15) \;\text{K} = 258.65\;\text{K}$.

The volume of the gas is proportional to its temperature if both $n$ and $P$ stay constant.

• $n$ won't change unless the balloon leaks, and
• consider $P$ to be constant, for calculations that include $T$.

$V_1 = V_0 \cdot \dfrac{T_1}{T_2} = 5.57\times 10^{4}\;\text{L}\times \dfrac{258.65\;\textbf{K}}{305.15\;\textbf{K}} = 4.72122\times 10^{4}\;\text{L}$.

Now, keep the temperature at $T_1 =258.65\;\text{K}$ and change the pressure on the gas:

• $P_1 = 0.995\;\text{atm}$,
• $P_2 = 0.720\;\text{atm}$.

The volume of the gas is proportional to the reciprocal of its absolute temperature $\dfrac{1}{T}$ if both $n$ and $T$ stays constant. In other words,

$V_2 = V_1 \cdot\dfrac{\dfrac{1}{P_2}}{\dfrac{1}{P_1}} = V_1\cdot\dfrac{P_1}{P_2} = 4.72122\times 10^{4}\;\text{L}\times\dfrac{0.995\;\text{atm}}{0.720\;\text{atm}}=6.52\times 10^{4}\;\text{L}$

(3 sig. fig. as in the question.).

See if you get the same result if you hold $T$ constant, change $P$, and then move on to change $T$.