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You fill a balloon with helium gas to a volume of 2.68L at 23°C and 789mmHg. Then you release the balloon and it rises. What will the volume be when the pressure drops to 632mm Hg if the temperature remains the same?

Respuesta :

Answer : The final volume of balloon will be, 3.35 L

Explanation :

Boyle's Law : It is defined as the pressure of the gas is inversely proportional to the volume of the gas at constant temperature and number of moles.

[tex]P\propto \frac{1}{V}[/tex]

or,

[tex]P_1V_1=P_2V_2[/tex]

where,

[tex]P_1[/tex] = initial pressure = 789 mmHg

[tex]P_2[/tex] = final pressure = 632 mmHg

[tex]V_1[/tex] = initial volume = 2.68 L

[tex]V_2[/tex] = final volume = ?

Now put all the given values in the above equation, we get:

[tex]789mmHg\times 2.68L=632mmHg\times V_2[/tex]

[tex]V_2=3.35L[/tex]

Therefore, the final volume of balloon will be, 3.35 L

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