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Consider an experimental run at 273 K where the initial number of moles (n1) is actually 1.00 mol, and the final number of moles (n2) is 2.00 mol. Use the simulation to find the volume (V1) of 1.00 mol of helium at 273 K, and calculate the final volume (V2).

Express the volume to three significant figures, and include the appropriate units.

Respuesta :

Answer: The volume [tex](V_{1})[/tex] of 1.00 mol of helium at 273 K is 22.4 L and the final volume [tex](V_{2})[/tex] is 44.8 L.

Explanation:

Given: [tex]T_{1}[/tex] = 273 K,   [tex]n_{1}[/tex] = 1.00 mol

[tex]T_{2}[/tex] = 273 K,         [tex]n_{2}[/tex] = 2.00 mol

At the standard pressure, 1 atm the value of [tex]V_{1}[/tex] and [tex]V_{2}[/tex] is calculated as follows.

[tex]V_{1} = \frac{n_{1}RT_{1}}{P}\\= \frac{1.00 mol \times 0.0821 Latm/mol K \times 273 K}{1 atm}\\= 22.4 L[/tex]

Similarly,

[tex]V_{1} = \frac{n_{2}RT_{2}}{P}\\= \frac{2.00 mol \times 0.0821 Latm/mol K \times 273 K}{1 atm}\\= 44.8 L[/tex]

Thus, we can conclude that the volume [tex](V_{1})[/tex] of 1.00 mol of helium at 273 K is 22.4 L and the final volume [tex](V_{2})[/tex] is 44.8 L.

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