Answer :
Sure, let's solve this problem step-by-step.
1. Understand the Problem:
- We need to calculate the volume of hydrogen chloride (HCl) gas.
- Given data:
- Pressure ([tex]\(P\)[/tex]) = 1.20 atm
- Temperature ([tex]\(T\)[/tex]) = 298 K (assuming room temperature if not provided)
- Amount of substance ([tex]\(n\)[/tex]) = 6.15 moles
2. Find the Formula:
- We will use the Ideal Gas Law which is given by:
[tex]\[ PV = nRT \][/tex]
- Where:
[tex]\[ \begin{align*} P & = \text{Pressure} \\ V & = \text{Volume} \\ n & = \text{Amount of substance (in moles)} \\ R & = \text{Universal gas constant} = 0.0821 \, \text{L·atm·K$^{-1}$·mol$^{-1}$} \\ T & = \text{Temperature (in Kelvin)} \end{align*} \][/tex]
3. Rearrange the Formula to Solve for Volume ([tex]\(V\)[/tex]):
- Rearranging the Ideal Gas Law to solve for [tex]\(V\)[/tex]:
[tex]\[ V = \frac{nRT}{P} \][/tex]
4. Substitute the Given Values:
- Given:
[tex]\[ \begin{align*} P & = 1.20 \, \text{atm} \\ n & = 6.15 \, \text{moles} \\ T & = 298 \, \text{K} \\ R & = 0.0821 \, \text{L·atm·K$^{-1}$·mol$^{-1}$} \end{align*} \][/tex]
- Substitute these values into the formula:
[tex]\[ V = \frac{6.15 \, \text{mol} \times 0.0821 \, \text{L·atm·K$^{-1}$·mol$^{-1}$} \times 298 \, \text{K}}{1.20 \, \text{atm}} \][/tex]
5. Calculate the Result:
- You can calculate the volume ([tex]\(V\)[/tex]) with these values.
The volume of hydrogen chloride gas under the given conditions is approximately 125.39 liters.
1. Understand the Problem:
- We need to calculate the volume of hydrogen chloride (HCl) gas.
- Given data:
- Pressure ([tex]\(P\)[/tex]) = 1.20 atm
- Temperature ([tex]\(T\)[/tex]) = 298 K (assuming room temperature if not provided)
- Amount of substance ([tex]\(n\)[/tex]) = 6.15 moles
2. Find the Formula:
- We will use the Ideal Gas Law which is given by:
[tex]\[ PV = nRT \][/tex]
- Where:
[tex]\[ \begin{align*} P & = \text{Pressure} \\ V & = \text{Volume} \\ n & = \text{Amount of substance (in moles)} \\ R & = \text{Universal gas constant} = 0.0821 \, \text{L·atm·K$^{-1}$·mol$^{-1}$} \\ T & = \text{Temperature (in Kelvin)} \end{align*} \][/tex]
3. Rearrange the Formula to Solve for Volume ([tex]\(V\)[/tex]):
- Rearranging the Ideal Gas Law to solve for [tex]\(V\)[/tex]:
[tex]\[ V = \frac{nRT}{P} \][/tex]
4. Substitute the Given Values:
- Given:
[tex]\[ \begin{align*} P & = 1.20 \, \text{atm} \\ n & = 6.15 \, \text{moles} \\ T & = 298 \, \text{K} \\ R & = 0.0821 \, \text{L·atm·K$^{-1}$·mol$^{-1}$} \end{align*} \][/tex]
- Substitute these values into the formula:
[tex]\[ V = \frac{6.15 \, \text{mol} \times 0.0821 \, \text{L·atm·K$^{-1}$·mol$^{-1}$} \times 298 \, \text{K}}{1.20 \, \text{atm}} \][/tex]
5. Calculate the Result:
- You can calculate the volume ([tex]\(V\)[/tex]) with these values.
The volume of hydrogen chloride gas under the given conditions is approximately 125.39 liters.
