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 mr? h.)

Express your answer using two significant figures.

1 answer

To find the mass of neon gas contained in the sign, we can use the Ideal Gas Law, which is represented as:

\[ PV = nRT \]

where:

  • \( P \) is the pressure in atmospheres,
  • \( V \) is the volume in liters,
  • \( n \) is the number of moles of gas,
  • \( R \) is the ideal gas constant (0.0821 L·atm/(K·mol)),
  • \( T \) is the temperature in Kelvin.

Step 1: Convert Pressure and Temperature

First, we need to convert the pressure from torr to atm and temperature from Celsius to Kelvin.

  • Pressure in atm:

\[ P = 1.78 , \text{torr} \times \left(\frac{1 , \text{atm}}{760 , \text{torr}}\right) = 0.0023421 , \text{atm} \]

  • Temperature in Kelvin:

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

Step 2: Calculate Moles of Neon (n)

Assuming we know the volume of the sign or cylinder, denoted as \( V \), we can plug the values into the ideal gas law. However, since volume was not given, let's represent \( V \) in the calculations assuming you have it.

Rearranging the Ideal Gas Law gives:

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

Step 3: Substitute Values

Plugging in the values for \( P, R, \) and \( T \):

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

Calculating the denominator:

\[ RT = 0.0821 \times 306.15 \approx 25.1872 \]

Now we calculate \( n \):

\[ n = \frac{0.0023421 \times V}{25.1872} \]

Step 4: Convert Moles to Grams

To find the mass of neon, we use the molar mass of neon:

  • Molar mass of neon (Ne) = 20.18 g/mol.

The mass \( m \) in grams can be found using:

\[ m = n \times \text{molar mass} \]

Thus:

\[ m = \left( \frac{(0.0023421 \times V)}{25.1872} \right) \times 20.18 \]

\[ m \approx \frac{0.0472398 , V}{25.1872} \]

Final Step: Presenting the Result

At this point without a specific volume \( V \), we cannot obtain a numeric solution. However, the final relationship expressing the mass of neon contained in the sign is found as follows:

\[ m \approx 0.0018746 , V , \text{grams} \]

To have a numeric answer, please provide the volume of the cylinder/signed space (in liters) to calculate the mass in grams.

If you assume a certain volume, plug it into (\( V \)) to get your grams of neon in the sign. Don’t forget to round your answer according to significant figures once you have \( V \).