A mixture of helium and neon gases at 317 K contains 3 times the number of helium atoms as neon atoms. The concentration (𝑛/𝑉) of the mixture is found to be 0.150 mol/L. Assuming ideal behavior, what is the partial pressure of helium?

Question

A mixture of helium and neon gases at 317 K contains 3 times the number of helium atoms as neon atoms. The concentration (𝑛/𝑉) of the mixture is found to be 0.150 mol/L. Assuming ideal behavior, what is the partial pressure of helium?

Answer 1

Calculate the total pressure of the 1 L as

PV = nRT
V = 1 L, n = 0.150, R = 0.08205, T = 317 K
Ptotal = 3.9 atm
P_He = X_He x P_total where X_He = 3/4 or 0.75
P_He = 0.75 x 3.9 atm = ?

Answer 2

Well, you've got a helium and neon party going on in that mixture! Let's break it down step by step.

First, let's assign some variables. Let 𝑛_𝐻 be the number of moles of helium and 𝑛_𝑁 be the number of moles of neon. Since the number of helium atoms is 3 times the number of neon atoms, we can say that 𝑛_𝐻 = 3𝑛_𝑁.

The total number of moles in the mixture is 𝑛_𝑡𝑜𝑡𝑎𝑙 = 𝑛_𝐻 + 𝑛_𝑁.

Now, the concentration (𝑛/𝑉) of the mixture is given as 0.150 mol/L, which means 𝑛_𝑡𝑜𝑡𝑎𝑙/𝑉 = 0.150. Let's assume a volume of 1 L for simplicity, so 𝑛_𝑡𝑜𝑡𝑎𝑙 = 0.150 mol.

Substituting 𝑛_𝐻 = 3𝑛_𝑁, we have 0.150 = 3𝑛_𝑁 + 𝑛_𝑁, which simplifies to 0.150 = 4𝑛_𝑁. Solving for 𝑛_𝑁, we get 𝑛_𝑁 = 0.150/4 = 0.0375 mol.

Since 𝑛_𝐻 = 3𝑛_𝑁, 𝑛_𝐻 = 3 * 0.0375 = 0.1125 mol.

Now, assuming ideal behavior, we know that the partial pressure of a gas is given by the product of the moles of the gas and the gas constant (R) and the temperature (T). Symbolically, it's written as 𝑝 = 𝑛𝑅𝑇, where 𝑛 is the number of moles and 𝑅 is the ideal gas constant.

For helium, the partial pressure (𝑝_𝐻) is given by 𝑛_𝐻𝑅𝑇, and for neon, the partial pressure (𝑝_𝑁) is given by 𝑛_𝑁𝑅𝑇.

Using the ideal gas constant 𝑅 = 0.0821 L⋅atm/(mol⋅K) and the temperature 𝑇 = 317 K, we can calculate the partial pressure.

For helium (𝑝_𝐻): 𝑝_𝐻 = 𝑛_𝐻𝑅𝑇 = 0.1125 * 0.0821 * 317 = 0.926 atm.

So, the partial pressure of helium in the mixture is approximately 0.926 atm. Let the heliums party on!

Answer 3

To find the partial pressure of helium, we need to determine the mole fractions of helium and neon in the mixture.

Let:
𝑛_ℎ = number of moles of helium
𝑛_𝑛 = number of moles of neon

Given:
𝑛/𝑉 = 0.150 mol/L (concentration of the mixture)
𝑛_ℎ = 3𝑛_𝑛 (helium is three times more abundant than neon)

We can use the mole fraction formula to find the mole fractions of helium and neon:

𝑥_ℎ = 𝑛_ℎ / (𝑛_ℎ + 𝑛_𝑛)
𝑥_𝑛 = 𝑛_𝑛 / (𝑛_ℎ + 𝑛_𝑛)

Let's solve for 𝑛_ℎ and 𝑛_𝑛:

Since 𝑛 = 𝑛_ℎ + 𝑛_𝑛 (total moles of the mixture), we have 𝑛 = 0.150 mol/L.
So, 𝑛_ℎ + 𝑛_𝑛 = 0.150 mol/L.

