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Need help with question 3!
*please answer correctly*


A student increases the temperature of a 300cm3 balloon from 40°C to 110°C.

What will the new volume of the balloon be? Round your answer to one decimal point.

(To convert Celsius to kelvin, add 273.15)

Thank you!!!

Need help with question 3 please answer correctly A student increases the temperature of a 300cm3 balloon from 40C to 110C What will the new volume of the ballo class=

Respuesta :

Answer:

[tex]V_2=367.06\ cm^3[/tex]

Explanation:

Charles law states that volume of the gas is directly proportional to the temperature i.e.

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

We ave,

V₁ = 300 cm³

T₁ = 40°C = 40 + 273.15 = 313.15  K

T₂ = 110°C = 110 + 273.15 = 383.15 K

Let V₂ be the new volume.

[tex]V_2=\dfrac{V_1T_2}{T_1}\\\\V_2=\dfrac{300\times 383.15 }{313.15}\\\\V_2=367.06\ cm^3[/tex]

So, the new volume of the balloon is [tex]367.06\ cm^3[/tex].

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