Question

The gas inside a balloon will always have a pressure nearly equal to atmospheric pressure since that is the pressure applied to the outside of the balloon. You fill a balloon with helium (a nearly ideal gas) to a volume of 0.600 L at 19.0°C. What is the volume of the balloon if you cool it to the boiling point of liquid nitrogen (77.3 K)?

Short answer

The volume of the balloon is $0.159\times {10}^{-3}{\text{\hspace{0.33em}m}}^{\text{3}}$if it gets cooled to the boiling point of liquid nitrogen.

Step-by-step solution

Ideal gas equation and given data

Given data:

$\begin{array}{c}{V}_1=0.6\times {10}^{-3}{\text{\hspace{0.33em}m}}^{\text{3}}\\ T_1=292\text{\hspace{0.33em}K}\end{array}$

The ideal gas equation is

$\mathit{p}\mathit{V}\mathbf{=}\mathit{n}\mathit{R}\mathit{T}$

Here, pressure is constant. So,

role="math" localid="1655807549892" $\mathit{T}\mathbf{\propto }\mathit{V}$

andas such, temperature depends on volume. As temperature decreases, volume decreases too.

Therefore, the equation is

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

Calculating the volume of the balloon

The boiling point of liquid nitrogen is $T_2=77.3\text{\hspace{0.33em}K}$.

The volume of the balloon can be calculated as

$\begin{array}{c}{V}_2=\frac{V_1\times T_2}{T_1}\\ =\frac{0.6\times {10}^{-3}\times 77.3}{292}\\ =0.159\times {10}^{-3}{\text{\hspace{0.33em}m}}^{\text{3}}\end{array}$

Hence, the volume of the balloon is $0.159\times {10}^{-3}{\text{\hspace{0.33em}m}}^{\text{3}}$if it gets cooled to the boiling point of liquid nitrogen.