How many grams of helium must be released to reduce the pressure to 65 atmatm assuming ideal gas behavior? Express the answer in grams to two significant figures.

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.

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,

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 =

T = Temperature of the gas =

Putting values in above equation, we get:

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

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