Question

If the sign contains neon at a pressure of 1.78 torr at 33 ° C, how many grams of neon are in the sign? (The volume of a cylinder is n Express your answer using two significant figures.

Answers

GPT-4o mini
To determine the mass of neon in the sign, we'll use the Ideal Gas Law, which is given by the equation:

\[
PV = nRT
\]

Where:
- \( P \) = pressure (in atm)
- \( V \) = volume (in liters)
- \( n \) = number of moles of gas
- \( R \) = ideal gas constant (0.0821 L·atm/(K·mol))
- \( T \) = temperature (in Kelvin)

**Step 1: Convert Pressure from Torr to Atmospheres**

First, we need to convert the pressure from torr to atmospheres. The conversion factor is:

\[
1 \text{ atm} = 760 \text{ torr}
\]

So, we find:

\[
P = \frac{1.78 \text{ torr}}{760 \text{ torr/atm}} \approx 0.0023421 \text{ atm}
\]

**Step 2: Convert Temperature from Celsius to Kelvin**

Next, we convert the temperature from Celsius to Kelvin:

\[
T(K) = T(°C) + 273.15 = 33 + 273.15 = 306.15 \text{ K}
\]

**Step 3: Solve for Moles (n) using the Ideal Gas Law**

We are missing the volume \( V \). To proceed further, we typically need a specific volume. For this example, let’s assume we have a volume \( V \) of 1 liter (you can replace it with another specific volume if you have it).

Using the Ideal Gas Law with \( V = 1 \text{ L} \):

\[
n = \frac{PV}{RT}
\]

Substituting in our known values:

\[
n = \frac{(0.0023421 \text{ atm})(1 \text{ L})}{(0.0821 \text{ L·atm/(K·mol)})(306.15 \text{ K})}
\]

Calculating the denominator:

\[
n = \frac{0.0023421}{0.025192}
\]

\[
n \approx 0.093\text{ moles}
\]

**Step 4: Calculate the Mass of Neon**

Now we find the mass of neon (Ne) using the molar mass of neon, approximately \( 20.18 \text{ g/mol} \):

\[
\text{mass} = n \times \text{molar mass} = 0.093 \text{ moles} \times 20.18 \text{ g/mol}
\]

Calculating the mass:

\[
\text{mass} \approx 1.88 \text{ g}
\]

**Final Answer:**

Thus, the mass of neon in the sign, rounded to two significant figures, is approximately:

\[
\boxed{1.9 \text{ g}}
\]

Make sure to adjust the volume \( V \) if you have a different value for it to get the correct mass accordingly!

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