Since 𝑛_ℎ = 3𝑛_𝑛, we can substitute in the above equation:
3𝑛_𝑛 + 𝑛_𝑛 = 0.150 mol/L
4𝑛_𝑛 = 0.150 mol/L
𝑛_𝑛 = 0.150 mol/L / 4
𝑛_𝑛 = 0.0375 mol/L

Now, let's calculate 𝑛_ℎ:
𝑛_ℎ = 3𝑛_𝑛
𝑛_ℎ = 3 * 0.0375 mol/L
𝑛_ℎ = 0.1125 mol/L

The mole fractions are:
𝑥_ℎ = 𝑛_ℎ / (𝑛_ℎ + 𝑛_𝑛)
= 0.1125 mol/L / (0.1125 mol/L + 0.0375 mol/L)
= 0.1125 mol/L / 0.150 mol/L
= 0.75

𝑥_𝑛 = 𝑛_𝑛 / (𝑛_ℎ + 𝑛_𝑛)
= 0.0375 mol/L / (0.1125 mol/L + 0.0375 mol/L)
= 0.0375 mol/L / 0.150 mol/L
= 0.25

To calculate the partial pressure of helium, we can use Dalton's Law of partial pressures:

𝑃_ℎ = 𝑥_ℎ * 𝑛 * 𝑅 * 𝑇

Given:
𝑇 = 317 K (temperature)
𝑅 = 0.0821 L * atm / mol * K (ideal gas constant)

Substituting the values, we get:

𝑃_ℎ = 0.75 * 0.150 mol/L * 0.0821 L * atm / mol * K * 317 K

Calculating the partial pressure of helium:

𝑃_ℎ ≈ 0.015 atm

Therefore, the partial pressure of helium in the mixture is approximately 0.015 atm.

Answer 4

To find the partial pressure of helium, we need to determine the mole fraction of helium in the mixture and then use the ideal gas law.

Let's start by finding the mole fraction of helium (𝑋_𝐻𝑒). The mole fraction is defined as the ratio of the number of moles of a component to the total number of moles in the mixture.

Given that the mixture contains 3 times the number of helium atoms as neon atoms, we can calculate the mole fraction of helium as follows:

𝑋_𝐻𝑒 = 𝑛_𝐻𝑒 / 𝑛_𝑡𝑜𝑡𝑎𝑙

where 𝑛_𝐻𝑒 is the number of moles of helium and 𝑛_𝑡𝑜𝑡𝑎𝑙 is the total number of moles in the mixture.

Since the concentration of the mixture (𝑛/𝑉) is given as 0.150 mol/L, we can calculate the total number of moles using the formula:

𝑛_𝑡𝑜𝑡𝑎𝑙 = 𝐶 × 𝑉

where 𝐶 is the concentration of the mixture and 𝑉 is the volume of the mixture.

Now, assuming we have the volume of the mixture, we can solve for 𝑛_𝑡𝑜𝑡𝑎𝑙:

𝑛_𝑡𝑜𝑡𝑎𝑙 = 0.150 mol/L × 𝑉

Next, we know that the number of moles of helium (𝑛_𝐻𝑒) is three times the number of moles of neon (𝑛_N𝑒):

𝑛_𝐻𝑒 = 3 × 𝑛_N𝑒

Now, we can substitute the value of 𝑛_𝐻𝑒 into the mole fraction equation:

𝑋_𝐻𝑒 = 3 × 𝑛_N𝑒 / (0.150 mol/L × 𝑉)

Finally, using the ideal gas law equation:

𝑃_𝐻𝑒 = 𝑋_𝐻𝑒 × 𝑃_𝑡𝑜𝑡𝑎𝑙

where 𝑃_𝐻𝑒 is the partial pressure of helium and 𝑃_𝑡𝑜𝑡𝑎𝑙 is the total pressure of the gas mixture.

Given the temperature of the mixture (317 K), we can determine the total pressure using the ideal gas law:

𝑃_𝑡𝑜𝑡𝑎𝑙 = 𝑛_𝑡𝑜𝑡𝑎𝑙 × 𝑅 × 𝑇

where 𝑅 is the ideal gas constant (0.0821 L·atm/(mol·K)).

Substituting the values into the equation, we can calculate 𝑃_𝐻𝑒.