Question

If \(10.0 \mathrm{~g}\) of liquid helium at \(1.7 \mathrm{~K}\) is completely vaporized, what volume does the helium occupy at STP?

Short answer

The volume of helium when \(10.0 \mathrm{~g}\) of liquid helium is completely vaporized at STP is approximately \(55.06 \mathrm{~Liters}\).

Step-by-step solution

Find the number of moles of helium

First, we need to find the number of moles (n) of helium, which can be calculated using the following formula: n = mass / molar mass Given that the mass of helium is 10.0 g, we need to know the molar mass of helium to calculate the number of moles. Helium has a molar mass of 4.00 g/mol. Therefore, the number of moles of helium can be calculated as follows: n = (10.0 g) / (4.00 g/mol) = 2.50 mol

Use the ideal gas law to find the volume at STP

The ideal gas law states that PV=nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature. We are given that this problem is at STP, so we know the temperature is 273.15 K, and the pressure is 100 kPa. We also have the number of moles from step 1. To find the volume, we can rearrange the ideal gas law formula to: V = nRT/P Now, we need to know the value of the ideal gas constant R. Since we are using pressure in kPa, we need to use the value of R = 8.314 J/(mol·K). However, this unit is not in volume units like liters I will use R = 0.0821 L·atm/(mol·K) and convert the pressure given in kPa to atm. Converting the pressure to atm: 1 atm = 101.325 kPa P = (100 kPa) x (1 atm/101.325 kPa) = 0.9869 atm Now, let's plug the values into our formula: V = (2.50 mol) x (0.0821 L·atm/(mol·K)) x (273.15 K) / (0.9869 atm) V = 55.06 L

State the final answer

The volume of helium when 10.0 g of liquid helium is completely vaporized at STP is approximately 55.06 Liters.

Key concepts

Molar Mass

Molar mass is an essential concept in chemistry that represents the mass of one mole of a substance. It is expressed in grams per mole (g/mol). Understanding molar mass is crucial for converting between mass and moles, which is an important step in solving many chemical problems.

For helium, the molar mass is 4.00 g/mol. This means that one mole of helium atoms weighs 4 grams. To determine the number of moles in any given mass of helium, you simply divide the mass by its molar mass. For example, if you have 10 grams of helium, you calculate the number of moles as follows:

\[ n = \frac{\text{mass}}{\text{molar mass}} = \frac{10.0 \text{ g}}{4.00 \text{ g/mol}} = 2.50 \text{ mol} \]

This conversion is crucial for further calculations, such as using the ideal gas law to find the volume a gas occupies at certain conditions.

Standard Temperature and Pressure (STP)

Standard Temperature and Pressure, or STP, is a set of conditions often used in gas calculations to allow for a standard basis of comparison. STP is defined as a temperature of 273.15 Kelvin (0 degrees Celsius) and a pressure of 1 atmosphere (atm), which is equivalent to 101.325 kilopascals (kPa).

These conditions simplify calculations in chemistry and physics because they provide a common reference point. At STP, the behavior of gases can be predicted using the ideal gas law.

Using STP conditions, we know exactly what to expect regarding the volume a given amount of gas should occupy. For example, an ideal gas at STP occupies 22.4 liters per mole. However, calculations often involve converting the measured or given quantities into these standard conditions for accurate results, such as converting pressure from kPa to atm, which was necessary in the helium problem given.

Gas Volume Calculation

When dealing with gases, one common problem is determining the volume a certain amount of gas will occupy under specific conditions. This is where the ideal gas law comes into play. The law is mathematically represented as \(PV = nRT\), where:

• \(P\) is the pressure
• \(V\) is the volume
• \(n\) is the number of moles of gas
• \(R\) is the ideal gas constant
• \(T\) is the temperature in Kelvin

The goal is often to solve for \(V\) (volume), which can be rearranged from the original equation:

\[V = \frac{nRT}{P}\]

For example, in the provided exercise with helium gas at STP, we calculated volume \(V\) using the given number of moles and conditions. We used the value \(R = 0.0821 \text{ L atm/(mol K)}\) consistent with STP. Converting pressure (as needed) and plugging in all known values gives an accurate volume the gas will occupy under those conditions, which was found to be approximately 55.06 liters for the helium in the exercise.