Question

If $2.01\mathrm{g}$ of helium gas occupies a volume of $12.0\mathrm{L}$ at ${25}^{\circ }\mathrm{C},$ what volume will $6.52\mathrm{g}$ of helium gas occupy under the same conditions?

Short answer

Under the same conditions, 6.52 g of helium gas will occupy a volume of approximately 38.9 L. This is determined by using the direct proportionality of mass and volume: $V_2=V_1\frac{m_2}{m_1}$, where $V_1$ = 12.0 L, $m_1$ = 2.01 g, and $m_2$ = 6.52 g.

Step-by-step solution

Understanding the relationship between gas mass and volume

If the pressure and temperature remain constant, the amount of gas and its volume are directly proportional. This means that if we increase the amount of gas (mass), the volume will also increase proportionally.

Set up the proportionality equation

Since mass and volume are directly proportional under constant pressure and temperature, we can set up a proportionality equation: $$\frac{m_1}{V_1}=\frac{m_2}{V_2}$$ where $m_1$ = 2.01 g, $V_1$ = 12.0 L, and $m_2$ = 6.52 g. We need to find $V_2$.

Rearrange the equation and solve for the new volume

Rearrange the proportionality equation to solve for $V_2$: $$V_2=V_1\frac{m_2}{m_1}$$ Now, substitute the given values and calculate $V_2$: $$V_2=12.0\mathrm{L}\times \frac{6.52\mathrm{g}}{2.01\mathrm{g}}$$

Calculate the new volume

Perform the calculation: $$V_2=12.0\mathrm{L}\times \frac{6.52\mathrm{g}}{2.01\mathrm{g}}=12.0\mathrm{L}\times 3.243\approx 38.9\mathrm{L}$$ So, 6.52 g of helium gas will occupy a volume of approximately 38.9 L under the same conditions.

Key concepts

Helium Gas

Helium is an inert gas and belongs to the group of noble gases in the periodic table. Its representation is simply 'He', and it is characterized by being both colorless and odorless. Helium is the second lightest element in the universe and has remarkable properties that make it unique, such as very low boiling and melting points.
This lightness allows helium balloons to float in air. The low density of helium compared to air gives it the ability to rise. Helium is used for various purposes, including breathing mixtures for deep sea diving, inflating airships, and in cryogenics due to its low reactivity and non-flammable nature. In the context of gas laws, helium often demonstrates predictable behavior under constant temperatures and pressures, making it a useful subject for understanding gas proportionality.

Proportionality in Gases

When discussing gases, particularly under the gas laws, an essential concept to grasp is proportionality. At constant temperature and pressure, the volume occupied by a gas is directly proportional to the amount (mass or moles) of the gas. This relationship is supported by Avogadro’s Law, which states that equal volumes of gases at the same temperature and pressure contain equal numbers of molecules.
So, if we increase the mass of a gas while keeping temperature and pressure unchanged, its volume will also increase proportionally. This is precisely what is happening when you calculate how much space a larger mass of helium will take up under constant conditions, as seen in the original exercise.

• Direct proportionality means doubling the mass results in doubling the volume.
• The foundation of this relationship allows for equations like: $$\frac{m_1}{V_1}=\frac{m_2}{V_2}$$

Understanding proportionality helps comprehend more intricate gas behaviors and predict outcomes under specified conditions.

Volume Calculation at Constant Conditions

Calculating volume at constant conditions involves using the gas law that deals with proportionality between the mass and volume of the gas, as explained earlier. Given certain values of mass and volume at a fixed temperature and pressure, we can use a simple mathematical relationship to determine the unknown volume for a different mass.

First, identify the initial and final mass and volume of the gas involved in the problem. In the original exercise, the initial mass ($m_1$) is 2.01 g, and the initial volume ($V_1$) is 12.0 L.

• The proportionality equation used is:$$V_2=V_1\frac{m_2}{m_1}$$
• Substitute the known values into the equation for calculating the final volume ($V_2$):$$V_2=12.0\mathrm{L}\times \frac{6.52\mathrm{g}}{2.01\mathrm{g}}$$
• Calculate:$$V_2=12.0\mathrm{L}\times 3.243\approx 38.9\mathrm{L}$$

This simple division and multiplication allow us to find the volume that the increased mass will occupy, given that temperature and pressure remained the same.