Question

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!!!

Answer 1

`V_2=367.06\ cm^3`

Step-by-step explanation:

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

`(V_1)/(T_1)=(V_2)/(T_2)`

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.

`V_2=(V_1T_2)/(T_1)\\\\V_2=(300* 383.15 )/(313.15)\\\\V_2=367.06\ cm^3`

So, the new volume of the balloon is
`367.06\ cm^3`.