Answer :
To solve this problem, we can use the Ideal Gas Law equation:
[tex]\[ PV = nRT \][/tex]
Where:
- [tex]\( P \)[/tex] is the pressure of the gas,
- [tex]\( V \)[/tex] is the volume of the gas,
- [tex]\( n \)[/tex] is the number of moles of gas,
- [tex]\( R \)[/tex] is the ideal gas constant, and
- [tex]\( T \)[/tex] is the temperature in Kelvin.
Given values:
- [tex]\( P = 1.20 \)[/tex] atm (pressure)
- [tex]\( V = 12.9 \)[/tex] L (volume)
- [tex]\( T = 315 \)[/tex] K (temperature)
The ideal gas constant [tex]\( R \)[/tex] is [tex]\( 0.0821 \)[/tex] L·atm·K⁻¹·mol⁻¹.
We need to solve for [tex]\( n \)[/tex], the number of moles. Rearranging the equation to solve for [tex]\( n \)[/tex], we get:
[tex]\[ n = \frac{PV}{RT} \][/tex]
Substituting the given values into the equation:
[tex]\[ n = \frac{(1.20 \, \text{atm}) (12.9 \, \text{L})}{(0.0821 \, \text{L·atm·K⁻¹·mol⁻¹})(315 \, \text{K})} \][/tex]
Now, calculate the numerator and the denominator:
[tex]\[ \text{Numerator} = (1.20 \times 12.9) = 15.48 \][/tex]
[tex]\[ \text{Denominator} = (0.0821 \times 315) = 25.8615 \][/tex]
Thus,
[tex]\[ n = \frac{15.48}{25.8615} \][/tex]
Calculating this:
[tex]\[ n \approx 0.5986 \][/tex]
Rounding this to 3 significant digits:
[tex]\[ n \approx 0.599 \][/tex]
Therefore, the sample contains approximately 0.599 moles of helium.
[tex]\[ PV = nRT \][/tex]
Where:
- [tex]\( P \)[/tex] is the pressure of the gas,
- [tex]\( V \)[/tex] is the volume of the gas,
- [tex]\( n \)[/tex] is the number of moles of gas,
- [tex]\( R \)[/tex] is the ideal gas constant, and
- [tex]\( T \)[/tex] is the temperature in Kelvin.
Given values:
- [tex]\( P = 1.20 \)[/tex] atm (pressure)
- [tex]\( V = 12.9 \)[/tex] L (volume)
- [tex]\( T = 315 \)[/tex] K (temperature)
The ideal gas constant [tex]\( R \)[/tex] is [tex]\( 0.0821 \)[/tex] L·atm·K⁻¹·mol⁻¹.
We need to solve for [tex]\( n \)[/tex], the number of moles. Rearranging the equation to solve for [tex]\( n \)[/tex], we get:
[tex]\[ n = \frac{PV}{RT} \][/tex]
Substituting the given values into the equation:
[tex]\[ n = \frac{(1.20 \, \text{atm}) (12.9 \, \text{L})}{(0.0821 \, \text{L·atm·K⁻¹·mol⁻¹})(315 \, \text{K})} \][/tex]
Now, calculate the numerator and the denominator:
[tex]\[ \text{Numerator} = (1.20 \times 12.9) = 15.48 \][/tex]
[tex]\[ \text{Denominator} = (0.0821 \times 315) = 25.8615 \][/tex]
Thus,
[tex]\[ n = \frac{15.48}{25.8615} \][/tex]
Calculating this:
[tex]\[ n \approx 0.5986 \][/tex]
Rounding this to 3 significant digits:
[tex]\[ n \approx 0.599 \][/tex]
Therefore, the sample contains approximately 0.599 moles of helium.
