Answered

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A student decreases the temperature of a 641 cm3 balloon from 435 K to 200 K.Assuming constant pressure, what should the new volume of the balloon be?Round your answer to one decimal place.Please help

Sagot :

Answer:

The new volume of the balloon is 294 cm^3.

Explanation:

The given information from the exercise is:

- Initial volume (V1): 641 cm^3

- Initial temperature (T1): 435 K

- Final temperature (T2): 200 K

To calculate the final volume (V2) of the balloon, we can use the Charles's law formula, by replacing the values of V1, T1 and T2:

[tex]\begin{gathered} \frac{V_1}{T_1}=\frac{V_2}{T_2} \\ \frac{641cm^3}{435K}=\frac{V_2}{200K} \\ 1.47\frac{cm^3}{K}=\frac{V_{2}}{200K} \\ 1.47\frac{cm^3}{K}*200K=V_2 \\ 294cm^3=V_2 \end{gathered}[/tex]

So, the new volume of the balloon is 294 cm^3.

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