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A sealed vessel contains $1.5 \times 10^{21}$ atoms of helium gas at S.T.P - Leaving Cert Chemistry - Question c - 2001

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A sealed vessel contains $1.5 \times 10^{21}$ atoms of helium gas at S.T.P. What is the volume of the vessel in cm$^3$?

Worked Solution & Example Answer:A sealed vessel contains $1.5 \times 10^{21}$ atoms of helium gas at S.T.P - Leaving Cert Chemistry - Question c - 2001

Step 1

Calculate Total Moles of Helium

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Answer

To find the total number of moles ( ( n )) of helium gas, we use Avogadro's number, which states there are approximately 6.022×10236.022 \times 10^{23} atoms per mole.

The formula is: n=NNAn = \frac{N}{N_A} where:\

  • ( N = 1.5 \times 10^{21} ) atoms\
  • ( N_A = 6.022 \times 10^{23} \text{ atoms/mole} Substitutinginthesevalues:Substituting in these values: n = \frac{1.5 \times 10^{21}}{6.022 \times 10^{23}} \approx 0.00249 \text{ moles} $$

Step 2

Calculate Volume Using Ideal Gas Law

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Answer

Using the ideal gas law, we can find the volume ( ( V )) at standard temperature and pressure (S.T.P).

The ideal gas law is: PV=nRTPV = nRT At S.T.P., the pressure ( P = 1 \text{ atm} ) and the temperature ( T = 273.15 \text{ K} ). The gas constant ( R = 0.0821 \text{ L atm/(mol K)} ) may also be converted as needed.

Rearranging for volume: V=nRTPV = \frac{nRT}{P} Substituting the values: V=0.00249 moles×0.0821 L atm/(mol K)×273.15 K1 atm≈0.056 LV = \frac{0.00249 \text{ moles} \times 0.0821 \text{ L atm/(mol K)} \times 273.15 \text{ K}}{1 \text{ atm}} \approx 0.056 \text{ L} Now converting to cm3^3 (1 L = 1000 cm3^3): V≈0.056×1000=56 cm3V \approx 0.056 \times 1000 = 56 \text{ cm}^3

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