Question

How many grams of He gas are in a 33.2L container at 1.13atm and 652K?

Answer 1

Step 1

The He gas is assumed to be ideal. Therefore, the next equation is applied:

$p\text{ x V = n x R x T \lparen1\rparen}$

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Step 2

Information provided:

p = pressure = 1.13 atm

V = volume = 33.2 L

n = number of moles = unknown

T = absolute temperature = 652 K

R = gas constant = 0.082 atm L/mol K

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Step 3

Firstly, n is calculated from (1):

$\begin{gathered} \frac{p\text{ x V}}{R\text{ x T}}=\text{ n} \\ \frac{1.13\text{ atm x 33.2 L}}{0.082\text{ }\frac{atm\text{ L}}{mol\text{ K}}x652\text{ K}}=0.702\text{ moles} \end{gathered}$

n = 0.702 moles

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Step 4

The mass of He is calculated as:

n = mass/the molar mass He

The molar mass of He = 4.00 g

So, n = mass/4.00 g/mol

n x 4.00 g/mol = mass

0.702 moles x 4.00 g/mol = 2.81 g

Answer: mass = 2.81 g