Question

Please help A student increases the temperature of a 579cm^3 balloon from 215 K to 575 K Assuming constant pressure, what should the new volume of the balloon be? Round your answer to one decimal place.

Answer 1

Answer:

1548.5cubic centimeters.

Explanations:

According to Charles law, the volume of a given mass is directly proportional to the temperature provided that the pressure is constant. Mathematically;

$\begin{gathered} V\alpha T \\ V=kT \\ k=\frac{V_1}{T_1}=\frac{V_2}{T_2} \end{gathered}$

where

V1 and V2 are the volumes

T1 and T2 are the temperatures

Given the following parameters

$\begin{gathered} V_1=579cm^3 \\ T_1=215K \\ T_2=575K \end{gathered}$

Required

New volume V2

Substitute

$\begin{gathered} V_2=\frac{V_1T_2}{T_1} \\ V_2=\frac{579\times575}{215} \\ V_2=\frac{332925}{215} \\ V_2=1548.48cm^3 \end{gathered}$

Hence the new volume rounded to nearest one decimal place is 1548.5cubic centimeters.