To determine the mass of neon gas in the sign, we will use the Ideal Gas Law:
\[ PV = nRT \]
where:
- \( P \) = pressure in atmospheres (atm),
- \( V \) = volume in liters (L),
- \( n \) = number of moles of gas,
- \( R \) = ideal gas constant \((0.0821 , \text{L} \cdot \text{atm} / \text{mol} \cdot \text{K})\),
- \( T \) = temperature in Kelvin (K).
Step 1: Convert the pressure from torr to atm.
- The conversion factor is: \(1 , \text{atm} = 760 , \text{torr}\).
\[ P = \frac{1.78 , \text{torr}}{760 , \text{torr/atm}} \approx 0.00234 , \text{atm} \]
Step 2: Convert the temperature from Celsius to Kelvin.
\[ T = 33 , \text{°C} + 273.15 = 306.15 , \text{K} \]
Step 3: We need the volume of the cylinder.
Assuming the volume \( V \) of the cylinder is given (since it’s not specified in the problem). The formula is:
\[ V = \pi r^2 h \]
You must provide the radius \( r \) and height \( h \) of the cylinder to proceed with calculations.
Step 4: Combine the equations.
To find the number of moles \( n \):
\[ n = \frac{PV}{RT} \]
Substituting in the values for \( P \) and \( T \) that we found, and using an arbitrary volume \( V \) (if not provided, specify a volume):
\[ n = \frac{(0.00234 , \text{atm}) \cdot V}{(0.0821 , \text{L} \cdot \text{atm} / \text{mol} \cdot \text{K})(306.15 , \text{K})} \]
Step 5: Calculate the mass of neon.
The molar mass of neon (Ne) is approximately \( 20.18 , \text{g/mol} \).
Then the mass \( m \) of neon can be calculated:
\[ m = n \times \text{molar mass of Ne} \]
Putting it all together:
\[ m = \left(\frac{(0.00234) \cdot V}{(0.0821)(306.15)} \right) \cdot 20.18 \]
Let's assume an arbitrary value for volume \( V \) (e.g., 1 L) to give a concrete answer, then you can adjust for actual values.
For \( V = 1 , \text{L} \):
\[ n = \frac{(0.00234) \cdot 1 , \text{L}}{(0.0821)(306.15)} \approx 0.000872 , \text{mol} \]
Now calculating the mass:
\[ m = 0.000872 , \text{mol} \cdot 20.18 , \text{g/mol} \approx 0.0176 , \text{g} \]
Final Answer:
If \( V = 1 , \text{L} \), then the mass of neon in the sign is:
\[ 0.0176 , \text{g} \quad \text{(to 2 significant figures: } 0.018 , \text{g)} \]
Note: Please replace \( V \) with the actual measured volume of your specific cylinder to get the precise answer.