What pressure (in atm) would be exerted by 76 g of fluorine gas in a 1.50-liter vessel at -37oC
Good
Well, if you make the gas laugh, it might exert some pressure from laughing too hard! But in all seriousness, we can use the Ideal Gas Law to solve this.
The Ideal Gas Law states that PV = nRT, where P is the pressure, V is the volume, n is the number of moles of gas, R is the ideal gas constant, and T is the temperature in Kelvin.
To solve for pressure, we need to convert the temperature from Celsius to Kelvin. So, -37°C + 273.15 = 236.15 K.
Next, we need to calculate the number of moles of fluorine gas. We can use the molar mass of fluorine, which is approximately 19 g/mol.
76 g / 19 g/mol = 4 moles of fluorine gas.
Finally, we plug in the values into the Ideal Gas Law equation:
P * 1.50 L = 4 moles * 0.0821 atm L/mol K * 236.15 K.
P * 1.50 L = 78.372 atm L.
Dividing both sides by 1.50 L:
P ≈ 52.2 atm.
So, the pressure exerted by 76 g of fluorine gas in a 1.50-liter vessel at -37°C would be approximately 52.2 atm. Now, let's make the gas laugh to see if we can increase the pressure even more!
To find the pressure exerted by the fluorine gas, we need to use the ideal gas law equation, which is given by:
PV = nRT
Where:
P is the pressure (in atm)
V is the volume (in liters)
n is the number of moles of gas
R is the ideal gas constant (0.0821 L·atm/(mol·K))
T is the temperature (in Kelvin)
First, we need to convert the given mass of fluorine gas to moles. To do this, we will use the molar mass of fluorine (F), which is approximately 19 grams/mol.
moles of F = mass of F / molar mass of F
moles of F = 76 g / 19 g/mol
moles of F = 4 moles
Next, we need to convert the temperature from Celsius to Kelvin. The Kelvin temperature (T) can be calculated by adding 273.15 to the Celsius temperature.
T = -37 oC + 273.15
T = 236.15 K
Now, we can substitute the given values into the ideal gas law equation:
PV = nRT
P * 1.50 L = 4 moles * 0.0821 L·atm/(mol·K) * 236.15 K
Now, solve for P:
P = (4 moles * 0.0821 L·atm/(mol·K) * 236.15 K) / 1.50 L
Calculating this expression will give us the pressure (P) exerted by the fluorine gas in atm.
since 1 mole = 22.4L at STP, and PV=kT, you want P such that
P*1.50/(273-37) = 1*22.4/273