A student increases the temperature of a 300 cm3 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, add 273.15)

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367.1 cm³

Explanation:

From the question given above, the following data were obtained:

Initial volume (V₁) = 300 cm³

Initial temperature (T₁) = 40 °C

Final temperature (T₂) = 110 °C

Final volume (V₂) =?

Next, we shall convert celsius temperature to Kelvin temperature. This can be obtained as follow:

T(K) = T(°C) + 273.15

Initial temperature (T₁) = 40 °C

Initial temperature (T₁) = 40 °C + 273.15

Initial temperature (T₁) = 313.15 K

Final temperature (T₂) = 110 °C

Final temperature (T₂) = 110 °C + 273.15

Final temperature (T₂) = 383.15 K

Finally, we shall determine the final volume (i.e the new volume) of the balloon. This can be obtained by using the Charles' law equation as illustrated below:

Initial volume (V₁) = 300 cm³

Initial temperature (T₁) = 313.15 K

Final temperature (T₂) = 383.15 K

Final volume (V₂) =?

V₁/T₁ = V₂/T₂

300 / 313.15 = V₂ / 383.15

Cross multiply

313.15 × V₂ = 300 × 383.15

313.15 × V₂ = 114945

Divide both side by 313.15

V₂ = 114945 / 313.15

V₂ = 367.1 cm³

Therefore, the new volume of the balloon is 367.1 cm³

"standard molar enthalpy:

h2o ( liquid water ) : - 285.8 kj/mole

h2o ( water vapor ) : - 241.8 kj / mole.

δhrxn = - 241. 8 - ( - 285.8 )= -241.8 + 285.8 = 44.0

answer: 44.0 kj"

2b2t

explanation:

2b2t