Answer :
To determine the volume of the gas given the conditions, we will use the Ideal Gas Law, which is represented by the 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 the gas,
- [tex]\( R \)[/tex] is the ideal gas constant, and
- [tex]\( T \)[/tex] is the temperature in Kelvin.
Given values:
- Number of moles, [tex]\( n \)[/tex] = 3.27 moles
- Pressure, [tex]\( P \)[/tex] = 125000 Pa
- Temperature in Celsius, [tex]\( T_{\text{Celsius}} \)[/tex] = 15.0°C
First, convert the temperature from Celsius to Kelvin using the formula:
[tex]\[ T_{\text{Kelvin}} = T_{\text{Celsius}} + 273.15 \][/tex]
So,
[tex]\[ T_{\text{Kelvin}} = 15.0 + 273.15 = 288.15 \text{ K} \][/tex]
Next, we use the Ideal Gas Law equation, rearranged to solve for volume [tex]\( V \)[/tex]:
[tex]\[ V = \frac{nRT}{P} \][/tex]
The ideal gas constant [tex]\( R \)[/tex] has a value of 8.314 J/(molK).
Plug in the known values:
[tex]\[ V = \frac{(3.27 \text{ moles}) \cdot (8.314 \text{ J/(molK)}) \cdot (288.15 \text{ K})}{125000 \text{ Pa}} \][/tex]
Calculating this:
[tex]\[ V \approx \frac{(3.27 \times 8.314 \times 288.15)}{125000} \][/tex]
[tex]\[ V \approx \frac{7844.7491}{125000} \][/tex]
[tex]\[ V \approx 0.062670965256 \text{ m}^3 \][/tex]
Therefore, the volume of the gas is approximately [tex]\( 0.0627 \text{ m}^3 \)[/tex].
[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 the gas,
- [tex]\( R \)[/tex] is the ideal gas constant, and
- [tex]\( T \)[/tex] is the temperature in Kelvin.
Given values:
- Number of moles, [tex]\( n \)[/tex] = 3.27 moles
- Pressure, [tex]\( P \)[/tex] = 125000 Pa
- Temperature in Celsius, [tex]\( T_{\text{Celsius}} \)[/tex] = 15.0°C
First, convert the temperature from Celsius to Kelvin using the formula:
[tex]\[ T_{\text{Kelvin}} = T_{\text{Celsius}} + 273.15 \][/tex]
So,
[tex]\[ T_{\text{Kelvin}} = 15.0 + 273.15 = 288.15 \text{ K} \][/tex]
Next, we use the Ideal Gas Law equation, rearranged to solve for volume [tex]\( V \)[/tex]:
[tex]\[ V = \frac{nRT}{P} \][/tex]
The ideal gas constant [tex]\( R \)[/tex] has a value of 8.314 J/(molK).
Plug in the known values:
[tex]\[ V = \frac{(3.27 \text{ moles}) \cdot (8.314 \text{ J/(molK)}) \cdot (288.15 \text{ K})}{125000 \text{ Pa}} \][/tex]
Calculating this:
[tex]\[ V \approx \frac{(3.27 \times 8.314 \times 288.15)}{125000} \][/tex]
[tex]\[ V \approx \frac{7844.7491}{125000} \][/tex]
[tex]\[ V \approx 0.062670965256 \text{ m}^3 \][/tex]
Therefore, the volume of the gas is approximately [tex]\( 0.0627 \text{ m}^3 \)[/tex].
