Question

a 3-liter balloon containing 4 grams of helium at room temperature 25 Celsius is placed in a freezer at -3 Celsius calculate the new volume assuming that the pressure inside the freezer is the same as outside the freezer

Answer 1

Given:

Initial volume:

$v_1=3\text{ L}$

Initial temperature:

$\begin{gathered} T_1=25^{\circ}C \\ =25+273 \\ =298\text{ K} \end{gathered}$

Final temperature:

$\begin{gathered} T_2=-3^{\circ}C \\ =-3+273 \\ =270\text{ K} \end{gathered}$

Mass of Helium gas:

$m=4\text{ g}$

Therefore, the number of moles of Helium gas is given as,

$\begin{gathered} n=\frac{mass\text{ of helium gas}}{Molecular\text{ mass of helium gas}} \\ =\frac{4\text{ g}}{4\text{ g}} \\ =1\text{ mole} \end{gathered}$

The ideal gas equation is given as,

$PV=\text{nRT}$

Here, P is the pressure and R is the universal gas constant.

Therefore,

$\frac{P_1V_1}{P_2V_2}=\frac{nRT_1}{nRT_2}$

As, the pressure is same. Therefore,

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

Therefore, the new volume is given as,

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

Substituting all known values,

$\begin{gathered} V_2=(3L)\times\frac{270\text{ K}}{298\text{ K}} \\ =2.718\text{ L} \end{gathered}$

Therefore, the new volume of the gas is 2.718 L.