Respuesta :
Answer: The volume of given amount of argon is [tex]0.394m^3[/tex] and [tex]13.91ft^3[/tex]
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Given mass of argon = 3.5 kg = 3500 g (Conversion factor: 1 kg = 1000 g)
Molar mass of argon = 40 g/mol
Putting values in above equation, we get:
[tex]\text{Moles of argon}=\frac{3500g}{40g/mol}=87.5mol[/tex]
To calculate the volume of gas, we use the equation given by ideal gas equation:
[tex]PV=nRT[/tex]
where,
P = pressure of the gas = 550 kPa
V = Volume of gas = ?
n = number of moles of argon = 87.5 moles
R = Gas constant = [tex]8.31\text{L kPa }mol^{-1}K^{-1}[/tex]
T = temperature of the gas = [tex]25^oC=[25+273]=298K[/tex]
Putting values in above equation, we get:
[tex]550kPa\times V=87.5mol\times 8.31\text{L kPa }mol^{-1}K^{-1}\times 298K\\\\V=394L[/tex]
Converting volume from liters to cubic meters and cubic foot, we use the conversion factor:
[tex]1m^3=1000L\\\\1ft^3=28.32L[/tex]
Converting the given volume into cubic meters:
[tex]\Rightarrow 394L\times (\frac{1m^3}{1000L})=0.394m^3[/tex]
Converting the given volume into cubic foot:
[tex]\Rightarrow 394L\times (\frac{1ft^3}{28.32L})=13.91ft^3[/tex]
Hence, the volume of given amount of argon is [tex]0.394m^3[/tex] and [tex]13.91ft^3[/tex]
Answer: The volume of argon will be [tex]0.394m^3[/tex] and [tex]14ft^3[/tex]
Explanation:
According to the ideal gas equation:
[tex]PV=nRT[/tex]
P = Pressure of the gas = 550 kPa = 5.43 atm (1 kPa= 0.0098 atm)
V= Volume of the gas = ?
T= Temperature of the gas = 25°C = 298 K [tex]0^00C=273K[/tex]
R= Gas constant = 0.0821 atmL/K mol
n= moles of gas
Moles of gas=[tex]\frac{\text{ given mass}}{\text{ molar mass}}= \frac{3.5\times 1000g}{40g/mole}=87.5moles[/tex]
[tex]V=\frac{nRT}{P}=\frac{87.5\times 0.0821\times 298}{5.43}=394L[/tex]
Thus volume of argon will be 394 Liters
1. Converting L to [tex]m^3[/tex]
[tex]1L=0.001m^3[/tex]
[tex]394L=\frac{0.001}{1}\times 394=0.394m^3[/tex]
Thus volume of argon will be [tex]0.394m^3[/tex]
2. Converting L to [tex]ft^3[/tex]
[tex]1L=0.035ft^3[/tex]
[tex]394L=\frac{0.035}{1}\times 394=14ft^3[/tex]
Thus volume of argon will be [tex]14ft^3[/tex]
