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,

`(P_1V_1)/(P_2V_2)=(nRT_1)/(nRT_2)`

As, the pressure is same. Therefore,

`(V_1)/(V_2)=(T_1)/(T_2)`

Therefore, the new volume is given as,

`V_2=V_1(T_2)/(T_1)`

Substituting all known values,

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

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