Answer :
The question is incomplete, here is the complete question:
How many grams of helium must be released to reduce the pressure to 65 atm assuming ideal gas behavior. Note : 334-mL cylinder for use in chemistry lectures contains 5.209 g of helium at 23°C.
Answer: The mass of helium released is 1.6 grams
Explanation:
We are given:
Mass of helium in the cylinder = 5.209 g
To calculate the number of moles, we use the equation given by ideal gas equation:
PV = nRT
Or,
[tex]PV=\frac{w}{M}RT[/tex]
where,
P = Pressure of the gas = 65 atm
V = Volume of the gas = 334 mL = 0.334 L (Conversion factor: 1 L = 1000 mL)
w = Weight of the gas = ?
M = Molar mass of helium gas = 4 g/mol
R = Gas constant = [tex]0.0821\text{ L atm }mol^{-1}K^{-1}[/tex]
T = Temperature of the gas = [tex]23^oC=[23+273]K=296K[/tex]
Putting values in above equation, we get:
[tex]65atm\times 0.334L=\frac{w}{4g/mol}\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 296K\\\\w=\frac{65\times 0.334\times 4}{0.0821\times 296}=3.573g[/tex]
Mass of helium released = (5.209 - 3.573) g = 1.636 g = 1.6 g
Hence, the mass of helium released is 1.6 grams