What is the pressure of a 3.00 L gas vessel that has 18.0 grams of helium at
?
(R= 0.0821 L atm/ mol K)

Question

What is the pressure of a 3.00 L gas vessel that has 18.0 grams of helium at 
?
(R= 0.0821 L atm/ mol K)

Quizplus · Accepted Answer

First, we need to calculate the number of moles of helium present in the vessel:

n = m/M = 18.0 g / 4.003 g/mol = 4.496 mol

Next, we can use the ideal gas law to calculate the pressure of the gas:

PV = nRT

P = nRT/V

P = (4.496 mol) x (0.0821 L atm/ mol K) x (11 K) / (3.00 L)

P = 36.7 atm

Therefore, the answer is choice B, 36.7 atm.

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