mirandaj12
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A gas is contained in a 5.0–liter container at a pressure of 100 kPa. What size container would be required to contain the gas at a pressure of 50 kPa, assuming the temperature remains constant?

Respuesta :

Willchemistry
P1* V1 = P2 * V2

100* 5 =  50 * V2

500 = 50 V2

V2 = 500 / 50

V2 = 10.0 L

hope this helps!

Answer : The volume of gas is, 10 L

Explanation:

According top the Boyle's Law, the pressure of a gas is inversely proportional to the volume of the gas.

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

or,

[tex]\frac{P_1}{P_2}=\frac{V_2}{V_1}[/tex]

where,

[tex]V_1[/tex] = initial volume of gas = 5 L

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

[tex]P_1[/tex] = initial pressure of gas = 100 kPa

[tex]P_2[/tex] = final pressure of gas = 50 kPa

Now put all the given values in the above formula, we get the final volume of the gas.

[tex]\frac{100kPa}{50kPa}=\frac{V_2}{5L}[/tex]

[tex]V_2=10L[/tex]

Therefore, the volume of gas is, 10 L

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