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A steel tank is completely filled with 1.90 m3 of ethanol when both the tank and the ethanol are at 32.0c. when the tank and its contents have cooled to 18.0c, what additional volume of ethanol can be put into the tank?

Respuesta :

Kalahira
To answer this, you will need to use Charles law which relates volume to temperature in two proportional situations. V1/T1 = V2/T2 => V2 = V1/T1 * T2 Using the information given, V1 = 1.9 m3 T1 = 32.0 C T2 = 18.0 C we can get the following: V2 = 1.9/32 * 18 = 1.06875. Using the correct number of significant figures, the volume that would be taken up by the existing ethanol after the temperature change is 1.07. If the capacity is 1.9 m3, then the amount of additional that can be added is 1.90-1.07 which is 0.83 m3.

Answer: [tex]0.09m^3[/tex]

Explanation: Charles' Law: This law states that volume is directly proportional to the temperature of the gas at constant pressure and number of moles.

[tex]V\propto T[/tex]    (At constant pressure and number of moles)

[tex]\frac{V_1}{T_1}=\frac{V_2}{T_2}[/tex]

[tex]V_1[/tex]= initial volume= [tex]1.90m^3[/tex]

[tex]T_1[/tex] = initial temperature= [tex]32^0C=(32+273)K=305K[/tex]

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

[tex]T_2[/tex] = final temperature=[tex]18^0C=(18+273)K=291K[/tex]

[tex]\frac{1.90}{305}=\frac{V_2}{291}[/tex]

[tex]V_2=1.81m^3[/tex]

As now the volume has reduced and the total volume occupied can be [tex]1.90m^3[/tex], additional volume that can be put =[tex]( 1.90-1.81)=0.09m^3[/tex]

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