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.

1 answer

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!