Question

A helium balloon with a volume of 550mL is cooled from 305 to 265K. The pressure on the gas is reduced from 0.45 atm to 0.25 atm. What is the volume of the gas at the lower temperature and pressure

Answer 1

860 mL.

Explanation

Separate this process into two steps:

1. Cool the balloon from 305 K to 265 K.
2. Reduce the pressure on the balloon from 0.45 atm to 0.25 atm.

What would be the volume of the balloon after each step?

After Cooling the balloon at constant pressure:

By Charles's Law, the volume of a gas is directly related to its temperature in degrees Kelvins.

In other words,

$\dfrac{V_2}{V_1} = \dfrac{T_2}{T_1}$,

where

• $V_1$ and $V_2$ are volumes of the same gas.
• $T_1$ and $T_2$ are the temperatures (in degrees Kelvins) of that gas.

Rearranging,

$V_2 = V_1 \cdot \dfrac{T_2}{T_1}\\\phantom{V_2} = 550 \times \dfrac{265}{305}\\\phantom{V_2} = 478 \; \text{mL}$.

The balloon ended up with a lower temperature. As a result, its volume drops: $V_2 < V_1$.

After reducing the pressure on the balloon at constant temperature:

By Boyle's Law, the volume of a gas is inversely proportional to the pressure on this gas.

In other words,

$\dfrac{V_2}{V_1} = \dfrac{P_1}{P_2}$,

where

• $V_1$ and $V_2$ are volumes of the same gas.
• $P_1$ and $P_2$ are the pressures on this gas.

Rearranging,

$V_2 = V_1 \cdot \dfrac{P_1}{P_2}\\\phantom{V_2} = 478 \times \dfrac{0.45}{0.25}\\\phantom{V_2} = 860 \;\text{mL}$.

There's now less pressure on the balloon. As a result, the balloon will gain in volume: $V_2 > V_1$.

The final volume of the balloon will be $860 \; \text{mL}$.