Question

Calculate the temperature of a gas that originally occupied 3.45 L and is expanded to 5.25 L. The original temperature of the gas was 282K and the pressure remains constant

Answer 1

Answer:

156 °C  

Explanation:

This looks like a case where we can use Charles’ Law:  

$\dfrac{V_{1}}{T_{1}} =\dfrac{V_{2}}{T_{2}}$  

Data:

V₁  = 3.45 L; T₁ = 282 K

V₂ = 5.25 L; T₂ = ?  

Calculation:

$\begin{array}{rcl}\dfrac{V_{1}}{T_{1}}& =&\dfrac{V_{2}}{T_{2}}\\\\ \dfrac{\text{3.45 L}}{\text{282 K}} &=&\dfrac{\text{5.25 L}}{T_{2}}\\\\\dfrac{\text{0.012 23}}{\text{1 K}} & = & \dfrac{5.25}{T_{2}}\\\\0.01223T_{2} & = & \text{5.25 K}\\T_{2} & = & \textbf{429 K}\\\end{array}$

(c) Convert the temperature to Celsius

T₂ = (429 – 273.15) °C = 156 °C