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 mr? h.)
Express your answer using two significant figures.
Express your answer using two significant figures.
Answers
GPT-4o mini
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 \( V \).
\[
PV = nRT
\]
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 \( V \).