Answer to Question #150566 in General Chemistry for Lexi

Question

A chemist determined that at constant pressure the volume of a gas is 3 L at a temperature of 304 Calvin for the volume of the gas to decreased 2 L the kelvin temperature must be approximately equal.

Answer 1

Charles’s law relates a gas’s volume and temperature at constant pressure and amount:

$\frac{V_1}{T_1}= \frac{V_2}{T_2}$

First, rearrange the equation algebraically to solve for $T_2$ :

$T_2=\frac{T_1 \times V_2}{V1}$

Second, substitute knowing quantities:

$T_2=\frac{304 K \times 2L}{3L} = 203 K$