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A 4 L sample of gas at 298 K and 2 atm contains 0.250 mol of gas. If we add another 0.250 mol of gas at the same pressure and temperature, what is the final total volume of gas?

Sagot :

Answer:

[tex]V_2=8L[/tex]

Explanation:

Hello there!

In this case, considering the Avogadro's gas law, which treats the volume and moles in a directly proportional way via:

[tex]\frac{V_1}{n_1}=\frac{V_2}{n_2}[/tex]

Which can be solved for the final volume, V2, as shown below:

[tex]V_2=\frac{V_1n_2}{n_1}[/tex]

Thus, by plugging in the given data, we obtain:

[tex]V_2=\frac{4L*(0.250mol+0.250mol)}{0.250mol}\\\\V_2=8L[/tex]

Regards!

The final total volume of gas is equal to 8 Liters.

Given the following data:

  • Initial volume = 4 Liters.
  • Initial temperature = 298 K.
  • Initial pressure = 2 atm.
  • Initial volume = 0.250 mol.
  • Final volume = 0.250 mol.

To determine the final total volume of gas, we would apply the Avogadro's equation for an ideal gas:

How to calculate the final total volume.

At constant temperature and pressure, Avogadro's equation for an ideal gas is given by this formula:

[tex]\frac{V_1}{n_1} = \frac{V_2}{n_2}[/tex]

For the final total moles:

[tex]n_2=n_1+n_1\\\\n_2 = 0.250+0.250\\\\n_2=0.50\;mol[/tex]

Making [tex]V_2[/tex] the subject of formula, we have:

[tex]V_2 = \frac{V_1n_2}{n_1}[/tex]

Substituting the given parameters into the formula, we have;

[tex]V_2 = \frac{4 \times 0.5}{0.25}[/tex]

Final total volume = 8 Liters.

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