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A 334-mL cylinder for use in chemistry lectures contains 5.363 g of helium at 23 ∘C. You may want to reference (Pages 404 - 407) Section 10.4 while completing this problem. Part A How many grams of helium must be released to reduce the pressure to 65 atm assuming ideal gas behavior?

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Final answer:

To find how many grams of helium must be released, the initial amount of helium in moles is calculated. Using the ideal gas law under constant volume and temperature, the moles needed for the desired pressure are derived. The difference in initial and final mass gives the amount to release.

Step-by-step explanation:

To determine how many grams of helium must be released to reduce the pressure to 65 atm in the cylinder, we can use the ideal gas law which is PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the gas constant, and T is temperature. Since the question assumes ideal gas behavior, we can use the initial conditions to find the number of moles of helium initially present and then calculate the number of moles required to reach the desired pressure of 65 atm, keeping the volume and temperature constant.

We can calculate the initial number of moles (ni) using the initial mass (mi) and molar mass (MHe) of helium:

ni = mi / MHe

The final pressure (Pf) is defined, so we calculate the final number of moles (nf) with the same volume (Vi) and temperature (Ti):

Pf * Vi = nf * R * Ti

Finally, we find the final mass (mf) using:

mf = nf * MHe

The mass of helium to release (mrelease) is:

mrelease = mi - mf

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