Question

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.

Answer 1

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,

$PV=\frac{w}{M}RT$

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 = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = Temperature of the gas = $23^oC=[23+273]K=296K$

Putting values in above equation, we get:

$65atm\times 0.334L=\frac{w}{4g/mol}\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 296K\\\\w=\frac{65\times 0.334\times 4}{0.0821\times 296}=3.573g$

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