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If 3.25 mol of ar occupies 100 l at a particular temperature and pressure, what volume does 14.15 mol of ar occupy under the same condition?

Sagot :

SOLUTION:

Step 1: List the given values.

[tex]\begin{aligned} V_1 & = \text{100 L} \\ n_1 & = \text{3.25 mol} \\ n_2 & = \text{14.15 mol} \end{aligned}[/tex]

Step 2: Calculate the final volume of the gas by using Avogadro's law.

[tex]\begin{aligned} \frac{V_1}{n_1} & = \frac{V_2}{n_2} \\ n_1V_2 & = V_1n_2 \\ \frac{n_1V_2}{n_1} & = \frac{V_1n_2}{n_1} \\ V_2 & = \frac{V_1n_2}{n_1} \\ & = \frac{(\text{100 L})(\text{14.15 mol})}{\text{3.25 mol}} \\ & = \text{435.3846 L} \\ & \approx \boxed{\text{440 L}} \end{aligned}[/tex]

Hence, 14.15 mol of Ar occupies a volume of 440 L under the same condition.

[tex]\\[/tex]

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