what pressure (in atm) would be exerted by 76g of fluorine gas in a 1.5 litre vessel at -37 degrees Celsius?
The volume of a sample of nitrogen is 6.00 liters at 35oC and 740 torr. What volume will it occupy at STP?
To find the pressure exerted by the fluorine gas in the given conditions, we can use the ideal gas law equation:
PV = nRT
Where:
P is the pressure (in atm)
V is the volume of the vessel (in liters)
n is the number of moles of the gas
R is the ideal gas constant (0.0821 L·atm/mol·K)
T is the temperature (in Kelvin)
We can start by converting the given values to the appropriate units:
- Temperature: -37 degrees Celsius = -37 + 273 = 236 Kelvin
- Volume: 1.5 liters
- Mass of fluorine gas: 76g
Next, we need to calculate the number of moles of the gas. We can use the molar mass of fluorine (F) to do this.
Molar mass of fluorine (F): 19.0 g/mol
Number of moles (n) = mass/molar mass
n = 76g / 19.0 g/mol
n = 4 moles
Now, we can substitute the values into the equation:
PV = nRT
P * 1.5 L = 4 mol * 0.0821 L·atm/mol·K * 236 K
P * 1.5 = 78.7664
Now, solve for P:
P = 78.7664 / 1.5
P ≈ 52.5 atm
Therefore, the pressure exerted by 76g of fluorine gas in a 1.5 liter vessel at -37 degrees Celsius would be approximately 52.5 atm.
13
fluorine gas F₂=>
μ = 36 g/mol =36•10⁻³ kg/mol,
m= 0.076 kg,
V=1.5 L = 1.5•10⁻³ m³,
t= -37℃ = 236 K,
pV=mRT/ μ =>
p =mRT/ Vμ =0.076•8.31•236/1.5•10⁻³•36•10⁻³=
=2.76•10⁶ Pa = 27.23 atm