An inflatable toy starts with 1.05 moles of air and a volume of 5.17 liters. When fully inflated, the volume is 8.00 liters. If the pressure and temperature inside the toy don’t change, how many moles of air does the toy now contain?

Respuesta :

Answer:-  1.62 moles

Solution:- At constant temperature and pressure, volume is directly proportional to the moles of the gas.

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

from given data, [tex]V_1[/tex] = 5.17 L, [tex]n_1[/tex] = 1.05 moles

[tex]V_2[/tex] = 8.00 L, [tex]n_2[/tex] = ?

Let's plug in the values in the formula:

[tex]\frac{5.17L}{8.00L}=\frac{1.05moles}{n_2}[/tex]

On cross multiply:

[tex]n_2=\frac{8.00L(1.05moles)}{5.17L}[/tex]

[tex]n_2[/tex] = 1.62 moles

So, now the toy contains 1.62 moles of the air.


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