Question

 gas occupies 900.0 mL at a temperature of 27.0 °C. What is the temperature if the same gas now has a volume of 1.75 L?

Answer 1

By using Charles Law,

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

Given in question, $V_1=900\ ml,T_1=27\degree C=(273+27)K=300K$

$V_2=1.75\ L=1750\ ml, T_2=?$

Substituting these values,

$\frac{900}{300}=\frac{1750}{T_2}\implies T_2=583.33\ K$

Or, $T_2=(583.33-273)\degree C=310.33\degree C$