A gas at 89.0 °C occupies a volume of 0.67 L. At what Celsius temperature will the volume increase to 15.9 L?

The answer to your question is T2 = 5755.8 °K

Explanation:

Data

temperature 1 = 89 °C

Volume 1 = 0.67L

temperature 2 = ?

Volume 2 = 15.9 L

Process

Use Charles' law to solve this problem

V1/T1 = V2/T2

-Solve for T2

T2 = V2T1/V1

-Convert temperature to °K

temperature 1 = 89 + 273 = 362°K

-Substitution

T2 = (15.9 x 362) / 0.67

-Simplification

T2 = 5755.8 / 0.67

-Result

T2 = 5755.8 °K